stuctural steel project ( ĐỒ Án kẾt cẤu thÉp ĐẠi hỌc xÂy dỰng )

STRUCTURAL STEEL PROJECT. NATIONAL UNIVERSITY OF CIVIL ENGINEERING (NUCE)

STRUCTURAL STEEL PROJECT

Member of Group 2: Tutor: TS. Dinh Van Thuat

- Trịnh Đức Duyên

- Nguyễn Hoàng Tùng

- Đỗ Duy Khang.

I. Design data and requirement

1) Design data

- Frame span : L = 30 (m)

- Frame bay : B = 6 (m)

- Number of frames : n = 15

- Height from the ground to the top of rail : H1 = 8(m)

- Crane capacity : Q = 7 (T)

- Number of traveling crane average acting levels : 2

- Wind zone (TCVN 3727:1995) : IIIB

- Topography types of the construction site (TCVN 3727: 1995) : B

- Type of roof cladding : Single- skin trapezoid sheeting system with insulating foam

- Height of crane runway girder : hr = 0,15 (m)

- Height of travelling crane : hdct = 0,65 (m)

- Height of skylight : Hct = 2 (m)

- Skylight bay : Lct = 5 (m)

- Materials : Steel CCT34 ; Concrete B15

- Surrounding covered structure : the height of brick wall from the ground is 1,5 m ; the above is

covered by sheet iron

2. Requirements


II. Calculation for design

1. Structural diagram of Portal Frame

Portal Frame Consist of solid columns and I- shaped beam, the columns have uniform cross–

section and fixed at the foundation. According to requirements of architecture and drainage,

we choose rafter with the slope of 10o (equal to i = 17%). Because a horizontal frame is

mainly subjected to self-weight loads and wind force, the internal force in rafter at its armpit

is much greater than that at the middle of rafter. The component of rafter has a changing

cross-section located at the distance of about 0,35- 0,4 of half of the rafter from column. The

rest of the rafter is constant.

The Skylight run along the length of building for ventilation and lighting. We choose

preliminarily the height of skylight is 2 m and its width is 5 m

Fig.1.1 -Portal Frame

1.1 The size of structural frame in vertical direction

The height of lower column : Hd = H1 – (hdct + hr) + hch

With, H1 = 8(m) : the elevation of the top of rail

+8.00

+9.46

+14.74

-1.00

Q= 7T

Cladding

Purlin

Rafter

Brick wall

wall cladding

I=17% I=17%


hdct = 0,65 (m) : the height of the crane girder

hr = 0,15 (m) : the height of the rail

hch = 0,5 (m) : embedded depth of the foundation

Hd = 8 –( 0,65+0,15) +1 = 8,2(m)

The height of upper column : Ht = (hdct + hr) + K1 + 0,5

With, K1 = 0,96 : the distance from the top of rail to the highest location of crane car.

This value is taken from the catalog of crane ( it depends on crane capacity Q = 8 (T) and

frame span L = 30 (m)).

0,5 (m) : safe distance from the highest location of crane car to portal frame

Ht = ( 0,65 + 0,15) + 0,96 + 0,5 = 2,26 (m)

So, the height of the column H = Ht + Hd = 8,2 + 2,26 = 10,46 (m)

1.2 Choose preliminary the size in horizontal direction

Frame span (positioning axis is considered at outer edge of column) : L = 30 (m)

, we choose approximately S = 28 (m) from catalog of the crane

safe distance from the axis of rail to inner edge of column : Zmin = 180 (mm)

a. the size of the column cross-section

- the height of the cross-section : h = 1 1 1 1

( ) ( )10,4610 15 10 15

H = 0,697 1,046 (m)

We choose h = 0,8 (m) = 80 (cm)

- The width of the cross-section : b = (0,3 0,5) (0,3 0,5).0,8h = 0,24 0,4 (m)

We choose b = 0,34 (m) = 34 (cm)

-The thickness of the web tw 1 1

( )70 100

h in order to satisfy the condition of corrosion

resistance and 6( )wt cm

1 1 1 1

( ) ( )80 1,14 0,870 100 70 100

wt h (cm)

So , we choose tw = 1,0 (cm)


-The thickness of flange tf : tf = 1 1 1 1

( ) ( )34 1,21 0,9728 35 28 35

b (cm)

So, we choose tf = 1,4 (cm)

Recheck safe distance from the axis of rail to the inner edge of column

Z = 1

( 2. )2

L h S

With, L : frame span ; S : crane span ; h: the height of the column cross-section

Z = 1

(30 2.0,8 28)2

0,2 (m) > Zmin = 0,18

Satisfy the condition of safety

b. The size of rafter cross-section

-The height of the cross-section at the armpit of the frame : h11 1

3040 40

L = 0,75 (m)

we choose h1 = 0,8 (m)

-The width of the cross-section at the armpit of the frame : b = 1

1 1

2 5h

and 180b mm . We

usually choose the width of the flange equaling to that of column

1

1 1 1 1.80 15 37,5

2 5 2 5b h

(cm)

we choose b = 34 (cm)

-The height of the constant cross-section of rafter h2 = (1,5 2)b

2 (1,5 2) (1,5 2).34h b 51 68 (cm)

we choose h2 = 50 (cm)

- The thickness of the web tw 1

1 1( )70 100

h in order to satisfy the condition of corrosion

resistance and 6( )wt cm


1

1 1 1 1( ) ( )80 1,14 0,870 100 70 100

wt h (cm)

So , we choose tw = 1,0 (cm)

-The thickness of flange tf : tf = 1 1 1 1

( ) ( )34 1,21 1,3628 35 28 35

b (cm)

So, we choose tf = 1,2 (cm)

-The location, has changing cross-section, be at the distance of about 0,35- 0,4 of half of the

portal frame from column : Ltd = (0,35 0,4).15 5,25 6 (m)

So, we choose Ltd = 5,5 (m)

c. The size of console cross-section

The size of console cross-section depends on crane loading ( concentrated forces made by

vertical pressure of overhead crane , self-weight of crane girder, rail , braking girder and live

load on overhead crane) and the span of console (the distance from the point put on by

concentrated force to the edge of the column). We choose preliminarily the size of console

cross-section as following:

-The distance from positioning axis to the axis of rail: =(L-S)/2=(30-28)/2 = 1 (m)

- The length of console (from the inner edge of the column to the outer edge of console)

LV = - hc + 0,15 = 1- 0,8 +0,15 = 0,35 (m)

( We choose the distance from the axis of rail to the outer edge of console is 150 mm

-Choose

-Choose the inclined angle between the bottom flange of console and horizontal direction is 15o

- We choose the height of a cross-section at fixed end : hdv= 60 (cm) (Z = 25)

We have: the height of console at the point put Dmax :

h = 60-20tan15o= 54,6(cm) cm)=> we choose: h = 55 (cm)

-The width of console cross-section : bf = 34 cm

-The thickness of console web : tw = 0,8 cm

-The thickness of console flange: tf = 1,2 cm


d. The size of skylight cross-section

-The height of the cross-section of skylight column : hc-ct = 20 cm

-The width of the cross-section of skylight column : bc-ct = 10 cm

- The thickness of web skylight column : tw = 0,6 cm

-The thickness of flange skylight column : tf = 1,0 cm

1.3 Bracing system

Bracing system is part of the building to connect horizontal frames together in order to form

space structural system leading to some following effect :

+ make sure geometrical stability in the direction of building length and space stiffness of

building

+ resist some loadings in the direction of building length and perpendicular to frame plan such

as wind load, seismic or braking forces from overhead crane

+ ensure stability of some compressive members such as : truss, column ....

