Đồ án Kết cấu nhà thép

Project: Structural Steel Design

Instructor: Ms.c Viet Hieu Pham

10 (m) 4.6 (m)

A) DATA I.> Initial Data Spans L1 = L2 = - Column spacing :B1 = B2 =

6 (m) 6 (m) 3.3 (m)

65 (daN/m2)

- Hight of floor: Ht = - Design frame: Frame 4th - Minh Hóa - Quảng Bình -> Zone of wind I A -> W0 = - 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=

× 2 = 0.07 = 7 () ≥ ℎ = 5 ()

× ≥ ℎ = 5 () 1.4 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:

Types of loading

No.

Table II.1 Dead Load of Slab Load (daN/m2)

Factor of Safety n

Factored Load (daN/m2)

1.1

17.6

1.3

Layer of ceramic tile, t = 8 mm Layer of mortar, t = 15 mm 2000x0.015

208

1.1

228.8

The concrete slab, t = 80 mm 2500x0.08

-> gs

285.4 (daN/m2)

- Determine the Dead Load of roof slab:

Types of loading

No.

Table II.2 Dead Load of Slab Load (daN/m2)

Factor of Safety n

Factored Load (daN/m2)

1.3

Layer of the sealing, t =20 mm 2000x0.02 Layer of the insulation 10 mm

208

1.1

228.8

The concrete slab, t = 80 mm 2500x0.08

-> gs

306.8 (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

Factor loads:

- Determine the Loading and Internal force: + × 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 13.5546

(daN/m) (daN/cm)

715609 (daN.cm)

4403.8 (daN)

340.77 (cm3)

× 8 × 2 ×

From Wx = 340 (cm3) seaching table of I.6 appendix I [2], Use I-Shape , I 27:

Fig III.1.2 Dimension of beam

40.2 (cm2) 12.5 (cm) 0.6 (cm)

A = b = d = gc =

31.5 (kN/cm)

371 (cm3) Wx = Ix = 5010 (cm4) h = t = S =

27 (cm) 0.98 (cm) 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)

2.8 m

Hv = Ht- Hdc = ->gv = 107.8 (daN/m)

- The self-weight of girder:

150 daN/m

Asumme the self-weight of girder is g = 1.5 Kn/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.

Factord Load (daN)

219.6

3024

1712.4

Types of load The self-weight of beams have gc = 37.1 (daN/m) -> (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) -> (180(daN/m2)x2.8(m)x6/2(m))x2 The self-weight of slab with L = 6(m) -> (285.4(daN/m)x(6/2)x(2/2))x2(m)

4956

GA =

Table IV.1.1a

Student: Thanh Nguyen Ngo - 172216544

Page:..

Project: Structural Steel Design

Instructor: Ms.c Viet Hieu Pham

GB = GC

Factord Load (daN)

No.

219.6

3681.6

Types of load 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) -> gs.(6/2(m))x(2/2)+gs.(6/2)x(2.3/2)

3901.2

GB =

Table IV.1.1b GBC > GAB

Factord Load (daN)

No.

219.6

107.80

3938.52

Types of load The self-weight of beams have gc = 37.1 (daN/m) -> (37.1(daN/m)x6/2)x2(m) The self-weight of gypsum partition tile with the hight of girder : 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) -> 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

Factored Load

No.

219.6

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)

1840.8 2060.4

GAm =

Table IV.1.2a GBm = GCm

Types of load

Factored Load

No.

219.6

3983.52

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)

4203.12

GBm =

Table IV.1.2b

Student: Thanh Nguyen Ngo - 172216544

Page:..

Project: Structural Steel Design

Instructor: Ms.c Viet Hieu Pham

GBCm > GABm

Loại tải trọng

Kết quả (daN)

Số TT

219.6

4261.44

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)

4481.04

GBCm =

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

No.

Types of load P1 = pc x 6x2/2x1.3

Factored Load 1560 1560

P1 =

Factored Load

No.

3120

Types of load P2 = pc x 6.x1x1x1.3x2 -> 200(daN/m)x6(m)x1x1x1.3x2

3120

P2 =

Factored Load

No.

2484

Types of load P3 = pc x 6x1x2,3/2x1.3x2 -> 300(daN/m)x6(m)x1.15x1/2x1.3x2

2484

P3 =

Student: Thanh Nguyen Ngo - 172216544

Page:..

Project: Structural Steel Design

Instructor: Ms.c Viet Hieu Pham

Factored Load

No.

4968

Types of load P3 = pc x 6x1x2.3x1.3x2 -> 300(daN/m)x6(m)x2.3x1.3x2

9936

P4 =

2.2> Roof-Live Load 1

Table IV.2.2 Calculate Roof-Live Load 1

P1m

Factored Load

No.

