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Project: Structural Steel Design Instructor: Ms.c Viet Hieu Pham - Determine the Dead Load of slab: Table II.1 Dead Load of SlabNo.. Project: Structural Steel Design Instructor: Ms.c Vie

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

- 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

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- Determine the Dead Load of slab:

Table II.1 Dead Load of SlabNo

1

2

3

- Determine the Dead Load of roof slab:

Table II.2 Dead Load of SlabNo

1

2

3

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:

= 11.71 (daN/cm)

Load(daN/m2)

Factored Load(daN/m2)

Factor of Safety nTypes of loading

Layer of ceramic tile, t = 8

(daN/m2)

Factor of Safety n

Factored Load(daN/m2)Layer of the sealing, t =20

mm2000x0.02

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= (daN/cm)

715609 (daN.cm)4403.8 (daN)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

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

1355.4613.5546

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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

1

2

3

GA =Table IV.1.1a

3024The self-weight of slab with L = 6(m)

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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

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

107.80

2060.4GBm = GCm

GB = GCTypes of load

Types of load

GAm =GDmTypes of loadThe self-weight of beams have gc = 37.1 (daN/m)

The self-weight of slab with L = 6(m) 1840.8

The self-weight of beams have gc = 37.1 (daN/m)

The self-weight of slab with L = 6(m)-> gsm.(6/2(m))x(2/2)+gsm.(6/2)x(2.3/2) 3983.52

4203.12Table IV.1.2a

Table IV.1.2b

Student: Thanh Nguyen Ngo - 172216544 Page:

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219.6The self-weight of slab with L = 6(m)

-> gsm.(6/2(m))x(2.3/2)+gsm.(6/2)x(2.3/2) 4261.44

4481.04

GBCm > GABmLoại tải trọng

The self-weight of beam have gc = 37.1 (daN/m)

-> 37.1(daN/m)x6/2(m)

P3Types of loadP3 = pc x 6x1x2,3/2x1.3x2-> 300(daN/m)x6(m)x1.15x1/2x1.3x2

P1Types of load

1560P2

Types of loadP2 = pc x 6.x1x1x1.3x2

3120

24842484

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No

1

P4 = 2.2> Roof-Live Load 1

Table IV.2.2 Calculate Roof-Live Load 1

P2mTypes of loadP2m = pc x 6x1xx1.3x2

P1m

9936

P4Types of loadP3 = pc x 6x1x2.3x1.3x2

Types of loadP1m = pc x 6x1x1/2x1.3

585Factored Load

Factored Load

Types of loadP1 = pc x 6x2/2x1.3

1560

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Types of loadP3 = pc x 6x1x2.3x1.3x2

1345.51345.5P4m

49689936

Factored Load(daN)Factored Load (daN)

P2Types of loadP2 = pc x 6.x1x1x1.3x2

31203120

Types of Load

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3.> Determine the Wind Load act to Frame

Table IV.3.1 Calculate Wind Load

238.12263.95284.4563.21

66.5268.89

39.69 317.491243.99

47.4149.8951.67

351.936

Wđ (daN/m2)

Wh (daN/m2) qđ (daN/m)

qh(daN/m)52.92

58.66

299.33310.00

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Fig IV.3.2 Wind Right

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Project: Structural Steel Design Instructor: Msc Viet Hieu Pham

V THE COMBINATION OF INTERNAL FORCE

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INTERNAL FORCE OF ELEMENTS

N Kn.

Shear Force

V K.n

Moment Kn.m

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TABLE V.1.3 INTERNAL FORCE OF ELEMENTS

27

Student: Thanh Nguyen Ngo- 172216544 Page:

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0 Min -27.57 -34.37 -79.812.3 Min -27.57 -30.75 -20.362.3 Min -27.57 20.12 -20.364.6 Min -27.57 23.74 -79.61

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

9/24/15 0:16:14

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

9/24/15 0:17:08

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

9/24/15 0:16:49

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C> DIMENSION AND CONNECTION DESIGN

I.> Design No.1 column

1 The dimension of column design( Uniform Cross-Section ):

*From diagram of moment envelope we have:

M = 112.63 (kN.m)

V = 62.63 (kN)

N = 1017.6 (kN)

* The height of storey : ht= 4.2 (m) = 420 (cm)

*The effective length with Major Axis :

4.2 (m) = 420 (cm) *The effective length with Minor Axis:

2.94 (m) = 294 (cm)

* The shape of column is H-Shape( Symmetry)

Based on Required: có l = 420 (cm), Choose h = 48 (cm)

* The eccentricity and required area:

Required: 24 (cm)

*The thickness of the web be choose:

1.2 (cm) *The thickness of the flange be choose:

0.66 (cm)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:

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A = 121.44 cm2

11.4 (cm)

(cm4)(cm4)19.4048 (cm)

5.15896 (cm)21.644156.9882

The dimension of column is satisfy with slenderness

*The checking condition for general stability inside of the flexuaral plane :

The value of interpolation 0.607

Check left-side of expression:

13.804 (kN/cm2) Check right-side of expression

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= 0.71With: mx = 0.7055 < 1

0.66941With: 1.8021 < 2.5, we have equation:

0.83677

45727.72483232.1088

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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

17.083The flange is statisfy with the local stability condition

*With the web:

37.66670.70549 < 1

43.3318

Local stability is not problem

II.> Design No.6 Column

1 The dimension of column design( Uniform Cross-Section ):

M = 130.17 (kN.m)

V = 56.34 (kN)

N = 1345.3 (kN)The height of storey 4.2 (m) = 420 (cm)