+ Create favorable condition for construction as fabrication, erection and transportation

Bracing system consist of roof bracing system and column bracing system

Column bracing system

Column bracing system ensure geometrical stability and stiffness of entire building in the

direction of the length of building. It’s subjected to the loading applying in the direction of the

length of building and make sure that the column is stable. Along the length of building, column

bracing system is placed at the middle and the two side of the block in order to transmit wind

loading rapidly. The bracing system has two layers. One is arranged from braking gird to the top

of column, the other is from the ground to the console. According to the cross-section of the

column , the bracing system is placed at the middle of column flange . Because crane capacity

Q<10(T) , we choose the cross-section of bracing members made by rebar with diameter of

25mm .We layout shores along the building . The height of the column H = 10,46 (m) > 8m, so

we layout shores along the building at the locations where have elevation of +4.00m. We need

choose the cross-section of the shores satisfied the condition of slenderness : max 200 , we

choose 2C20


Fig.1.2 Column bracing system

Roof bracing system

Roof bracing system is layout at the location where have column bracing system. The

bracing system includes diagonal members and shores. We need to meet demand about

slenderness of shores max 200 . Diagonal members is made from rebar with 25, we

choose 2C20 for shores. According to the cross-section of portal frame, roof brace is

layout up (in order to make portal frame stable when they are subjected loadings – the

top flange of portal frame is subjected to compressive loads). When the frame is

subjected to wind loading , its bottom flange suffer compressive loadings , so we need to

strengthen by adding diagonal shores (connected to purlin) . They are spaced 3-bay

purlin distance centre . We choose the cross-section of shores L50X5. The location

connected to portal frame is at the distance of 800mm from portal frame. Besides, we

need to layout shores run along the ridge with cross-section of 2C20 in order to create

favourable condition for construction

Fig.1.3 Roof bracing system

2. Determine the loading applying on the frame

+8.00

+4.00

-1.00

5500

1

84000

6000 6000 6000 6000 6000 6000 6000 6000 6000 6000 6000 6000 5500

2 3 4 5 6 7 8 9 10 11 12 13 14 15

500 500

+9.46

0.00

Longitudinal bracing

Column bracing

Crane runway girder

5500

1

84000

6000 6000 6000 6000 6000 6000 6000 6000 6000 6000 6000 6000 5500

2 3 4 5 6 7 8 9 10 11 12 13 14 15

500 500

30000

Longitudinal Bar

Roof bracing


* 2.1 Continuous loadings

- Continuous loading caused by iron sheet , bracing system, purlin, skylight : gtc = 15 (daN/m2)

(distributed according to the slope of roof)

-The load factors ng = 1,1:

qtc = gtc . B = 6.15 = 90 (daN/m)

qt t = ng. qtc = 90.1,1 = 99 (daN/m)

-The loading caused by surrounding covered structure : gtc = 12 (daN/m2)

qt t = ng.gtc. B = 1,1.12.6= 79,2 (daN/m)

- Self-weight loading of crane bridge :

Gdct = 2

dct dctL = 30.62 = 1080 (daN)

With cdt = 30 : factor of self-weight

- Self-weight loading of braking girder: Gdh = 500 (daN) ( according to experience)

Table 1.1-Roof dead load

STT Types of loading

Characteristic load

(daN/m2)

Load factor

Factored load

(daN/m2)

The bay of frame (m)

Loadings (daN/m)

1 Iron sheet 8 1,1 8,8 6 52,8

2 purlin 7 1,1 7,7 6 46,2

3 99

-The load factors ng = 0,9:

qtc = gtc . B = 6.15 = 90 (daN/m)

qt t = ng. qtc = 90.0,9 = 81 (daN/m)

-The loading caused by surrounding covered structure : gtc = 12 (daN/m2)

qt t = ng.gtc. B = 0.9.12.6= 64,8 (daN/m)

- Self-weigh loading of crane bridge :


Gdct = 2

dct dctL = 30.62 = 1080 (daN)

With cdt = 30 : factor of self-weight

- Self-weight loading of braking grider : Gdh = 500 (daN) ( according to experience)

Table 1.2-Roof dead load

STT Types of

loading

Characteristic

load (daN/m2)

Load

factor

Factored

load (daN/m2)

The bay of

frame (m)

Loadings

(daN/m)

1 Iron sheet 8 0,9 7,2 6 43,2

2 purlin 7 0,9 6,3 6 37,8

3 81

* 2.2 Live loads of repairing roof

- Load factor of the live load : np = 1,3

- According to TCVN 2737 – 1995 , we have the magnitude of live loads caused by repairing

roof : 30 (daN/m2) distributed on the plan of the building, so the live load distributed on portal

frame is determined as following : p = 30.B

- we need to refer to the loading distributed on the portal frame as following :

ptc = 30.B. cos = 30.6.0cos10 = 177,26 (daN/m)

pt t = np.ptc = 1,3.177,26 = 230,45 (daN/m)

* 2.3 Wind loads

Wind pressure acting on the frame is determined according to TCVN 2737-1995

q = n.W0.k.C.B (daN)

With, q: wind pressure distributed along the frame

W0 : characteristic wind pressure ; wind zone IIIB : W0 = 125 (daN/m2)

n = 1,2 : wind load factor

k : the factor considering the wind pressure varying along the height of structure


C : the wind dynamic coefficient depends on the shape of structure

B : the distance between in-plan frame structure of the building

a) In case of wind direction blowing in horizontal direction of the building

* Determine the wind dynamic coefficient Ce

Fig.1.4-

Dimension of diagram:

Frame span :Lo= 30 (m),

Height of column :Hce = 10,46+0,67-0,5=10,63 (m) , Hm1= (Lo/2).tan10-(Lct/2)tan10=2,2(m),

Hm2= 2(m),

Hm3=(Lct/2)tan10 =0,44(m)

Follow the table 8 of TCVN 2737-1995 which is depends on , /B L ,and H/L => Ce

We have : Ce1=-0,8; Ce2=-0,8 ; Ce3=-0,5; Ce4=-0,324

- Determine k-factor:

K depends on height construction and topography type, Construction site is topography type B.

250

800

49

010°

800

67

0


Fig.1.5- the wind dynamic coefficient Ce

Table 1.3 Win load in transverse direction

ordinal Load type Wind

load (daN/

m)

Z

(m)

K C n B Total

wind load

(daN/m)

1 Cột đón gió 125 10,63 1,009 0,8 1,2 6 726,48

2 Mái đón gió 125 12,83 1,045 -0,324 1,2 6 -304,72

3 Cột cửa trời đón gió

125 14,83 1,077 0,7 1,2 6 678,51

4 mái cửa trời đón gió

125 15,27 1,082 -0,8 1,2 6 -779,04

5 Mái cửa trời

hút gió

125 15,27 1,082 -0,8 1,2 6 -779,04

6 Cột cửa trời hút gió

125 14,83 1,077 -0,6 1,2 6 -581,58

7 Mái hút gió

125 12,83 1,045 -0,5 1,2 6 -469,65

8 Cột hút gió 125 10,63 1,009 -0,5 1,2 6 -454,05

Minus (-) show wind load direction outside of the frame

b) In case of wind direction blowing in longitudinal direction of the building


Fig.1.6- the wind dynamic coefficient Ce

* Determine the wind dynamic coefficient Ce


Fig.1.7

In this the win dynamic coefficient both two side of the building roof and both side of the roof

light window, Ce =-0,7.

Ce3 depends on both L/ B and H/ B .

Building has : 30 0,75

0,3514.6

L

B

< 1

10, 45 2, 2 2 0, 440,18

14.6

H

B

< 0,5

=>Ce3=-0,4

Table 1.4 Win load in along direction

ordinal Load type

Wind load

(daN/m)

H K C n B Total wind load(daN/m)

1 Column 125 10,45 1,007 -0,4 1,2 6 -362.52

2 Roof 125 12,65 1,042 -0,7 1,2 6 -656,46

3 Column of roof

125 14,65 1,074 -0,4 1,2 6 -386,64

250

800

49

010°

800

67

0


window

4 Roof window

125 15,09 1,081 -0,7 1,2 6 -681,03

*2.4 Crane loads

a, Vertical loads

ZLK double girder crane , Load capacity 8 tons, Crane span 28 meters

=> The width of girder crane Bct=5,3m

The distance between two wheels : R=4,6m

The maximum pressure under each wheel : ax

c

mP 7020daN

The minimum pressure under each wheel : min

cP 2950daN

b, Vertical load on Console

max max. . .c iD n n P y ; min min. . .c iD n n P y

Where:

n : is load factor n=1,1

nc =0,85 combination factor ( when 2 girder cranes work in normal – not hard)

iy Total of ordinate of influence line

y1=1 ; y2=0,233 ; y3= 0,883 ; y4= 0,117

=> 2,233iy

STT Loads Ptc iy n nc Total

(daN)

1 Dmax 7020 2,233 1,1 0,85 14657

2 Dmin 2950 2,233 1,1 0,85 6159


Fig.1.8

c, Horizontal loadsWhen the hoist trolley brakes, this acts generate horizontal inertial force. The

inertial force transmits through the wheels of hoist trolley to the column ( T forces).Calculating T

forces also bases on influence line. Transverse force T is applied to one column( on the right or

left side of the frame structure in both right and left directions and it is located at the top level of

the crane runway girder.