585

Types of load P1m = pc x 6x1x1/2x1.3 -> 75(daN/m)x6(m)x1x1/2x1.3

585

P1m =

P2m

Factored Load

No.

1170

Types of load P2m = pc x 6x1xx1.3x2 -> 75(daN/m)x6(m)x1x1x1.3x2

1170

P2m =

Fig IV.2.1 Live Load 1

2.3> Live Load 2

Table IV.2.3 Calculate Live Load 2

Factored Load (daN)

No.

1560

Types of load P1 = pc x 6x2/2x1.3 -> 200(daN/m)x6(m)x1x1/2x1.3

1560

P1 =

Student: Thanh Nguyen Ngo - 172216544

Page:..

Project: Structural Steel Design

Instructor: Ms.c Viet Hieu Pham

No.

Factored Load (daN) 3120

Types of load P2 = pc x 6.x1x1x1.3x2

3120

P2 =

No.

Types of load P3 = pc x 6x1x2,3/2x1.3x2

Factored Load (daN) 2484 2484

P3 =

No.

Types of load P3 = pc x 6x1x2.3x1.3x2

Factored Load (daN) 4968 9936

P4 =

2.4> Roof-Live Load 2

Table IV.2.4 Calculate Roof-Live Load 2

P3m

No.

Types of Load P3m = pc x 6x2,3/2x1.3x2

Factored Load(daN) 672.75 672.75

P3m =

P4m

Factored Load(daN)

No.

1345.5

Types of Load P3m = pc x 6x2,3x1.3x2 -> 75(daN/m)x6(m)x2.3x1.3x2

1345.5

P4m =

Fig IV.2.2 Live Load 2

Student: Thanh Nguyen Ngo - 172216544

Page:..

Instructor: Ms.c Viet Hieu Pham

Project: Structural Steel Design

3.> Determine the Wind Load act to Frame

3.1> Calculating formulas

đ= × × × đ

= × × ×

đ=đ × = ×

+ + 2 2

With

1.2

W0 =

65 (daN/m2)

n =

0.6

Ch =

Cđ =

0.8

6 (m)

qđ (daN/m)

Floors

Wđ (daN/m2) 52.92 58.66 63.21 66.52 68.89

Wh (daN/m2) 39.69 43.99 47.41 49.89 51.67

317.4912 351.936 379.2672 399.1104 413.3376

qh (daN/m) 238.12 263.95 284.45 299.33 310.00

6 + 6 + 2 2 3.2> Calculate Wind Load Table IV.3.1 Calculate Wind Load Z (m) 4.2 7.5 10.8 14.1 17.4

0.848 0.94 1.013 1.066 1.104

Ht (m) 4.2 3.3 3.3 3.3 3.3

1 2 3 4 5

Fig IV.3.1 Wind Left

Student: Thanh Nguyen Ngo - 172216544

Page:..

Project: Structural Steel Design

Instructor: Ms.c Viet Hieu Pham

0 25 0 25

Fig IV.3.2 Wind Right 6 6 6 6 86 87 477 02 128 7 10 2 554 99 ả ả

ℎ ℎ3 86 87 3 1247 128 7 ℎ ℎ ℎ ℎ 1 h 1 ℎ ℎ 25 25 ly=μ lx=μ ậ 15 ậ ậ 15 ậ ậ 15 ậ ậ 15 ậ ∑ ∑

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

Name

No.

Position

Column

Axial Force N Kn. -626.14 -626.14 -1017.55 -1017.55 -498.38 -498.38 -805.19 -805.19 -368.90 -368.90 -576.45 -576.45 -237.63 -237.63 -363.25 -363.25 -104.19 -104.19 -134.24 -134.24 -706.23 -706.23 -1345.34 -1345.34 -574.18 -574.18 -1066.05 -1066.05 -439.86 -439.86 -771.92 -771.92 -300.10 -300.10 -493.45 -493.45 -155.24 -155.24 -199.58 -199.58

Shear Force V K.n 7.58 -4.40 -62.63 -53.63 -36.20 -47.82 -101.35 -93.51 -32.67 -45.18 -90.49 -82.14 -32.43 -44.26 -86.35 -81.01 -61.87 -67.08 -103.11 -101.48 56.34 56.34 -16.89 -16.89 91.70 91.70 11.80 11.80 75.86 75.86 15.55 15.55 67.43 67.43 20.55 20.55 74.14 74.14 47.90 47.90

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

Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min

0 4.2 0 4.2 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 4.2 0 4.2 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3

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

Name

No.

