Splices & Connections

7/28/2019 Splices & Connections

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Splices and connections

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1- Design of connections of vertical and diagonal: ECP 130

a- At the connection away from X.G.:

1-vertical and diagonal

Number per one side = n =P

F

s

max / 2 distributed on 2 flanges

Fmax = Maximum allowable stress * total area

Maximum allowable stress

= Fc = 1.6 ( 8.5 * 10-5

) for Compression member

= Ft =

max

min1T

TFsr

= 0.58Fy for tension member

2-flange

Arrange bolts in the 2 flanges.

( 7 rows ))(

.

b Fl 2*3 + t w case of 2 columns



1.5d1.5d

1.5d3d

1.5d


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ButtW

eldbet.theT

wo

diff

erentThick


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3-:

1- Centerline 2- Upper or lower chord

3- Vertical 4- Diagonal

4- )diagonal(edge and pitch

gusset

Min e,p

Arranging bolts on min edge and pitch


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Formation of gusset plate according to

arrangment of bolts of diagonal (mostly critical)

Very important note:

The gusset plate has only 3 shapes no more. If there is no diagonal,

the angle of gusset plate will be 150

3 differnt shapes of Gusset plate

Min e,pMin e,pMin e,p

vertical

edge = 2 d


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l memberwith vertica.(G.Design of connection at position of X-b

i- If the connection is notsubjected to moment:

i.e. All connections of deck bridge & all connections of through bridge

except at position of end portal frame bracing.

This means that every face is designed alone.



F

s


2-flange


3-:




gusset

Min e,p

Arranging bolts on min edge and pitch


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5- Draw head plate and XG vertical

(Height of XG is designed from loads on XG, while height of head plate

= hxg + 4cm (2cm from each side)

a- If hhead plate < hgusset plate:


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1XGGusset plate

XGvertical

16

2head plateedge &

pitchXGn1head

plateGusset plate

C Check n2 of XG from drawing. From drawing in this page n2

Connecting head plate with Gusset plate) = 16 bolt

sXG Pn

R 2

:

n1verticalGusset plate

XG

n1 of vertical in this drawing = 20 bolt


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b- If hhead plate > hgusset plate

1XGGusset plate

2n120Gusset

plate)(gussethead

platehead platevertical

gusset plate

3packingGusset platehead plate

Gusset platehead plateXG


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Min e,p

Draw the connection between the vertical & upper or lower chord first,

then draw X.G.

vertical member

Cross-sectionofDeck bridgeCross-section ofthroughbridge

Stringer n2

1n

n2

n

11

n

cross girder

cross girder

lower chord

vertical

member

diagonal of X-frame


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ii) If the connection is subjected to moment:

i.e. All connections of Pony Bridge and connections of through bridge at

position of portal frame bracing.

The inner face of bolts is designed the maximum strength of the vertical

While the outer face of bolts designed on M & Q where Q = RXG

M = hC

100+ w

2

2hfor pony bridge where w = 0.1*0.5*S

M = hRu *3

1*

2

1for through bridge with closed vertical frame

(Loaded case because of RXG

Pony bridge

Through bridge

CL of upper beam

M

uVloftruss

Vloftruss

CL of XG

h

CL of XG

2h/3

h/3

CL of upper beamRu

M

uR

2h/3

R /2

h/3

R /2u

R /2u

CL of XG

R /2u

Vloftruss

Vloftruss

M = lRu *2

1for through bridge with inclined open frame

Every face is designed alone. Inner face for bolts between vertical

and gusset plate. And outer face between head plate and gusset



F

s


2-flange


3-:




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gusset



5- Draw head plate and XG vertical

(Height of XG is designed from loads on XG, while height of head plate

= hxg + 4cm (2cm from each side)

haunchXG


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n1gusset plate

n 1 is number of bolts in one side (one flange) of the I-beam = 20 bolt

n 2 : check : sXG Pn

R

2

n2 = 8*4 = 32bolts

MbextT ,, 0.8 T where

MbextT ,, = +

XBff

n**

2*

1 21 0.8 T

n is the no. of bolts in one row

Where 11 yM

f = , where12

3BHI = ,

21

Hy =

22 yI

Mf = ,

22

Hy = - x

The diagonal is designed as before


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Cross section of pony bridge

rail

upper chord

1

lower chordcross girder

2n

n

bracket

n1

vertical member

1n

n1

backing

ss girder

B

H

f1

f2X


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Design of CHORD Spli ce

The field splice for upper and lower chord will be used each 16m (max).

There is no field splice for diagonal and vertical

The splice for vertical and diagonal is used only when number of

bolts in one row is more than 7.