The total characteristic transverse force of the crane :

1

0

0,05.( )c overheadQ GT

n

In which :

Q is the maximum load of overhead crane

800overheadG daN is the total self weight of the overhead crane

n0 =2 is the number of wheels on one side of the traveling crane

1

0,05.(7000 800)195

2

cT daN

1. . .c

c iT n n T y

In which :

n =1,1 ( load factor)

nc = 0,85 ( conbination factor)

4600350 350 4600350 350

Pmax Pmax Pmax Pmax

6000 6000

Y1 Y3

Y2 Y4


2,233iy

1. . . 407,132c

c iT n n T y daN

3.Design purlin

3.1 Design hot – rolled purlin

C12 shape purlin

Purlin and bracing layout

Properties of C12 purlin

Type of section

hxg bxg Ix Iy Wx Wy G

(mm) (mm) (cm4) (cm4) (cm3) (cm3) (daN/m)

C12 120 52 340 31,2 50,6 8,52 10,4

a, The load distribute on purlins cause by dead load , live load and selfweight of purlins.

ur

.( . ). .

os

c

claddingc c

p p lin g

g nq p n d g n

c

Where

g = 8(daN/m ) : is the self weight of roof elements; pc = 30 (daN/m2): is the roof live load;

d: is the spacing between purlin centers projected in the ground plan d = 1,5cosα = 1,5cos10 = 1.477(m)

ng; np – is the load factors, ng = 1.1 và np = 1.3

ur 10,4( / )c

p ling daN m : is the self weight of purlin


Un-factored load (standard load) 0

8(30 ).1,477 10,4 66,72( / )

os10

cq daN mc

Factored load( design load) 0

8.1,1(30.1,3 ).1,477 10,4.1,1 82,25( / )

os10q daN m

c

Un-factored load in the x direction and the y direction

.sin 11,585 /c c

xq q daN m

. os 65,703 /c c

yq q c daN m

Factored load in the x direction and the y direction

.sin 14, 283 /xq q daN m

. os 81,002 /yq q c daN m

Using wire ropes ф18 brace at midspan position purlins.

Fig 2.0Analytical model for roof purlin systems

The maximum bending moment :

2 2. 70,02.6315,09( . )

8 8

y

x

q BM daN m

2 2. 12,38.613,93( . )

32 32

xy

q BM daN m


+ Check strength condition follow formula :

.W W

yxx y c

x y

MMf

2 2315,09.100 12,38.100498,08( / ) 2100 /

77,8 13,3daN cm daN cm

+ Check the deflection of purlins

22

yx

B B B B

Deflection allowed of purlins under cladding 1

200B

In case of using a bracing purlin at mid-span, need to check the deflection at the mid-span point

(Dy max) and at the point is about z =0,21B=1,26m distance from the tip of purlin.

The relative deflection the mid-span point in the y direction 3 2 3

3

6

5. . 5.57,47.10 600 11,558.10

384. . 348.2,1.10 .545 200

c

y y

x

q B

B E I

The relative deflection at the point is about z =0,21B=1,26m distance from the tip of purlin.

3 2 35

6

. 10,3.10 .6006,24.10

2954. . 2954.2,1.10 .57,5

c

x x

y

q B

B E I

3 2 34

6

3,1. . 3,1.57,47.10 .6008,76.10

384. . 384.2,1.10 .545

c

y y

x

q B

B E I

5 2 4 2 4 1(6,24.10 ) (8,76.10 ) 8,78.10

200B

b, Wind load apply on purlins

The wind load transmit to purlin include dead load and wind

Factored wind load :

,w ur.W . . . 0,9.( . ). osos os

c c

y ind e o m p lin

d dq C k n g g c

c c


0

,w 0 0

1,477 1,4770,8.125.1,082.1,2. 0,9.(8. 13,3). os10 172,31( / )

os10 os10y indq c daN m

c c

Standard wind load

,w ur.W . . 0,9.( . ). osos os

c c c

y ind e o m p lin

d dq C k g g c

c c

0

,w 0 0

1,477 1,4770,8.125.1,082. 0,9.(8. 13,3). os10 139,85( / )

os10 os10

c

y indq c daN mc c

Dead load in the x direction :

ur0,9.( . ).sinos

c c

x m p lin

dq g g

c

0

0

1,1820,9.(8. 13,3).sin10 3,58( / )

os10daN m

c

Check strength condition :

2 2,w

w

. 135,54.6609,93( . )

8 8

y ind

ind

q BM daN m

2 2. 3,58.64,03( . )

32 32

xqx

q BM daN m

2 22w 609,93.10 4,03.10

814,27 2100 /W W 77,8 13,3

qxind

x y

MMdaN cm

Check deflection of the purlin : 3 2 3

,w ,w 3

6

3,1. . 3,1.109,57.10 .6001,67.10 1/ 200

384. . 384.2,1.10 .545

c

y ind y ind

x

q B

B E I

3.2 Design cold-formed shapes purlin

+ Vertical factored load (excluding selfweight of purlin) :

0

. 8.1,1q=(p . ). (30.1,3 ).1,182 56,65( / )

cos os10

cc m

p

g nn d daN m

c

+ Wind factored load ( excluding self-weight of purlin )


,w

0

0 0

q = .W . . . 0,9.( . ). osos os

1,477 1,4770,7.125.1,081.1,2. 0,9.(8. ). os10 127,59( / )

os10 os10

y ind e o c

d dC k n g c

c c

c daN mc c

Choosing the purlin follow wind loads qwind=1,276 kN/m (direction out of the roof ,the length of

bay B = 6000 mm) and in case of using a bracing purlin at midspan ,

looking up table 3.3 => Purlin Z20019, bearing capacity is 1,68(KN/m)

5.Check for selected cross-sections

5.1 Check for cross-section of column

5.1.1 Properties of the cross-section

The maximum internal forces in the columns ‘re at the connection between the column and

foundation.

Table 5.1 Internal forces


characteristics M (daN.m) N (daN) V (daN)

Combine 1 ax , ,m tu tuM N V 40853,58 -11012,1 7112,51

Combine 2 ax , ,m tu tuN M V 23378,92 -24491,6 5590,43

Combine 3 ax , ,m tu tuM N V -56178,9 -7023,1 -11095,8

Using CCT34 steel : f = 2100 daN/cm2

E = 2.1x106daN/cm2

Table 5.2 The dimension of cross-section (mm)

H Top flange Web bottom flange

bf tf hw tw bf tf

800 340 14 772 10 340 14

Table 5.3 geometric properties of the cross section

Jx Wx rx Jy Wy ry A

(cm4) (cm3) (cm) (cm4) (cm3) (cm) (cm2)

185392.63 4634.82 32.79 9177.37 539.85 7.926 172.4

The length of column :

-In frame plane buckling : .xL H

determine base on : .

.

cT

xa

b IG

H I ( b = 15 m the half of length of rafter

4185392.63cI cm , 457578.68cmrafterI , H=10,46 m)

.