Position

Column

Axial Force N Kn. -706.23 -706.23 -1344.81 -1344.81 -574.18 -574.18 -1065.57 -1065.57 -439.86 -439.86 -771.55 -771.55 -300.10 -300.10 -493.19 -493.19 -155.24 -155.24 -199.46 -199.46 -626.14 -626.14 -980.61 -980.61 -498.38 -498.38 -783.84 -783.84 -368.90 -368.90 -555.07 -555.07 -237.63 -237.63 -357.45 -357.45 -104.19 -104.19 -128.41 -128.41

Shear Force V K.n 16.89 16.89 -56.35 -56.35 -11.80 -11.80 -91.72 -91.72 -15.55 -15.55 -75.89 -75.89 -20.55 -20.55 -67.46 -67.46 -47.90 -47.90 -74.18 -74.18 62.63 53.64 -7.58 4.40 101.37 93.53 36.20 47.82 90.52 82.17 32.67 45.18 86.38 81.04 32.43 44.26 103.16 101.52 61.87 67.08

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

Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min Max Max Min Min

0 4.2 0 4.2 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 4.2 0 4.2 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3 0 3.3

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

Name

No.

Position

Dầm

Axial Force N Kn. 56.76 56.76 56.76 28.25 28.25 28.25 1.68 1.68 1.68 -18.16 -18.16 -18.16 4.30 4.30 4.30 -12.75 -12.75 -12.75 33.27 33.27 33.27 9.66 9.66 9.66 -67.08 -67.08 -67.08 -101.48 -101.48 -101.48 11.71 11.71 11.71 11.71 4.30 4.30 4.30 4.30 -0.70 -0.70 -0.70 -0.70 -2.32 -2.32 -2.32 -2.32

Shear Force V K.n -77.99 8.23 150.53 -147.70 -6.43 79.78 -79.72 6.50 149.46 -147.81 -5.49 80.72 -81.50 4.72 149.40 -147.85 -3.73 82.49 -83.67 2.65 149.19 -148.08 -2.16 84.05 -83.37 3.14 112.13 -107.57 1.06 87.37 1.63 5.25 66.79 70.41 -70.43 -66.81 -5.25 -1.63 -2.38 1.24 63.14 66.76 -66.83 -63.21 -1.24 2.38

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

Max Max Max 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 Max Max Max Max Min Min Min Min Max Max Max Max Min Min Min Min

0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 0 5 10 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

Student: Thanh Nguyen Ngo- 172216544

Page:........

Project: Structural Steel Design

Instructor: Msc. Viet Hieu Pham

Dầm

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

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

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

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

Name

No.

Position

Dầm

Axial Force N Kn. -67.08 -67.08 -67.08 -101.52 -101.52 -101.52

Shear Force V K.n -87.37 -1.06 107.59 -112.10 -3.12 83.37

Moment Kn.m -157.05 135.20 -136.40 -204.88 99.72 -183.68

Max Max Max Min Min Min

0 5 10 0 5 10

Student: Thanh Nguyen Ngo- 172216544

Page:........

9/24/15 0:16:14

SAP2000

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

9/24/15 0:17:08

SAP2000

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

9/24/15 0:16:49

SAP2000

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) 4.2 (m) = 420 (cm)

4.2 (m) = 420 (cm)

2.94 (m) = 294 (cm) * The height of storey : ht= *The effective length with Major Axis : lx=μ×H=1×4.2= *The effective length with Minor Axis: ly=μ×H=0.7×4.2= * The shape of column is H-Shape( Symmetry)

Based on Required: có l = 420 (cm), Choose h = 48 (cm) ≤ ≤ , 1 15 1 10 h * The eccentricity and required area:

The eccentricity e: 0.11 (m) = 11.1 (cm) = = Grade of steel: CCT34 with f = 21 E = (kN/cm2) 21000 (kN/cm2)

= × 1.25 + 2.2 ÷ 2.8 × × ℎ

= = × 1.25 + 2.8 × 91.85 (cm2) × 1017.6 21 × 1 11.26 62.3

*Determine bf, tf and tw:

Required: 24 (cm) b = ÷ = 1 20 1 30 *The thickness of the web be choose:

1.2 (cm) tw = ÷ ℎ ≥ 0.6 = 1 60 1 120 *The thickness of the flange be choose:

0.66 (cm) tf ≥ × = 21 × = 21 21000

tf ≥ = 1.2 (cm) (cm) 1.4 => Choose tf = *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 121.4 cm2 The area of column is satisfy * The area of colum is: A = Check: So Act< A therefore : 2> Calculate index property and check in dimension of column:

Page:

SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham

11.4 (cm) =

45727.7248 (cm4) − 2 = =

3232.1088 (cm4) + 2 = = = × ℎ 12 ℎ × 12 A = 121.44 cm2 − 2 × ℎ 12 × 12 19.4048 (cm) = / =

5.15896 (cm) = / =

= < = 120 21.6441 =

< = 120 56.9882 = =

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 1.16 < 20 So: me =η mx= *The checking condition for general stability inside of the flexuaral plane :

≤ ×

= Have ̅ × = 0.747 à = 1.41 ả . 2 ℎụ ụ ó 0.607 The value of interpolation = 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

0.71 *The checking condition for general stability outside of the flexuaral plane : According to the flexuaral plane, we have: mx = m= With: 0.7055 < 1

= 56.988 < 99 1 = 3.14 × √ 0.7 = mx = = =

= = 0.66941 1 +

With: 1.8021 < 2.5, we have equation: =

0.83677 × ̅ × ̅ = = 1 − 0.073 − 5.53 ×

Page:

SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham

= We have: 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:

≤

= 8.14286 Left-side of expression

17.083 = 0.36 + 0.1̅ × = Right-side of expression

≤

Left-side expression 37.6667 = The flange is statisfy with the local stability condition *With the web: ℎ ℎ ℎ

0.70549 < 1 = 0.74 < 2 = Right-side experssion

= 43.3318 = 1.3 + 0.15 × ̅ × ℎ

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) 4.2 (m) = 420 (cm)

4.2 (m) = 420 (cm)

2.94 (m) = 294 (cm) The height of storey *The effective length with Major Axis : lx=μ×H=1×4.2= *The effective length with Minor Axis: ly=μ×H=0.7×4.2= * The shape of column is H-Shape( Symmetry)

≤ ≤ , Based on Required: có l = 420 (cm), chọn h = 50 (cm) 1 15 1 10 h * The eccentricity and required area:

The eccentricity e: 0.10 (m) = 9.68 (cm) = =

Grade of steel CCT34 with: 21 f = E = (kN/cm2) 21000 (kN/cm2)

= × 1.25 + 2.2 ÷ 2.8 × × ℎ ×

114.8 (cm2) = × 1.25 + 2.8 × = 896.03 21 × 1 13.2 42 *Determine bf, tf and tw:

Based on Required: 25 (cm) b = ÷ = 1 20 1 30 *The thickness of the web be choose:

Page:

SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham

1.2 (cm) tw = ÷ ℎ ≥ 0.6 = 1 60 1 120

*The thickness of the flange be choose:

0.66 (cm) tf ≥ × = 21 × = 21 21000

1.2 (cm) tf ≥ = (cm) 1.4 => Choose tf = *The dimension of column be choose:

The flange: (1.4x25) cm The web : (1.2x47.2) cm

Fig II.1 Dimension of No.6 column

126.6 cm2 The area of column is satisfy * The area of colum is: A = Check: So Act< A therefore : 2> Calculate index property and check in dimension of column: A = 126.64 cm2

= 11.9 (cm) = − 2

51861.1381 (cm4) − 2 = =

× ℎ 12 × 12 20.2365 (cm)

3652.63013 (cm4) + 2 = = × ℎ 12 ℎ × 12 = / =

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 = × ×

Page:

SVTH: Ngô Thanh Nguyên -172216544

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 0.977 < 20 Vậy me =η mx= *The checking condition for general stability inside of the flexuaral plane :

≤ ×

= Have ̅ × = 0.651 à = 0.704 ả . 2 ℎụ ụ ó 0.72 The value of interpolation = Check left-side of expression:

= 14.75 (kN/cm2) = × Check right-side of expression:

) × = 21 × 1 = 21 ( The dimension is statisfy with the general stability conditon

0.59 *The checking condition for general stability outside of the flexuaral plane : According to the flexuaral plane, we have: mx = m= With: 0.5907 < 1 mx =

= 54.743 < 99 = = 3.14 × √ 1 0.7 = =

= = 0.70748 1 + With: < 2.5, we have equation: 1.7311 =

0.84632 × ̅ × ̅ = = 1 − 0.073 − 5.53 ×

= We have: 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:

≤

= = 8.5 Left-side of expression

= 0.36 + 0.1̅ × = 16.8585 Right-side of expression

≤

39.3333 =

0.59067 < 1 = 0.651 < 2 The flange is statisfy with the local stability condition *With the web: ℎ ℎ Left-side of ℎ expression Right-side of = expression

43.1528 = = 1.3 + 0.15 × ̅ ×

ℎ 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)

Page:

SVTH: Ngô Thanh Nguyên -172216544

Instructor: Msc. Viet Hieu Pham Project: Structural Steel Design

56.76 (kN) N = 1>.Choose dimension of the girder: *The height of girder: (h): Assume: Chọn tw = 0.8 (cm) ~ 6 ÷ 22 , About economic value:

= 50.77 (cm) ℎ = 1.15 ÷ 1.2 × × ×

(cm) 46 Choose h = *Check the thickness of web according to the resistance shear condition:

× = 0.43 (cm) ≥ 3 2 ℎ × ×

The web is statisfied *The thickness of the flange:

× ℎ 12

× − × × ≥ ℎ 2 × 2 ℎ

44.8 (cm) 29.97 (cm2)

1.00 (cm)

ℎ, ÷ ℎ ≤ , = ≥ 180; 1 5 1 2 1 10

(cm), bf = 23 (cm) ℎ = ℎ − 15 ÷ 20 → ℎ = × ≥ *About detailing: = 10 ÷ 24 ≥ = ≤ 30 Choose tf = 1.2 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) = =

× ℎ = 12 1504.81 (cm3)

34610.5973 (cm4) − 2 = − 2 × ℎ 12

Wx =2 Ix/h = A = 98.80 (cm2)

618.2 (cm3) = = × × ℎ 2 *Check strength condition when M and N simultaneously support to girder :

20.56 (kN/cm2) + = ≤ ×

≤ 1.15 × ×

× = 21 × 1 = 21( Left-side of expression Right-side of expression ) Strength condition is not problem *Check the equivalent stress condition when M and N simultaneously support to girder:

+ 3

18.6813 2.69 = × × = = = Left-side of expression ℎ ℎ

+ 3

19.25 =

Page:

24.15 Right-side of expression 1.15 × × = SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham 1 15 × ×

The equivalen stress condition is not problem * Check the local stability condition of girder: *The flange

15.81 ≤ 0.5 × =

9.17 = The local stability of flange is not problem * Check the local stability condition of the web when supported by normal stress:

173.93 ≤ 5.5 × = ℎ

54.5 = ℎ 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:

× ≤ 3.2 ℎ

1.72 × =

ℎ The local stability when supported by shear stress is not problem *Check the general stability :

21.78 ≤ 0.41 + 0.0032 + 0.73 − 0.016 × × × = ℎ

= 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) 70.43 (kN) V = 11.71 (kN) N = 1>.Choose dimension of the girder: *The height of girder: (h): Assume: Chọn tw = 0.8 (cm) ~ 6 ÷ 22 , About economic value:

34.61 (cm) = ℎ = 1.15 ÷ 1.2 × × × (cm) 32 Choose h = *Check the thickness of web according to the resistance shear condition:

× = 0.30 (cm) ≥ 3 2 ℎ × ×

The web is statisfied *The thickness of the flange:

× ℎ 12

× − × × ≥ ℎ 2 × 2 ℎ

30.8 (cm) ℎ = ℎ − 15 ÷ 20 → ℎ = 21.18 (cm2)

0.80 (cm) × ≥ *About detail: = 10 ÷ 24 ≥ =

Page:

SVTH: Ngô Thanh Nguyên -172216544

Instructor: Msc. Viet Hieu Pham 24 ≥

÷ ℎ, ℎ ≤ , = ≥ 180; 1 5 1 2 1 10

(cm), bf = 20 (cm) Project: Structural Steel Design 10 ≤ 30 Choose tf = 1.2 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) =

× ℎ 12

13118.3957 (cm4) − 2 = − 2 × ℎ 12 = 819.9 (cm3) Wx =2 Ix/h = A = 71.68 (cm2)

= 369.6 (cm3) = × × ℎ 2 *Check strength condition when M and N simultaneously support to girder :

+ ≤ ×

+ = 17.21 (kN/cm2)

≤ 1.15 × ×

× = 21 × 1 = 21( Left-side of expression Right-side of expression ) Strength condition is not problem *Check the equivalent stress condition when M and N simultaneously support to girder:

+ 3

= × × = = = 15.449 2.48 Left-side of expression ℎ ℎ

+ 3

16.035257 =

24.15 Right-side of expression 1.15 × × =

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:

173.93 ≤ 5.5 × = ℎ

= 37 ℎ 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:

Page:

SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham

× ≤ 3.2 ℎ

1.17 × × = =

ℎ ℎ The local stability when supported by shear stress is not problem *Check the general stability :

24.17 ≤ 0.41 + 0.0032 + 0.73 − 0.016 × × × = ℎ

= = 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 (kN.m)

M = V = N = 289.4 128.53 (kN) 44.244 (kN)

5.8 , The diameter of bold is d= 36 (mm)