The splice may be either single shear or double shear

a- Splice with plate from one side (single shear):

Key drawing


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(1) Dimensioning:

For flange: Takefsp

f

bb

tt

=

=1

For web: Take)*{*

2

2 webwebsp

websp

tHtH

cmHH

=

=

(2) Number of bolts:

For flange:shear)(singleP

)For(F*)t*(b

s

cff1

tn =

For web: shear)(singleP

)For(F*)t*(H

s

tcwweb2 =n

FC for upper chord and 0.58Fy for lower chord.

If the number of bolts in one row is greater than 7, so we can use 2 splice

plates.

b- Splice with plates from both sides (double shear):

(1)Dimensioning of splice plates:

For flange:2

1b ft f = b f t 1 t 1 =

2

1t f

2

1b ft f= 2 * 0.4 b ft 2 get t 2

For web: h w t w = 2 * h s.pl * t 3

t min = 8 mm(2)Number of bolts:

For flange:shear)(double2P

)For(F*)t*(b

s

cff1

tn =

For web:shear)(double2P

)For(F*)t*(H

s

tcwweb2 =n

FC for upper chord and 0.58Fy for lower chord.


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1

3t

1n

n2

t2

t

c- Splice If there is stiffener in the upper or lower chord:We must make splice for the stiffener

3n

b2

h

b1


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n 3 (no. of bolts of stiffener)s

tcstst

P

orFFtbn

)(**3 =

Look at the drawing of the stiffener

For pony & through bridges:

stiffenerupper chordX.G

lower chordX.G

For deck bridge: vise versa

:diagonalvertical andSplice for

diagonal7vertical

XGXG7XG

XGhaunch.

1diagonalverticalXG

77gusset

plate:-



2720%

1.2 *[n required 7 * no of bolts in one row]

3splice platediagonal

:

Assume after design number of bolts for D1 = 40,

For D2 = 18 bolts and for v = 17 bolt and 4 bolts in each row

1404

10


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D1

D2

V

3verticaldiagonalXG7

D120D220verticalgusset3

gussethaunch

D1

D2

V


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4

ndiff= (40 4 * 7 ) * 1.2 = 14.4 Taken 16 bolts

4 rows

5Splice plate4agonaldigusset4

gussetpacking

In case of using splice plate, packing will be used at the position of

connection as shown.

D1

D2

V

vertical7


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Example 1:

Design & draw to Scale 1:10 the given joint using H.S.B M24 (Ps = 6.29

t). Ldiagonal = 3.5m Double Track Pony bridge St 44

-985t-945t

+144t

-76t

16

388

16

22

376

22

10 10

480 320

450

Vertical Diagonal

Solution:

Since the vertical and diagonal are subjected to both tension and

compression, so we will use the bigger of Fc and Ft .

For diagonal:

23

)2

6.1

2

38.8(6.1*32*2

12

8.38*1++=xI = 46651 cm

4

12

32

*6.1*2

3

=yI = 8738 cm4

A = 32 * 1.6 * 2 + 38.8 * 1 = 141.2 cm2

2.141

46651=xr = 18.2 cm

2.141

8738=yr = 7.9 cm

9.7

350*7.0=in = 31

2.18

350*2.1=out = 23.1

FC = 1.6 8.5 * 10

-5

* 31

2

= 1.52 t / cm

2


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Diagonal, class II, n = 200000 srF = 2.8 t / cm2

144

761

8.2max

=F = 1.83 t / cm2 > 1.6 t / cm2

designF = 141.2 * 1.6 = 225.92 t

29.6

92.225=n = 35.9 bolts per 2 flanges i.e 18 bolts per flange

b flange = 32 cm , (2 bolts) 6 d + t w = 6 * 2.4 + 1 = 15.4 cm < 32

(4 bolts) 12 d + t w = 12 * 2.4 + 1 = 29.8 cm < 32

So we have 4 bolts each row No. of rows =4

18 = 4.5

Take 5 rows (40 bolts in 2 flanges)

For vertical:

23

)2

2.2

2

37.6(2.2*48*2

12

6.37*1++=xI = 88067 cm

4

12

48*2.2*2

3

=yI =40550 cm4

A = 48 * 2.2 * 2 + 37.6 * 1 = 248.8 cm2

8.248

88067=xr = 18.8 cm

8.248

40550=yr = 12.8 cm

L vertical = 3.5 sin 450

= 2.47 m

89.12

247*7.0=in = 13.5

8.18

247*2.1=out = 15.8

FC = 1.6 8.5 * 10-5

* (15.8)2

= 1.58 t / cm2

srF as before

So designF = 248.8* 1.6 = 398.1 t

29.6

1.398=n = 63.3 bolts per 2 flanges i.e 31.6 bolts per flange

b flange = 48 cm , (4 bolts) = 12 d + t w = 12 * 2.4 + 1 = 29.8 cm < 48


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(6 bolts) = 6*3 d + t w = 18 * 2.4 + 1 = 44.2 cm < 48