.

cT

xa

b IG

H I 4.617 => 1.489 =>

15.575 m

.xL H


- Out of frame plane buckling : Ly The distance between two points at two adjacent columns

which bracing that columns ( the distance between two points which prevent displacements of

two adjacent columns)

Ly =5m

5.1.2 Check for condition about slenderness of the columns

The column slenderness ratio:

47, 495xx

x

L

r , 68,53

y

y

y

L

r

Độ mảnh quy ước:

1,502x x

f

E , 2,167y y

f

E

The slenderness limit:

180 60. ; . . .e c

N

A f

180 60.0,5 150

ax( , )x yM 68,53 <

5.1.3 Check for strength condition

a, Combine 1

M= 40853,58 daN.m ; N = - 11012,1 daN ; V = 7112,51 daN

The relative eccentricity ratio : Wx

e M Am

N 13,8

5x , m >5 , w

47,50,615

77,2

fA

A => 1,25

The reference eccentricity ratio : . 17,249 20em m ( Don’t need to check for strength

condition)

b, Combine 2


M (daN.m) N(daN) V(daN)

23378.92 -24491.56 5590.43

The relative eccentricity ratio : Wx

e M Am

N 3,55

5x , m < 5 , w

54,40,591

92,16

fA

A =>

(1,75 0,1 ) 0,02 (5 ) (1,75 0,1.3,55) 0,02.(5 3,55).1,502 1,351m m

The reference eccentricity ratio : . 4,798 20em m ( Don’t need to check for strength

condition)

c, Combine 3

M (daN.m) N(daN) V(daN)

-56178.86 -723.1 -11095.8

The relative eccentricity ratio :

Wx

e M Am

N 288,988

5x , m >5 , w

54,40,591

92,16

fA

A => 1,25

The reference eccentricity ratio : . 361,235 20em m

Check according to the condition following : .W

c

n x

N Mf

A ( . cf 2100 daN/cm2)

1 1216,3 daN/cm2

2 1207,9 daN/cm2

Satisfy all

5.1.3 Check for overall buckling in frame plane buckling

a, Combine 1


20em The reduction factor for column buckling e ,which is determined from table D10

TCXDVN 338:2005 ( 17,249em and 1,502x )

=> 0.081e

2788,58 /.e

NdaN cm

A < 2. 2100 /cf daN cm

a, Combine 2

20em The reduction factor for column buckling e ,which is determined from table D10

TCXDVN 338:2005 ( 4,798em and 1,502x )

=> 0.246e

2577,49 /.e

NdaN cm

A < 2. 2100 /cf daN cm

C, combine 3

20em Check according to the condition : ..W

c

b c

Mf

b is determined from appendix E TCXDVN 338:2005

23

w

3

. .8. . 1

. .

o f

fk f f f

L t a t

h b b t

Where : fkh 78,6cm ;2

fkha 39,3cm ; Lo =5m

=> 0,78 => 0,1 40 From appendix E1 TCXDVN 338:2005

=> 11,14 1,14.(2,25 0.07. ) 2,627

=>

2

1 .y

x o

I h E

I L f

3,329 > 0,85 Therefore 10,68 0,21 1,24b >1

Choose b 1

.Wb c

M

1212,106 <

2. 2100 /cf daN cm

5.1.4 Check for overall buckling out of frame plane buckling


.. .

c

y

Nf

c A

Coefficient c to account for the effect of the moment Mx (applied in the frame plane ) on the

column buckling out of frame plane buckling

a, Combine 1

M= 40853,58 daN.m ; N = - 11012,1 daN ; V = 7112,51 daN

The maximum bending moment at the middle segment 1/3 length of the column Mx =

292.753kN.m > M/2

Choose Mx = 29275,3 daN.m

W

xx

x

M Am

N 9,89 5 => 5< mx<10

=> 5 10(2 0.2 ) (0.2 1)x xc c m c m 0,141

( 5 0,0921 . x

cm

; 10

10,143

1y

x

b

c

m

)

. .y

N

c A 742,584 daN/cm2 < 2. 2100 /cf daN cm

a, Combine 2

The maximum bending moment at the middle segment 1/3 length of the column Mx = 14059,7

daN.cm >M/2 => Choose Mx = 14059,7 daN.cm

W

xx

x

M Am

N 2,135 => mx< 5

1 . x

cm

0,382 ( 0,757 ; 1 Table 16 TCXDVN 338-2005)

. .y

N

c A 611,346 daN/cm2 <

2. 2100 /cf daN cm ( Satisfy )

a, Combine 3

The maximum bending moment at the middle segment 1/3 length of the column


Mx = => / 2xM M Choose Mx = 381.7719 daN.cm

W

xx

x

M Am

N 196,386 => mx >10

1

1y

x

b

c

m

0,0065

(y

y

y

L

r 68,53 ;

2210 /f N mm Table D.8 TCXDVN 338-2005 => y 0,782; b 1 )

. .y

N

c A 829,068 daN/cm2 < 2. 2100 /cf daN cm ( Satisfy )

5.1.6 Check for local flange buckling of the column

Check follows : 0f 0

f

b b

t t

Where : w

02

f

f

b tb

16,5cm => 0f

f

b

t 11,786

min( , )x y 1,502 => 0.8 4 Choose 1,499

0f (0.36 0.1 )f

b E

t f

16,134 > 0f

f

b

t(satisfy)

5.1.7 Check for local web buckling of the column

local web buckling condition : w w

w w

h h

t t

; w

w

h

t 77,2

+) Combine 1 : M= 40853,58 daN.m ; N = - 11012,1 daN ; V = 7112,51 daN

me < 20

Total buckling of the column denpends on total buckling out of the frame plane buckling, so

need to compute coefficient α and average shear stress τ

1

x

N My

A I 986,208 daN/cm2 ( w

1 22

hy y 38,6 cm )


1 2

x

N My

A I -714,99 daN/cm2

w w.

V

t h 727,705 daN/cm2

1

1,725 > 1 ; 1.4(2 1)

2,531

=> w

2 2w

(2 1)4,35

(2 4

h E

t

132,502

3,8E

f 120,67 > 132,502 => choose w

w

h

t

132,502

w w

w w

h h

t t

the web doesn’t local buckling

2,3 72,32E

f < w

w

h

t

Need ribbed stiffener for the column

+) Combine 2 similar to combine 1

+) Combine 3 : me > 20 Check for local web buckling condition same as structural bending :

3,8E

f 120,67 > w

w

h

t 77,2

2,3 72,32E

f < w

w

h

t

Need ribbed stiffener for the column

5.1.8 Calculation of fillet weld connections between flange and web of the column

The fillet weld connections resist shear forces V which are caused by buckling. Pick combine

internal forces which contain maximum shear force Vmax

The height of weld lines : w min

.

2.( ) . .

V f

f

x c

V Sh

f I


VV 11095,8 daN ;1

. .2

f f f fxS b t h 1870,68 cm3

The strength of weld lines when using electrode N42 : w min( . )f 1260

=> 0, 444fh cm

=> choose 6fh mm

5.2 Check for cross-sections of rafter

5.2.1 Check for the cross-section of rafter which is at the join between column and rafter

Internal force :

M (daN.m) N (daN) V (daN)

-31323 -6195 -6378




bf tf hw tw bf tf

800 340 12 776 10 340 12


Jx Wx Jy Wy Sx Sfx A

(cm4) (cm3) (cm4) (cm3) (cm3) (cm3) (cm2)

129932.658 3248.32 5406.05 318 2360.24 1607.02 144.6

The maximum internal forces

The length of rafter : - in frame plane lx = 30m

- Out of frame plane ly = 1,477 m ( the distance between two adjecient

purlins)

The relative eccentricity ratio : W

x

x

e M Am

N 49,5cm

b, check for strength condition :

+) Check for the normal stress resistance of the cross-section with axial force and bending

moment about the x axis is required as follow :

. c

n xn

N Mf

A W

n xn

N M

A W -1007,126 daN/cm2 => IσI<

2. 2100 /cf daN cm

( The cross sectional area with considering the reduction An =144,6cm2

The elastic section modulus about the x with considering reduction Wx= 3248.32 cm3 )

+) Check for the normal stress resistance of the cross-section with shear force

max

.

.

x

x w

V S

I t -115,857 daN/cm2 => axm <

2. 1200 /v cf daN cm


+) Check for the normal stress resistance of the cross-section with axial force bending and shear

force

1 . w

x

hN M

A W h -978,198 daN/cm2

1

.