1. Determine Bolts: Use Bold have grade of bolts Arrangement bolts into 2 line, with distance of bolds accrording to table I.13 appendix I, [2]

163.2 (kN) *Tensile capacity of a bold: = =

With ftb is Tensile strength of a bold ( table I.9 appendix I, [2]) 20 (kN/cm2) ftb =

Abn = With Abn is the net area of a bolt ( Table I.11, appendix I, [2]) 8.16 (cm2) * The lateral resistance of the bolt :

44.89 (kN) × 1 = = = 0.7 × 95 × 4.59 × 1 × 0.25 1.7 fub - Tensile strength of hight strength bolts in frictional connection , fub = 0.7*fub

95 (kN/cm2), grade of steel 40Cr

= 0.25, = 1.7

Friction ratio and safety ratio. fub -The break tension strength ( Table I.12 , appendix I, [2]) fub = Ƴb1 = 1 μ,Ƴb2 -

nf -The quatity of frictional face, nf =1 * The tension force were applied the bolt in the farthest:

Page:

SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham

161 (kN) = + = , = 289.4 × 10 × 47 2 × (12 + 24 + 36 + 47) 44.24 10 ℎ1 2 ∑ ℎ So all of bolts were satisfacted * Check condition of the bolt be bearing shear force

= = 12.85 (kN) < 45 (kN) × = 128.53 10

2> Determine the collar * Determine the thickness of collar according to bending codition

1.76 (cm) = 1.1 × = 1.1 × = 12.5 × 161 23 + 12.5 21 , +

2.03 (cm) = 1.1 × = 1.1 × = 10 × 161(12 + 24 + 36 + 47) 37 23 + 47 21 ∑ + ℎ Chọn t = 2.1 (cm)

3. Determine the weld be connected between the collar and girder *The total length of welds at outside flange

65 (cm) = 2 × 23 − 1 + 2 × 11.5 − 1 =

* The tension force in outside plate according to the applied mement and axial force

651.253 (kN) ± = ℎ 2

= We have: Stell CCT34: fu Weld N42 34 18 0.7 = (kN/cm2) (kN/cm2) = 1 → × = × ; × = × ; × 0.45 × 12.6 (kN/cm2)

= = 0.80 (cm) → × = = ℎ 571.591 54 × 12.6 × 1 ∑ ×

1 (cm) Choose the height of weld at the web is hf= The requirement height of weld be connected the web and collar:

= ℎ

= = 0.17 (cm) 128.53 2 × (31 − 1) × 12.6 × 1 ∑ ×

0.6 (cm)

Choose the height of weld at the web is: hf= 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 = V = N = 139.77 (kN.m) 70.43 (kN) 11.71 (kN)

5.8 , The diameter of bold is d= 24 (mm)

1. Determine Bolts: Use Bold have grade of bolts Arrangement bolts into 2 line, with distance of bolds accrording to table I.13 appendix I, [2]

Page:

SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham

163.2 (kN) *Tensile capacity of a bold: = =

With ftb is Tensile strength of a bold ( table I.9 appendix I, [2]) 20 (kN/cm2) ftb =

Abn = With Abn is the net area of a bolt ( Table I.11, appendix I, [2]) 8.16 (cm2) * The lateral resistance of the bolt :

44.89 (kN) × 1 = = = 0.7 × 95 × 4.59 × 1 × 0.25 1.7 fub - Tensile strength of hight strength bolts in frictional connection , fub = 0.7*fub

95 (kN/cm2), grade of steel 40Cr

= 0.25, = 1.7

Friction ratio and safety ratio. fub -The break tension strength ( Table I.12 , appendix I, [2]) fub = Ƴb1 = 1 μ,Ƴb2 -

nf -The quatity of frictional face, nf =1 * The tension force were applied the bolt in the farthest:

152 (kN) = + = , = 139.77 × 10 × 32.8 2 × (10 + 20 + 32.8) 11.71 8 ℎ1 2 ∑ ℎ So all of bolts were satisfacted * Check condition of the bolt be bearing shear force

(kN) 7.04 < 45 (kN) = = × = 70.43 10

2> Determine the collar * Determine the thickness of collar according to bending codition

1.70 (cm) = 1.1 × = 1.1 × = 10 × 151 20 + 11.2 21 , +

1.87 (cm) = 1.1 × = 1.1 × = 11.2 × 151(10 + 20 + 32.8) 20 + 32.8 21 ∑ + ℎ Chọn t = 2 (cm)

3. Determine the weld be connected between the collar and girder *The total length of welds at outside flange

56.4 (cm) = 2 × 20 − 1 + 2 × 10.2 − 1 * The tension force in outside plate according to the applied mement and axial force