6 bolts per one row No. of rows =6

6.31= 6 rows < 7

USE 6 rows * 6 bolts * 2 flanges = 72 bolts

Upperchord

4075

7575

7540

70

70

140

140

140

140

140

480320

gL


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Example 2:

Design the given diagonal joint using H.S.B M24 (Ps = 6.29 t). Ldiagonal =

3.5m. Double Track. Pony bridge. Assume vertical as previous example.

-985t-945t3.6

388

16

14

Diagonal

360

450 tension

Solution:

A = 38.8 * 1.4 + 2 * 36 * 3.6 = 313.52 cm2

F t = 1.6 t / cm2

< F max of fatigue

F design = 313.52 * 1.6 = 501.6 t

29.6

6.501=n = 80 bolt (40 bolt each flange)

b flange = 36 cm , (4 bolts) = 12 d + t w = 12 * 2.4 + 1 = 29.8 cm < 36 cm

(6 bolts) = 18 d + t w = 18 * 2.4 + 1 = 44.2 cm >36 cm

Use 4 bolts per row No. of rows =4

40= 10 rows > 7 use Splice

Assume n 1 = 7 rows n1 = 7*4 = 28 bolts

n2 = (40-28)*1.2 = 14.4 bolts No. of rows = 14.4 / 4 = 4 rows


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Example 3:

Pony Bridge, steel 52, M 24, Maximum compression force in upper

chord is 900 t, Force is the shown vertical = 110t. hxg = 100cm

It is required to design & draw the connection between the vertical, lower

chord & X.G. St 52

3.6

388

16

14

Diagonal

360

A

110t

5m

5m

Solution:

A = 38.8 * 1 + 2 * 30 * 1.6 = 134.1 cm2

Maximum & minimum forces are not given

F max = 134.1 * 0.58 * 3.6 = 280 t

For M 24, steel 52, case I, PS = 6.94 t

h = 5 0.5 = 4.5m2900 4.5

* 4.5 0.1 * 5 * 0.5 *100 2

M= + = 43 m t

2/

94.6

2801 =n = 20.2 bolts each side

30 cm > 12 * 2.4 + 1 = 29.8 taken 4 bolts

each row

no. of rows =20.2

4= 6 rows

For n 2 : take haunch 20 cm

So n 2 = 11 * 4 = 44 bolts

B = 30 cm (b fof X.G)

10050

f1

2f8*100mm

100

10050

H

170

End of G.Pl.

Packing


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Important note: The maximum force in the vertical is the maximum

reaction of the X.G (vertical carry only X.G)

44

110, =bextQ = 2.5 < 6.94 t

MbextT ,, = XBff

**)2

(*4

1 21 +

B = 30 cm X = 2 + 5 + 10 =17 cm

H = 100 + 2 + 20 + 2 = 124 cm12

124*30 3=I = 4766560 cm4

1

4300 124*4766560 2f = = 0.055 t / cm

2

2

4300 124* ( 17)

4766560 2f = = 0.04 t / cm2

MbextT ,, = 17*30*)2

04.0055.0(*

4

1 += 6.06 t < 0.8 * 22.23 = 17.8 t

O.K safe we can use no haunch.


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cross girder

bracket

vertical member

lower chord

n = 6 row

n = 11 row2

1

upper chord

rail


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Example 4:

Design & draw to Scale 1:10 the given joint using H.S.B M24 (Ps = 6.29

t). Ldiagonal = 3.5m inclined at 450, Single track, deck bridge, st 52, the

section of X.G is IPE 600. The connection is at the position of X-frame.

RXG = 150 t

16

388

16

22

376

22

10 10

Vertical Dia onal

480 320A

Solution:

For diagonal: as given in example 1

Ix= 46651 cm4

Iy= 8738 cm4

A = 141.2 cm2

rx= 18.2 cm ry = 7.9 cmin

= 31out

= 23.1

FC = 1.6 8.5 * 10-5

* 312

= 1.52 t / cm2

Diagonal, class II, n = 200000 srF = 2.8 t / cm2

144

761

8.2max

=F = 1.83 t / cm2 > 1.6 t / cm2

designF = 141.2 * 1.6 = 225.92 t

29.6

92.225=n = 35.9 bolts per 2 flanges i.e 18 bolts per flange

So we have 4 bolts each row No. of rows =4

18= 4.5

Take 5 rows (40 bolts in 2 flanges)


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For vertical: n1 is the same as for example 1.