.

f

x

x w

V S

I t -78,884 daN/cm2

=>2 2

1 13td 987,69 daN/cm2 < 1,15 cf 2415 daN/cm2

c, Check for total buckling of the rafter at this cross-section.

ol = 1,477 => o

f

l

b 4,344

0.41 0.0032 (0.73 0.016 )f f fo

f f fk

b b bl E

b t t h f

19,607

( f

f

b

t28,33 ; fkh 78,8 cm )

d, Check for local buckling of flange and web of the rafter.

- The flange of rafter : 0f 0,5f

b E

t f

w

0f2

fb tb

165mm ; 0f

f

b

t 13,75 < 0,5

E

f 15,811

- The web of rafter : w

w

3,2h E

t f

w

w

h

t 77,6 < 3,2

E

f101,193

5.2.2 Check for the unchanged cross-section of rafter

Internal forces :

M (daN.m) N (daN) V (daN)


11444,43 1651,58 1278,42



bf tf hw tw bf tf

500 340 12 476 10 340 12


Jx Wx Jy Wy Sx Sfx A

(cm4) (cm3) (cm4) (cm3) (cm3) (cm3) (cm2)

57578.68 2303.147 7864.77 462.634 1278.74 995.52 129.2

The relative eccentricity ratio : W

x

x

e M Am

N 38,872 cm

b, check for strength condition :


+) Check for the normal stress resistance of the cross-section with axial force and bending

moment about the x axis is required as follow :

. c

n xn

N Mf

A W

n xn

N M

A W 509,687 daN/cm2 < 2. 2100 /cf daN cm

( the cross sectional area with considering the reduction An = 129,2 cm2

The elastic section modulus about the x with considering reduction Wx= 2959,646 cm3 )

+) Check for the normal stress resistance of the cross-section with shear force

max

.

.

x

x w

V S

I t 28,392 daN/cm2 < 2. 2100 /cf daN cm

+) Check for the normal stress resistance of the cross-section with axial force bending and shear

force

1 . w

x

hN M

A W h 485,836 daN/cm2

1

.

.

f

x

x w

V S

I t 22,104 daN/cm2

=>2 2

1 13td 487,342 daN/cm2 < 1,15 cf 2415 daN/cm2

5.2.3 Calculation of fillet weld connections between flange and web of the rafter

The height of weld lines : w min

.

2.( ) . .

V f

f

x c

V Sh

f I

VV 3493 daN ;1

. .2

f f f fxS b t h 995,52 cm3

The strength of weld lines when using electrode N42 : w min( . )f 1260

=> 0,024fh cm

=> choose 6fh mm


5.3 Check for displacement at the tip of column

The displacement must satisty the following condition :

300

H

- The displacement caused by combination of dead load and wind load:

1

0,00379 0.04020,0301

1,1 1,2

m < 0,0349m

- The displacement caused by combination of dead load and crane load:

2

0,00379 0.001180,0045

1,1 1,1

m > 2 < 0,0349m

- The displacement caused by combination of dead load ,wind load and crane load:

3

0,00379 0,0402 0,00240,5 0,0155

1,1 1,2 1,1

< 0,0349m

6. Calculation for some details of frame

6.1. Fixed connection between column and foundation

6.1.1 Internal force for calculation

The couple of internal forces has Nmax and Mtu , Vtu

N1 = - 24492 daN ; M1 = 23379 daN.m ; V1 = 5590 daN

The couple of internal force has max

M and Ntu , Vtu

N2 = -723 daN ; M2 = -56179 daN.m ; Q2 = -11096 daN

a) Calculate the base plate

- Assuming that the width of base plate is wider 20 cm that of column

So, the width of base plate : B = bc + 20 = 34 + 20 = 54 (cm)

- The length of base plate is calculated according to following formular :


2

, , ,

6.( )

2. . 2. . .b loc b loc b loc

N N ML

B R B R B R

With, Rb,loc : local compressive strength of concrete used for foundation

, . .b loc b bR R

We use B15 grade concrete : = 1 ; Rb = 85 (daN/cm2) ; Rbt = 7,5 (daN/cm2)

3 1,5mb

bd

A

A We choose temporarily b = 1,2

, . .b loc b bR R = 1.1,2.85 = 102 (daN/m2)

: the factor depend on the feature of distributing loads ; because of ununiform distributed load ,

so = 0,75

- For the first couple of internal force have N1 = - 24492 daN ; M1 = 23379 daN.m ; V1 = 5590

daN

We have : 2

224492 24492 6.23379.10( ) 61,31( )

2.54.0,75.102 2.54.0,75.102 54.0,75.102L cm

For the second couple of internal force have N2 = -723 daN ; M2 = -56179 daN.m ; Q2 = -11096

daN

22723 723 6.56179.10

( ) 90,24( )2.54.0,75.102 2.54.0,75.102 54.0,75.102

L cm

We choose L = 100 cm

We choose the second couple of internal force to calculate stress at base plate

22

ax .2 2

6. 723 6.56179.1062,3( / ) . 76,5

. . 54.100 54.100m b loc

N MdaN cm R

B L B L

22

min 2 2

6. 723 6.56179.1062,5( / )

. . 54.100 54.100

N MdaN cm

B L B L

The distance from the point having min to the point having 0


1

62,5100. 50,1( )

62,5 62,3y cm

We choose the thickness of base beam is the same thickness of the web of column tdd = 12 mm.

The thickness of stiffener tsd = 10 mm ; the distance between stiffeners is 110 mm . The detail

dimension is showed below :

In order to determine the thickness of base plate, we determine the maximum stress at the

boundary of each slab :

- Stress at the edge of column : 40.1

62,5. 5050.1

c (daN/cm2)

- Calculate bending moment in parts of base plate

* Slab 1 ( a slab supported by three position) :

We have : b2/a2 = 26,4/26 = 1,015 0,113b

2 2. . 0,113.50.26 3819,4bM d (daN.cm/cm)

* Slab 2 ( a slap supported by three position) :

We have : b2/a2 = 8,8/ 13= 0,677 0,085b

2 2. . 0,085.62,5.13 895,5bM d (daN.cm/cm)

* Slab 2 ( a slap supported by two position) :

We have : 2 2

2 8,8 12,5 15,2a (cm) ; 2

12,58,8. 7,23

15,2b

2

2

7, 230, 475

15.2

b

a we calculate as console with d = b2 = 7,2 0,5b

2 2. . 0,5.62,5.7, 2 1620bM d (daN.cm/cm)

So , we choose Mmax = 3819,4 (daN.cm/cm) to determine the thickness of base plate


6. 6.3819,4

3,38. 2000.1

bd

c

Mt

f (cm) = 33,8 (mm)

We select tbd = 35 (mm)

b) Calculate base beam

Base beam and the flange of column is welded each other. The thickness of the flange of

column is 16 mm, so the minimum height of weld line hfmin = 6 mm, so we choose hf = 7 mm.