204.307 (kN) ± = ℎ 2

Page:

= We have: Stell CCT34: fu Weld N42 34 18 (kN/cm2) (kN/cm2) SVTH: Ngô Thanh Nguyên -172216544

Instructor: Msc. Viet Hieu Pham Project: Structural Steel Design 0.7 1 = = → × = × ; × = × ; × 0.45 × 12.6 (kN/cm2)

= = 0.29 (cm) → × = = ℎ 204.307 56.4 × 12.6 × 1 ∑ ×

0.5 Choose the height of weld at the web is hf= (cm) The requirement height of weld be connected the web and collar:

= ℎ

0.09 (cm) = 70.43 56.4 × (31 − 1) × 12.6 × 1 ∑ × 0.5 (cm)

Choose the height of weld at the web is hf= VII> Detail of The base colum: No1 1. Determine the base plate of column From table V.1 we have M = V = N = 112.63 (kN.m) 62.63 (kN) 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 (cm) 36 1.15 (kN.m), , = × ×∝= 1.15 × 1.2 × 1 = Bbd = b + 2c1 = The grade of concrete B20 1.38 (kN/cm2) The length of base plate be determined follow the partial press of footing concrete

+ + = 2 × × × , 6 × 2 × × × , 2 × × × , 46.72 (cm) ≥

(cm) 74 = h + 2 + 2 = 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 Determine the stress reaction of footing concrete under the base plate:

+ 0.725 (kN/cm2) =

− 0.04 (kN/cm2) = × × 6 × 6 ×

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

Space b2 a2 b2/a2 αb Quatity edge 0.06 M1 = αbσd2 (kN.cm) 20.764 Space 1 2 edge 9.9 21.2 0.467

Page:

SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham 0.092 28.576 23.5 0.745 17.5 Space 2 3 edge The thickness of base plate be calculated follow the formular

2.857 (cm) = =

6 × × 3 (cm)

1 (cm) 36 (cm)

495.9 (kN)

Choose tbd 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 = Width: Bdd = 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 = 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 11.71 (kN/cm)

1793.4 (kN.cm) = = qs = 0.48 x (23.5)= × 2 187.392 (kN) Vs = qs x ls =

Determine the height of stiffener A

22.64 (cm) = ℎ ≥ 6 × × ×

24 Select hs= (cm) Check follow the equivalent stress

+ 3

23.1 (kN/cm2) + 3 × = = × ℎ

6 × × ℎ 24.15 (kN/cm2) 1.15 × × = Statisfaction < 1.15 ⟹

Follwing the design condition, select height of weld hf =1cm Area and Moment of weld Aw = 2 x 1 x (24-1) = 46 (cm)

176 (cm3) = = 2 ×

1 × 24 − 1 6 Check the ability force of weld

11 (kN/cm2) + = =

12.6 (kN/cm2) OK × = ậ < × ⟹

4> Determine the stiffnener B The width of the load tranfer to stiffener is 1.5ls=1.5x12=18cm

qs = 0.77 x (18)= 13.86 (kN/cm)

Page:

SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham

× 2

997.92 (kN.cm) = =

221.76 (kN) Vs = qs x ls =

Select ts = 1cm

Determine the height of stiffener B

15.41 (cm) = ℎ ≥ 6 × × ×

22 Select hs= (cm) Check follow the equivalent stress

+ 3

21.4 (kN/cm2) + 3 × = = × ℎ

6 × × ℎ 24.15 (kN/cm2) (Ok) 1.15 × × = ậ < 1.15 ⟹

Follwing the design condition, select height of weld hf =1cm Area and Moment of weld Aw = 2 x 1 x (22-1) = 42 (cm)

147 (cm3) = = 2 ×

1 × 22 − 1 6 Check the ability force of weld

8.6 (kN/cm2) + = =

12.6 (kN/cm2) (OK) × = ậ < × ⟹

5> Calculate the anchor bolt From Table V.1 we have M = V = N = 28.73 (kN.m) -62.63 (kN) -1018 (kN) 74 (cm) The length of the compression concrete zone under the base plate Select the margin of the base plate to center of anchor bolt is 6cm, we have:

12.3333 (cm) = − = 2 3

43.3333 (cm) − 6 = = − 3 The total tension force of the anchor bolt:

-223.31 (kN) = = − ×

= + -508.39 (kN) = 2 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 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

Page:

SVTH: Ngô Thanh Nguyên -172216544

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

57 (cm) − 1 + 2 × − 1 + 2 × − 1 = = 2 × 21 − 1 2 33 − 21 2

= ℎ

0.63 (cm) =

= ℎ Select hf= 0.6 cm.