Ix = 88067 cm4

Iy =40550 cm4

A = 248.8 cm2

rx= 18.8 cm ry = 12.8 cm L vertical = 3.5 sin 450

= 2.47 m

in = 13.5 out = 15.8

FC = 1.6 8.5 * 10-5

* (15.8)2

= 1.58 t / cm2

srF as before So designF = 248.8* 1.6 = 398.1 t

29.6

1.3981 =n = 63.3 bolts per 2 flanges i.e 31.6 bolts per flange

b flange = 48 cm ,(6 bolts) = 6*3 d + t w = 18 * 2.4 + 1 = 44.2 cm < 48

6 bolts per one row No. of rows =6

6.31= 6 rows < 7

USE 6 rows * 6 bolts * 2 flanges = 72 bolts

Design of n2: Assume n 2 = 6 * 4 = 24 bolts

For94.6

150:2n = 22 bolts < 24

Upperchord

4075

7575

7

540

70

140

140

140

170

100

480320

Lg

HeadPL.

StiffenerPL.

70


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Example 5:

Single track through bridge, RXG = 170 t, M 24, st 52, b = 40cm. L=80m.

For X.G. hxg = 100cm, tw = 1cm, bf= 30cm, tf= 2.5cm

Design & draw the marked connections in different views

2

110t

8m

8m

4

75t

144t

6m

8m 8m3

1

Upper plan

Solution:

For connection 1:

Estimation of diagonal:yF

A58.0

144= = 68.6 cm2

A f= (68.6 38 * 1) / 2 = 14.3 cm2

b f= 20 cm (min)

A = 20 * 2 + 38 = 78 cm2

F max = 78 * 2.1 = 163.8 t

2/

94.6

8.163=n = 11.8 bolts each flange

10

380

10

10

Diagonal

200


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Use 2 bolts each row no. of rows =2

8.11= 6 rows

For upper chord: There is no difference, so it will be designed on

maximum strength.

Force = 75 cos 45 = 53 t

Assume the section is as shown

6.3

3040=

tt = 2.6 cm

A = (60 + 2 * 40) * 2.6 = 364 cm2

y&&=364

)206.2(*206*40*23.1*6.2*60 ++ = 13.5

Ix = 60*2.6*(13.5-1.3)2

+2

+ 2

3

)5.136.22(6.2*4012

40*6.2 =

68177cm4

Iy = 2.6 *

12

6030+ 2*40*2.6*20

2= 130000 cm

4

Lin = lout = 0.85*8 = 6.8 m

Rx =364

68177= 13.7m in = 7.49

7.13

680=

Fc = 2.1 13.5 *10-5

* 49.72

= 1.77t/cm2

Fmax = 1.77* 364= 643t

lg = 114.5cm from drawing

5.114**4

643

S = 0.2 * 5.2 Smin = 1.4cm = 14mm

For vertical V1: No X.G but there is horizontal beam subjected to

moment.

F = 144 cos45 = 102 t (comp.)

t=2.6cm 40

40

60


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For the design of vertical and the connection of the vertical (n1)

we will use maximum force in the member, due to D + L + I, so use

loaded case of wind,

For design of horizontal beam and its connection (n2), we will

use unloaded case of wind, because the connection is subjected to wind

only.

Loaded case Unloaded case

4 3.1

5

4 4

For loaded case (for vertical): h/= 8 1.0 0.35 3.5 = 3.15 m

W w = 0.1 [0.5 * 4 + 0.5 * 3.15] = 0.36 t / m/

Ru = 0.36 * 80 / 2 = 14.4 t h = 8 -2

1= 7.5

M = hRu

3

2*

2=

14.4 2 * 7.5*

2 3= 36 m t

1h

2h3

2 3

Ru

u

2

RClosed frame

Estimation of vertical member: F = 102 t comp.

6.3

6440=

wtt w = 1.2 cm

Assume F C = 1.8 t / cm2 12

Vertical

400

20

20

360


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32/37

102 36

2 0.41.8

flangeA+

= = 78 cm2

Take b = 20 t, take t = 2 cm, b = 40 cm

A = 2 * 40 * 2 + 36 * 1.2 = 203.2 cm2

23

)2

2

2

36(*2*40*2

12

36*2.1++=xI = 62426 cm

4,

12

40*2*2

3

=yI = 21333 cm4

2.203

62426

=xr = 17.5 cm 2.20321333

=yr = 10.2 cm

2.10

800*7.0=in = 54.9 < 90

5.17

800*85.0=out = 38.9

F C = 2.1 13.5 * 10-5

* 54.92

= 1.69 t / cm2

n1 using maximum strength: 2/94.6

2.203*69.1= = 25 bolts each side

n1 (between Vl. and gusset) = 25 bolts each sideUse 4 bolts each row * 7 bolts each row.