The height of base beam according to condition of force transportation

d 2d

c

N MN

h

hc = 78,6 cm : the distance between the center of two flange of frame

d

723 561790071113

2 2 78,6d

c

N MN

h

(daN)

d

711131 1 40,3

2.( . ). 2.1260.0,7d

dd

w f

Nh

h (mm)

We have ( . w ) = 0,7.1800 =1260 (dan/cm2)

We choose hdd = 45 mm

Consider the loading causing bending as uniform one and having the value as following :

62,5.(10 26 / 2) 1437,5q (daN/cm)

Base beam is calculated similiarly to console fixed at the flange of column and had span :

L = (54 – 34 )/2 = 10 (cm)

Bending momment of base beam : 2 21437,5.10

718752 2

qlM (daN.cm

- Checking bend of beam : 2 2

6. 6.71875177

1, 2.45

dddd

dd dd

M

t h (daN/cm2) < 2100 daN/cm2

Satisfy the condition

c) Calculate bolts


* Determine bolts :

In order to determine bolt, we choose the couple of internal force that cause maximum

tensile force between base plate and foundation. From force combination table we have the most

dangerous force couple : N = -723 daN ; M = -56179 daN.m ; V = -11096 daN

With : Nt t = -8152,6 daN ; Nht = 7429,5 daN ; Mt t = 11841 daN.m ; Mht = -68020 daN.m

Internal force to determine bolt :

8152,6

.0,9 .0,9 7429,5 759,21,1 1,1ht

ttNN N

daN

11841

.0,9 .0,9 68020 583321,1 1,1ht

ttMM M daN.m

Stress under base plate :

2

ax 2 2

6. 759,2 6.58332.10064,67( / )

. . 54.100 54.100m

N MdaN cm

B L B L

2

min 2 2

6. 759,2 6. 58332.10064,95( / )

. . 54.100 54.100

N MdaN cm

B L B L

The distance between the edge of base plate ( the point having max ) and the point having 0

2

64.67100. 49,9

64,67 64,95y

(cm)

The distance from point that the axial force put on to the center of compressive stress diagram

1 100 49,933,4( )

2 3 2 3

yLa cm

We put bolts at the distance of 5cm from the edge of base plate, so the distance from bolt to the

center of compressive stress diagram : y = 5 + 100/2 + 33,4 = 88,4 (cm)

Tensile force apply on bolts : 5833200 759,2.33,4

66273,788,4

blN

(daN)

We use type of bolt MnSi with fba = 1900 daN/cm2 ; required area of bolt : Abl = 66273,7/1900 =

34,4 (cm2) We choose 2 bolts with 56mm ( Abl = 41 cm2 )

d) Calculate stiffener supporting bolts


Choose the thickness of stiffener supporting bolts : tsd = 10 mm ; the height of supporting

stiffener equal to the height of base beam hsd = hdd = 45 mm

Bending moment caused by tensile force of bolt ( the distance between the axis of bolt and

base beam is 13,8 cm)

M = 66273,7.13,8 = 914577 (daN.cm)

+) Check the thickness of stiffener supporting bolts :

2 2

6.( / 3) 2.9145770,43

. . 45 .2100sd

sd c

Mt

h f (cm)

So, with tsd = 1 cm sastify condition

+) Check weld connection between stiffener supporting bolts and base beam :

We choose the height of weld line hf = 9 mm ; there are three stiffener supporting bolts ,

one is right in the middle ane the others is put at two side.Firstly , we welded middle stiffener

with two weld line and then we weld two other stiffener with two weld line. However the

distance between stiffeners is near that caused difficulty in welding duration, so for more safe ,

we can consider weld lines is external weld line of two side stiffener and two weld lines of

middle stiffener that is subjected force.

We calculate the section modulus and the area of four weld line as following :

2 2( 1) (45 1)

4. . . 4.0,7.0,9. 813,126 6

sdwf f f

hW h

(cm3)

4. . .( 1) 4.0,7.0,9.(45 1) 110,88wf f f sdA h h (cm2)

Stress caused external force in the weld lines :

2 2 2 2914577 66273,7( ) ( ) ( ) ( ) 1273,7

813,12 110,88

blsd

wf wf

NM

W A (daN/cm2) < fwf. c = 1800

daN/cm2

The weld line satisfy condition

+) Check steel plate what e-cu of bolt is put on :

Because the tensile force in bolt is quite large, so we use two profile steel C12 ( with Wx =

2.50,6 cm3) put on two stiffener supporting bolts. The distance between two the stiffener is l =

140 mm = 14 cm


Bending moment caused by tensile force of bolt :

max

3. . 3.66273,7.14173968,5

16 16

blN lM (daN.cm)

Stress in the profile : max 173968,51719

2.50,6X

M

W (daN/cm2) < 2100 (daN/cm2)

satisfy

d) Calculate partition stiffener

Partition stiffener works as console and fix at the web of column with span l = 26,4 cm. The

partition stiffener is subjected to compressive stress from base plate. We can consider the stress

is uniform and has value that equal to the value at the centre of ô bản. The distance from this

point to the outer edge of base plate is : 10 + 26/2 = 23 cm

We have : 100 50,1 23

33,762,5 100 50,1

(daN/cm2)

The load apply on partition : 33,7.(100 50,1 23) 906,53q (daN/cm)

2906,53.26,4

315987,62

M (daN.cm)

906,53.26,4 23932,4V (daN)

- The height of partition stiffener : 6. 6.315987,6

30. . 1.2100.1

sn

sn c

Mh

t f (mm)

Choose hsn = 35 mm

- Check two weld line connecting partition to the web of column. We choose the height of

weld line hf = 8 mm ; lw = 35 -1 = 34 mm

Stress appear in weld line :

2 2 2 2 2 2

2 2

6.315987,6 23932,4( ) ( ) ( ) ( ) 1593,5( / ) 1800( / )2 . . 2 . . 2.0,7.0,8.34 2.0,7.0,8.34

sd

f f w f f w

M VdaN cm daN cm

h l h l

Satisfy condition

e) Calculate weld line connecting column to base plate


According to TCXDVN338:2005 ; the minimum height of weld line hfmin = 9 mm when the

maximum thickness of element is 35 mm

- The maximum height of weld line according to perimeter the flange of column with base plate

:

hfmax = 1,2.tmin = 1,2.16 = 19,2 mm

- The maximum height of weld line according to perimeter the web of column with base plate :

hfmax = 1,2.tmax = 1,2.12 = 14,4 mm

We choose the height of weld line : hf =10 mm

Ax = 2.0,7.1(34+31) + 2.0,7.1.75,2 = 196,28 (cm2)

Iw = 2.0,7.1.(34.40,52 + 31.38,12) + 2.0,7.1.75,22

12 = 141735,6 (cm4)

.2 141735,6.2

34572 80 2

xx

c

IW

h

(cm3)

The couple of internal force to calculate weld line :

N2 = -723 daN ; M2 = -56179 daN.m ; Q2 = -11096 daN

2 2 2 256179.100 723 11096

( ) ( ) ( ) ( ) 1629,73457 196,28 196,28

hl

x x x

M N Q

W A A

(daN/cm2)

So 1800hl (daN/cm2) satisfy condition

6.2 Calculate crane runway girlder and console

Choose crane runway girder has the dimension of cross – section as following :

hdct = 0,65 (m) ; bf = 0,3 (m) ; tf = 1,4 cm ; tw = 1 cm

Because the crane capacity is not large (Q = 7T) ; so we choose the dimension of section of

brake beam is the top flange of crane runway girder. We consider this beam as simple beam put

on console . To connect the top flange of the beam to the flange of column, we use the angle

steel L90X9 ; and steel plate 170x120x10.

a) Check crane runway girder according to strength condition :


- Internal force is caused vertical pressure

With simple beam , internal force is calculated based on Vinkple principal . It mean that

the maximum moment will happen if combination of all forces applying on the beam R is

symmetric through the middle beam point with force P that be closest to , maximum moment

appear at the point that put on.

Base on diagram of loading , we calculate maximum moment in beam as following :

b) console calculation

distance e between rail centroid and inner column edge e = hc = 100 – 80 =20 cm

- Shear force and bending moment in fix cross section Mv = (Dmax + Gdct).e ; M = (14657+1080)x20 =314740 . daN.cm

Vv = (Dmax + Gdct ); V = 14657+1080= 15737 daN

Select preliminary wing thickness tf=12 mm, the required thickness of console web:

max 15737

0,14( 2 ). . (30 2.1, 2).3460.1

dctw dct

f f c c

D Gt

b t f

cm

Choose tw = 8 mm

Choose the heigh of console at a point Dmax put on h = 40cm.