0.07 (cm) = 33 − 1 2 The requirement height of the weld at the column flange ∑ × × The requirement height of the weld at the column web ∑ × ×

VIII> Detail of The base colum: No.6 1. Tính toán bản đế 1. Determine the base plate of column 106.48 (kN.m) 56.34 (kN) 1345.3 (kN) M = V = N =

According to the dimension of column No.1, Choose 8 anchor bolts for base column Determin the dimension of base plate:

(cm) 37 1.15 (kN.m), , = × ×∝= 1.15 × 1.2 × 1 = Bbd = b + 2c1 = The grade of concrete B20 1.38 (kN/cm2) The length of base plate be determined follow the partial press of footing concrete

+ + = 2 × × × , 6 × 2 × × × , 2 × × × , 51.369 (cm) ≥

(cm) 76 = h + 2 + 2 = 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 Determine the stress reaction of footing concrete under the base plate:

+ 0.777 (kN/cm2) =

Page:

× 6 × 6 SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham

× × − 0.11 (kN/cm2) = × 6 ×

Với ≤ × , = 0.75 × 1.38 = 1.035 2

Fig VIII.1 Detail of base column

*According to the divide space of base plate and interpolated value, defind table: Table VIII.1

Ô sàn Bản kê b2 a2 b2/a2 αb

0.06 0.091 M1 = αbσd2 (kN.cm) 20.764 30.589 21.2 0.467 24.5 0.735 2 cạnh 3 cạnh 9.9 18 Ô 2 Ô 1 The thickness of base plate be calculated follow the formular

2.956 (cm) = =

6 × × 3 (cm)

1 (cm) 37 (cm)

509.7 (kN)

Choose tbd 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 Width: 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 = 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 17.89 (kN/cm)

2897.37 (kN.cm) = = qs = 0.73 x (24.5)= × 2 321.93 (kN) Vs = qs x ls =

Determine the height of stiffener A

28.77 (cm) = ℎ ≥ 6 × × ×

30 (cm) Chọn hs = Check follow the equivalent stress

Page:

SVTH: Ngô Thanh Nguyên -172216544

Project: Structural Steel Design Instructor: Msc. Viet Hieu Pham

+ 3

22.3 (kN/cm2) + 3 × = = × ℎ

6 × × ℎ 24.15 (kN/cm2) OK 1.15 × × = ậ < 1.15 ⟹

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)

280 (cm3) = = 2 ×

1 × 30 − 1 6 Check the ability force of weld

11.3 (kN/cm2) + = =

12.6 (kN/cm2) OK × = ậ < × ⟹

4> Determine the stiffnener B The width of the load tranfer to stiffener is 1.5ls=1.5x12=18cm

13.86 (kN/cm)

997.92 (kN.cm) = = qs = 0.78 x (18)= × 2 249.48 (kN) Vs = qs x ls =

Determine the height of stiffener B

15.41 (cm) = ℎ ≥ 6 × × ×

22 Select hs= (cm) Check follow the equivalent stress

+ 3

23.2 (kN/cm2) + 3 × = = × ℎ

6 × × ℎ 24.15 (kN/cm2) OK 1.15 × × = ậ < 1.15 ⟹

Follwing the design condition, select height of weld hf =1cm Area and Moment of weld Aw = 2 x 1 x (24-1) = 42 (cm)

176 (cm3) = = 2 ×

1 × 24 − 1 6 Check the ability force of weld

8.2 (kN/cm2) + = =

12.6 (kN/cm2) OK × = ậ < × ⟹

5> Calculate the anchor bolt From Table V.1 we have M = 28.73 (kN.m)

Page:

SVTH: Ngô Thanh Nguyên -172216544

Instructor: Msc. Viet Hieu Pham

Project: Structural Steel Design V = N = -56.34 (kN) -1345 (kN) 76 (cm) The length of the compression concrete zone under the base plate Select the margin of the base plate to center of anchor bolt is 6cm, we have:

11.6667 (cm) = − = 2 3

42.6667 (cm) − 6 = = − 3 The total tension force of the anchor bolt:

-300.53 (kN) = = − ×

= + -672.29 (kN) = 2 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 T2 < T1 The diameter of boltl is OK

± = -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

57 (cm) − 1 + 2 × − 1 + 2 × − 1 = = 2 × 21 − 1 2 33 − 21 2

= ℎ

0.85 (cm) =

= ℎ Select hf= 0.6 cm.

0.06 (cm) = 33 − 1 2 The requirement height of the weld at the column flange ∑ × × The requirement height of the weld at the column web ∑ × ×

Page:

SVTH: Ngô Thanh Nguyên -172216544