*The effective length with Major Axis :

4.2 (m) = 420 (cm) *The effective length with Minor Axis:

=

= 0.74 < 2

=ℎ

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1.2 (cm) *The thickness of the flange be choose:

0.66 (cm)1.2 (cm) => Choose tf = 1.4 (cm)

*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

* The area of colum is: A = 126.6 cm2

Check: So Act< A therefore : The area of column is satisfy

2> Calculate index property and check in dimension of column:

A = 126.64 cm2

11.9 (cm)

(cm4)(cm4)20.2365 (cm)

5.37053 (cm)20.754654.7432

The dimension of column is satisfy with slenderness

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* Seaching of apependix table IV.5, with the type of No.5 dimension, We have:

Với Af/Aw = 0.61794 >=1

1.654Vậy me =η mx= 0.977 < 20 <Therefore do not check strength of section>

*The checking condition for general stability inside of the flexuaral plane :

The value of interpolation 0.72

Check left-side of expression:

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

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= 0.59With: mx = 0.5907 < 1

0.70748With: < 2.5, we have equation:1.7311

*With the web:

39.33330.59067 < 1

43.1528Local stability is not problem

III Design No.21 Girder

M = 300.73 (kN.m)

V = 150.53 (kN)

Left-side of expressionRight-side of expression

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N = 56.76 (kN)1>.Choose dimension of the girder:

*The height of girder: (h):

About economic value:

50.77 (cm) Choose h = 46 (cm)

*Check the thickness of web according to the resistance shear condition:

0.43 (cm)The web is statisfied

*The thickness of the flange:

44.8 (cm)29.97 (cm2

20.56 (kN/cm2)

Strength condition is not problem

*Check the equivalent stress condition when M and N simultaneously support to girder:

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The equivalen stress condition is not problem

* Check the local stability condition of girder:

*The flange

15.81

9.17The local stability of flange is not problem

* Check the local stability condition of the web when supported by normal stress:

173.93

54.5 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:

1.72

The local stability when supported by shear stress is not problem

*Check the general stability :

21.78

8.6957The general stability do not check

IV Design No.6 Girder

M = 139.77 (kN.m)

V = 70.43 (kN)

N = 11.71 (kN)1>.Choose dimension of the girder:

*The height of girder: (h):

About economic value:

34.61 (cm) Choose h = 32 (cm)

*Check the thickness of web according to the resistance shear condition:

0.30 (cm)The web is statisfied

*The thickness of the flange:

30.8 (cm)21.18 (cm2)

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17.21 (kN/cm2)

Strength condition is not problem

*Check the equivalent stress condition when M and N simultaneously support to girder:

Left-side of expression

Right-side of expression

24.15 The equivalen stress condition is not problem

* Check the local stability condition of girder:

* The flange:

15.81

8.00The local stability of the flange is not problem

* Check the local stability condition of the web when supported by normal stress:

173.93

37 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:

=

SVTH: Ngô Thanh Nguyên -172216544 Page:

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1.17 The local stability when supported by shear stress is not problem

*Check the general stability :

24.1710

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

M = 289.4 (kN.m)

V = 128.53 (kN)

N = 44.244 (kN)

1 Determine Bolts:

Use Bold have grade of bolts 5.8 , The diameter of bold is d= 36 (mm)

Arrangement bolts into 2 line, with distance of bolds accrording to table I.13

appendix I, [2]

*Tensile capacity of a bold:

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

, fub = 0.7*fub

fub -The break tension strength ( Table I.12 , appendix I, [2])

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

Ƴb1 = 1

μ,Ƴb2 - Friction ratio and safety ratio

nf -The quatity of frictional face, nf =1

* The tension force were applied the bolt in the farthest:

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161 (kN)

So all of bolts were satisfacted

* Check condition of the bolt be bearing shear force

2> Determine the collar

* Determine the thickness of collar according to bending codition

1.76 (cm)

2.03 (cm)Chọn t = 2.1 (cm)

3 Determine the weld be connected between the collar and girder

*The total length of welds at outside flange

The requirement height of weld be connected the web and collar:

(cm)Choose the height of weld at the web is: hf= 0.6 (cm)

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 = 139.77 (kN.m)

V = 70.43 (kN)

N = 11.71 (kN)

1 Determine Bolts:

Use Bold have grade of bolts 5.8 , The diameter of bold is d= 24 (mm)

Arrangement bolts into 2 line, with distance of bolds accrording to table I.13

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*Tensile capacity of a bold:

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

, fub = 0.7*fub

fub -The break tension strength ( Table I.12 , appendix I, [2])

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

Ƴb1 = 1

μ,Ƴb2 - Friction ratio and safety ratio

nf -The quatity of frictional face, nf =1

* The tension force were applied the bolt in the farthest:

152 (kN)

So all of bolts were satisfacted

* Check condition of the bolt be bearing shear force

2> Determine the collar

* Determine the thickness of collar according to bending codition

1.70 (cm)

1.87 (cm)Chọn t = 2 (cm)

3 Determine the weld be connected between the collar and girder

*The total length of welds at outside flange

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12.6 (kN/cm2)

0.29 (cm)Choose the height of weld at the web is hf= 0.5 (cm)

The requirement height of weld be connected the web and collar:

(cm)Choose the height of weld at the web is hf= 0.5 (cm)

VII> Detail of The base colum: No1

1 Determine the base plate of column

From table V.1 we have

M = 112.63 (kN.m)

V = 62.63 (kN)

N = 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

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)With

Fig VII.1 Detail of base column

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

0.09

204.30756.4 × 12.6 × 1=

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