Select preliminary angle between lower console flange and horizontal direction = 150, hence

console fix section heigh is: hdv= 40+ 20.tg150 = 45,4 cm

we choose hdv = 50 cm

The required flange area of console :

3 3

2 2

. . 2 314740.50 0,8.50 2( ). ( ). 0,67

. 12 2100.1 12

yc v dv w dvf

c dv dv

M h t hA

f h h (cm2)

select console flange: bf x tf = 30x1.2 cm; Af = 36 cm2 > -0,67 cm2

* Check the selected cross section (neglect console selfweight)

The cross section of console fixed with column

Bảng 6.1.geometry properties of fix section

Ix

Wx A Sx Sf

(cm4) (cm3) (cm2) (cm3) (cm3)

50064,6 2002,6 110,1 1105 878,4


314740

157, 22002,6

v

x

M

W daN/cm2 < 2100 daN/cm2

. 15737.1105

434, 2. 50064,6.0,8

v x

x w

V S

I t daN/cm2 < 1200 daN/cm2

checking equivalent stress :

1

314740 47,6. . 149,6

2002,6 50

v w

x

M h

W h daN/cm2

. 15737.878,4

345,1. 50064,6.0,8

v f

x w

V S

I t daN/cm2

2 2 2 2

1 13. 149,6 345,1 376,1td daN/cm2 < 1,15.2100 = 2415 daN/cm2

- checking the shear stress at minor section

Bảng 6.2. geometry properties of console minor section at the point the crane

runway girlder put on

Ix Wx A Sx Sf

(cm4) (cm3) (cm2) (cm3) (cm3)

30658,4 1532,5 102.1 839,8 698,4

. 15737.839,8

538,8. 30658,4.0,8

v x

x w

V S

I t daN/cm2 < 1200 daN/cm2

- Check the local stability :

+ the flange : 630 2,1.10

25 31,61,2 2100

f

f

b

t ; so the flange satisfy the local

stability condition

+ the web : 647,6 2,1.10

59,5 2,2. 69,60,8 2100

w

w

h

t ; so the web satisfy the local

stability condition


- The heigh of flange- web weld connection :

min

. 15737.698,40,142

2.( . ) . . 2.1.1260.30658,4.1

v f

f

w x c

V Sh

f I (cm)

We choose hf = 6 mm

* calculate welding link between console and column flange

We choose the height of the welding connection hf = 6 mm ; we have :

Aw = 2.0,7.0,6.(29 + 26 + 46) = 84,84 (cm2)

Iw = 2.0,7.(29.0,6.252 + 26.0,6.23,82 )+ (0,6.463)/12 = 34489,6 (cm4)

34489,6

1379,6/ 2 25

xw

IW

h (cm3)

2 2 2 2314740 15737

( ) ( ) ( ) ( ) 294,831379,6 84,84

hl

x x

M V

W A

So , hl = 294,83 daN/cm2 < 1800 daN/cm2 satisfy condition

Check equivalent stress in column web

In column web, position connected to console flange, is subjected horizontal force. it will

appear complex stress state. therefore check the equivalent stresses according to the following

formula :

2 23. 1,15. .td cf with cot cot

M N

W A ;

( )

b

V H

A

The internal force of the column at position connecting to console :

M = - 22462 daN.m ; N = - 22488 daN ; V = 5590 (daN)

Wcot = 4634,8 (cm3) - elastic column section modulus

. Acot = 172,4 (cm2) ; Ab = 77,2 (cm2) - respectively the area of cross section of whole column

and the web of column.

- Horizontal force due to console effect on column : 314740

6294,850

H daN


cot cot

22462.100 22488615,9

4634,8 172,4

M N

W A

(daN/cm2)

( ) 5590 6294,8

178,377,2b

V H

A

(daN/cm2)

2 2 2 23. 615,9 3.178,3 689td (daN/cm2) < 1,15.2100 = 2415 daN/cm2

* The dimension of stiffener

- stiffen for console

The height of the stiffener : hs = h – 2.tf = 40 – 2.1,2 = 37,6 (cm)

The width : 376

40 40 52,5330 30

ss

hb (mm) we choose bs = 12 cm

The thickness of the stiffener is chosen based on condition of local buckling of stiffener

0,5.s

s

b E

t f ; so

6

12 2100. 0,753

0,5 2,1.10st

Choose ts = 0,8 cm

- stiffen for the web of column :

The height of the stiffener : hs = hwc = 76,8 cm

The width : 768

40 40 65,630 30

ss

hb (mm) we

choose bs = (bf – tw)/2= (34-1,2)/=16,4

The thickness : 6

16,4 2100. 1

0,5 2,1.10st ,1 (cm) We choose ts = 1,2 cm

6.3 Design the connection between column and rafter

The maximum internal forces combinations :

M = -31323,4 daN.m ; N = -7351,2 daN ; V = 4993,8 daN

M = 26289 daN.m ; N = 3620daN ; V = -3500daN


The combine :

M = -31323,4 daN.m ; N = -7351,2 daN ; V = 4993,8 daN

Regard as the connection rotates around edge bolt row (on the right) :

+) Design and check :

Choose 12 bolts ,the diameter of each bolt is 20mm.Bolts of grade 8.8

The actual area of each bolt : bnA 2,45 cm2 ; the area of each bolt blA 3,14 cm2

The tensile strength 24000 /tbf daN cm , The shear strength 23200 /vbf daN cm

- Check for the tensile resistance of bolts :

Neglect bolt rows near center of rotation , ( 1 78h cm ; 2 66h cm ; 3 54h cm ) :

1max 2 2 2 2

. 3132340.78

2. 2.(78 66 54 )b

i

M hN

h

9146,55 daN

The tensile strength of bolts : .bn tbtbN A f 9800daN > maxbN

- Check for the shear resistance of bolts :

Shear force applies to one bolt : 4993,8

416,1512

vbN daN

The shear strength of the bolt : 3,14.0,9.3200 9043,2vb

N daN > vbN


+) Calculation of joint plates:

-The thickness of joint plates :

1 22

1

. 9146,55.66

78

N hN

h 7739,38 daN

1 33

1

. 9146,55.54

78

N hN

h 6332,23 daN

max. 14.2.9146,551,1 1,1

2.( ). 2.(34 14).2100bb

bb

g Nt

b g f

1,24cm

. 14.2.(9146,55 7739,38 6332,23)1,1 1,1

2.( ). 2.(34 14).2100

i

bb

bb

g Nt

b g f

2 cm

Choose bbt 2 cm

Combine : M = 26289 daN.m ; N = 3620daN ; V = -3500daN

Regard as the connection rotates around edge bolt row (on the left) :

1max 2 2 2 2

( . ). (2628900 3620.45,5).78

2. 2.(78 66 54 )b

i

M N e hN

h

8157,44 daN<

tbN

b, Calculation of fillet weld connections between joint plates and column (rafter)

Follow table 43 TCXDVN 338 -2005 : the minimum height of fillet weld : 7mm


The maximum height of fillet weld between joint plates and the flange of column:

1,2tmin = 1,2tf = 14,4 mm => choose 8mm

The maximum height of fillet weld between joint plates and the web of column:

1,2tmin = 1,2tw = 12 mm => choose 8mm

42.0,7.0,8(33 31) 2.0,7.0,8.76 156.8WA cm cm2

3

2 2 0,8.762.0,7(33.0,8.40 31.0,8.38,4 ) 151303,82

12WI cm4

cm3

With the internal force combination at this :

M = -31323,4 daN.m ; N = -7351,2 daN ; V = 4993,8 daN

2 2

2 2 2 231323,4.100 7351,2 4993,8( ) ( ) 1736,5 / 1800 /

1854,2 156,8 156,8hl

W W W

M N VdaN cm daN cm

W A A

-Calculating the welding line between butt plate and column, rafter.

We choose : height of weld is 8mm

Axial force N and moment applied on the welding line of flange

7351,2 31323,4.1007650,65

2 2 788C

TK

N MNc

h daN

w

7650,65147,13

( 1) ( 1) 0,8.33 0,8.32

cc

f f f f

N

h b h b t

daN/cm2<0,7.1800=1260

daN/cm2

Welding line of column web subjected to shear force which is : V=4993,8 daN

Stress of welding line on web: 2 2

w

4993,841,1 / 1260 /

2.0,8.76t daN cm daN cm

It can be capable bearing!

w2. 151303,821854,2

2.0,8 80 2.0,8W

IW

hc


6.4 Connection details of rafter.

The Internal force of cross-section 1-1:

M=105,52 (KN.m)= 10552 (daN.m) N= 43,22 ( KN) =4322(daN) V= 26,51 (KN)=2651 (daN)

a, Interconnection Calculation according to conception 1:

+ Number of Bolts:

Choose 8 bolts with diameter is 20 ,kind of 8.8 and shown as the figure.

Net bolt area is Abn=3,14 (cm2), gross bolt area is Abl=2,45 (cm2), calculates tensile strength of

bolts ftb=4000 daN/cm2 and calculates shearing strength of bolts fvb=3200 daN/cm2

+ Check the pull of resistant bolts:

Safely, Bolts is viewed as rotate outside bolt row in cross-section of rafter. The maximum force

at outside bolt row. ( y= 31 cm, h1=62 cm , h2=49 cm )

1max 2 2 2

( . ). (1055200 4322.31).625944,95( )

2. 2(62 49 )b

i

M N y hN daN

h

Tensile Bearing capacity of bolts: [N]tb=Abn.ftb=2,45.4000=9800 (daN)

Condition: [N]tb =9800 (daN) > maxbN =5944,95 (daN) (satisfaction )

+ Shear force influence a bolt:

Nvb=2651/8=331,38 (daN)

Shear bear capacity of a bolt : [N]vb=3,14.0,9.3200=9043,2 (daN)

12

468

12

500

32 110110

1616

165 10 165

95 75 75 95340

50

07

10

11

54

76

11

54

51

30

36

01

30

45

6

6

8M20


Condition: [N]vb=9043,2 (daN) > 331,38 (daN) (satisfaction )

Butt plate calculation.

+ The thickness of butt plate. Tension force of each bolt

1 22

1

. 5944,95.494690,33

62

N hN

h (daN)

ax. 15.2.5944,951,1 1,1 1,02

2( ) 2(34 15).2100

M

bb

bb

g Nt cm

b g f

. 15.2.(5944,95 4690,33)1,1 1,1 1,37

2( ) 2(34 15).2100

i

bb

bb

g Nt cm

b g f

For safety, We choose : tbb=1,6 cm

Check couple internal force:

M=114,44 (KNm) = 1144400 (daN.m), N=18,53 (KN) = 1853 (daN), V=12,79(KN)=1279 (daN)

The maximum force of a bolt at outside bolt row.

1max 2 2 2

( . ). (1144400 1853.31).626010,76( )

2. 2(62 49 )b

i

M N y hN daN

h


Condition: [N]tb =9800 (daN) > maxbN = 6010,76( )daN (satisfaction )

The load applied on a bolt: Nvb=1279 /8=159,88 (daN)



b, Calculation of welding line for connecting between butt plate and rafter

According appendix 43 in TCVN 338:2005: The minimum height of automatic welding line is 5

mm when the maximum thickness of the butt plate was 16 mm.

The maximum height of welding line between the flange of rafter and the butt plate is:

1,2tmin=1,2tf=1,2.12=14,4 mm => we choose : 5mm

The maximum height of welding line between the web of rafter and the butt plate is:


1,2tmin=1,2tw = 1,2.10=12 mm=> we choose: 5 mm

Aw=2.0,7.0,5.(33+31) +2.0,7.0,5.45,6=76,72 (cm2 )

Iw=2.0,7(33.0,5.252 +31.0,5.23,32+0,5.45,63

12 )=31749,3 cm4

=> Ww=𝐼𝑤.2

ℎ𝑑+2.0,5=

31749,3 .2

50 +2.0,5 =1245,1 cm3

M=105,52 (KN.m)= 10552 (daN.m) N= 43,22 ( KN) =4322(daN) V= 26,51 (KN)=2651

(daN)

thl= 2 2 2 2

w w w

1055200 4322 2651( ) ( ) ( ) ( ) 904,5W A 1245,1 76,72 76,72

M N V

A daN/cm2

<1800daN/cm2

The welding line was sufficient bearing capacity

6.5 The top of rafters.

The Internal forces at top of rafter.

M=86,39 (KN.m)= 8639 (daN.m) N= -41,63 ( KN) =-4163 (daN) V= 7,34 (KN)=734 (daN)

=> consider the following General coordinates ( OXY) : M=86,39 (KN.m)= 8639 (daN.m)

V’= -4163.cos(80*)-734cos(10*)=-1445,75 (daN)

12

476

12

500

110 32 110

16 167

7

8M20

165 10 165

95 75 75 95340

500

710

115

476

115

45

130

360

130

45


a, Bolt Calculation

+ Number of Bolts:

Choose 8 bolts with diameter is 20 ,kind of 8.8 and shown as the figure.

Net bolt area is Abn=3,14 (cm2), gross bolt area is Abl=2,45 (cm2), calculates tensile strength of

bolts ftb=4000 daN/cm2 and calculates shearing strength of bolts fvb=3200 daN/cm2

+ Check the pull of resistant bolts:

Safely, Bolts is viewed as rotate outside bolt row in cross-section of rafter. The maximum force

at outside bolt row. (y= 31cm, h1=62 cm , h2=49 cm)

1max 2 2 2

( ). (863900).624288,37( )

2. 2(62 49 )b

i

M hN daN

h


Condition: [N]tb =9800 (daN) > maxbN = 4288,37 (daN) (satisfaction )

+ Shear force influence a bolt:

Nvb=1445,75 /8=180,72 (daN)



Butt plate calculation.

+ The thickness of butt plate. Tension force of each bolt

1 22

1

. 4288,37.493389,2

62

N hN

h (daN)

ax. 15.2.4288,371,1 1,1 0.87

2( ) 2(34 15).2100

M

bb

bb

g Nt cm

b g f

. 15.2.(4288,37 3389,2)1,1 1,1 1,2

2( ) 2(34 15).2100

i

bb

bb

g Nt cm

b g f

We choose : tbb=1,6 cm

Check internal force in combination 2:

M=82,39 (KNm) = 8239 (daN.m), N=-39,72 (KN) = -3972(daN), V=7,01(KN)=701 (daN)


Consider the following General coordinates ( OXY):

M=82,39 (KNm) = 8239 (daN.m), V=-3972.cos(80)-701cos(10=1380,1 (daN)

The maximum force of a bolt at outside bolt row.

1max 2 2 2

( ). (823900).624089,8( )

2. 2(62 49 )b

i

M hN daN

h


Condition: [N]tb =9800 (daN) > maxbN = 6105.19( )daN (satisfaction )

The load applied on a bolt: Nvb=1380,1 /8=172,51 (daN)


Condition: [N]vb=9043,2 (daN) >172,51 (daN) (satisfaction )

b, Calculation of welding line for connecting between butt plate and rafter

According appendix 43 in TCVN 338:2005: The minimum height of automatic welding line is 5

mm when the maximum thickness of the butt plate was 16 mm.

The maximum height of welding line between the flange of rafter and the butt plate is:

1,2tmin=1,2tf=1,2.12=14,4 mm => we choose : 5mm

The maximum height of welding line between the web of rafter and the butt plate is:

1,2tmin=1,2tw = 1,2.12=14,4 mm=> we choose: 5 mm

Aw=2.0,7.0,5.(33+31) +2.0,7.0,5.45,6=76,72 (cm2 )

Iw=2.0,7(33.0,5.252 +31.0,5.23,32+0,5.45,63

12 )=31749,3 cm4

Ww=𝐼𝑤.2

ℎ𝑑+2.0,5=

31749,3 .2

50 +2.0,5 =1245,1 cm3

M=86,39 (KN.m)= 8639 (daN.m) N= -41,63 ( KN) =-4163 (daN) V= 7,34 (KN)=734

(daN)

thl= 2 2 2 2

w w w

863900 4100 723( ) ( ) ( ) ( ) 747,3W A 1245,1 76,72 76,72

M N V

A daN/cm2

<1800daN/cm2

The welding line was sufficient bearing capacity

stuctural steel project ( ĐỒ Án kẾt cẤu thÉp ĐẠi hỌc xÂy dỰng )

Engineering

height of skylight

height of lower column

height of upper column

catalog of crane

crane safe distance

horizontal frame

frame bay

column h