For n2 (between horizontal beam and gusset)

There is no shear (neglect O W). For moment, unloaded case is critical

W w = 0.2* 0.5 * 4 *2 = 0.8 t / m/

Ru = 0.8 * 80 / 2 = 32 t

M = hRu32*

2= 32 2 * 7.5*

2 3= 80 m t

Horizontal beam:8000

1.8 1.8x

x

MS = = = 4570 cm3 use H E B 550

If we make beam without haunch so the arrangement of bolts will be as

shown


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upper chord

100

5*72=36060

36

114.5

36

Without haunchHEB550

H = 55 + 2 * 2 = 59 cm X = 2 +2.9+3 + 5 = 12.9 cm

12

59*30 3=I = 513447.5 cm4 1

8000 59*

513447.5 2f = = 0.46 t / cm2

2

8000 59* ( 12.9)

513447.5 2f = = 0.26 t /cm2

, ,

1 0.26 0.46( ) * 30 * 12.9

4 2ext b MT

+= = 35 t > 0.8 * 22.23 t = 17.8 t

So we have to use haunch. Take 7 rows (as vertical)

Take h haunch = 20cm and t fof stiff = 2 cm

upper chord

100

5*72=36

0

36

8065

60

114.5

36

HEB550

Packing

H = 2 + 55 + 20 + 2 + 2 = 81 cm X = 2 + 2 + 6.5 +4 = 14.5cm


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34/37

12

81*30 3=I = 1328603 cm4 1

8000 81*

1328603 2f = = 0.24 t / cm2

2

8000 81* ( 14.5)

1328603 2f = = 0.16 t / cm2

, ,

1 0.16 0.24( ) * 30 * 14.5

4 2ext b MT

+= = 21.8 t > 17.8 t unsafe

Increase one or 2 rows & recheck

For connection 2:

Since force in diagonal is smaller and we designed the previous

diagonal using min. section, (use same section)

For vertical, we have to estimate new section because the force

decreases and there is no moment.

Diagonal as before (same section), 2 bolts * 6 rows

Vertical: F = 75 cos45 = 53t

6.3

6440

=wt tw = 1.2 cm

Assume F C = 1.6 t / cm2

6.1

53= flangeA = 33 cm

2

A f= 33 40 * 1.2 = -ve

Take minimum flange 20 * 1 cm

A = 38 * 1.2 + 2 * 20 * 1 = 85.6 cm2

inl = 0.7 * 8 = 5.6 m outl = 0.85 * 8 = 6.8 m

23

)2

1

2

38(*20*2

12

38*2.1++=xI = 20697 cm

4,

6.85

20697=xr = 15.5 cm

2*12

20*1 3=yI = 1333.3 cm

4

6.85

3.1333=yr = 3.95 cm

95.3

800*7.0=

in = 142 > 90 unsafe

10

10

12

Vertical

200

380


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Use flange 30 * 2 cm note6.3

217.6

2

2/)1*22.130(


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36/37

lower chordcross girder

100cm

For connection 3:

The connection has X.G & subjected to moment because this connection

is in the vertical frame supporting the upper bracing.

The lower connection is subjected to moment & shear. The shear is the

maximum reaction of X.G = 170t

Diagonal: as before 6rows*2 bolts each row

For vertical n1 as in connection 1: (same member)

Use 4 bolts each row * 7 bolts each row.

114.5

5*72

95

10050

36

45

36

n 2 (for X.G): Assume n = 4 bolts per row in 7 rows (as n1) and add 3*4

bolts in the packing. n2 = 4 bolts per row in 10 rows n 2 = 40 bolt


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94.625.440

170

2

==n

Qt

B = 30 cm H = 100 + 2 * 2.5 + 2 * 2 = 109 cm

X = 2 + 2.5 + 5 + 10 = 19.5 cm12

109*30 3=I = 3237573 cm4

Ru (loaded case) = 11.44t M = hRu

3

1*

2=

14.4 1 * 7.5*

2 3= 18 m t

Note that: we used 1/3 because we are studying lower connection.

1

1800 109*

3237573 2f = = 0.03 t / cm2

2

1800 109* ( 19.5)

3237573 2f = = 0.019 t / cm2

, ,

1 0.03 0.019( ) * 30 * 19.5

4 2ext b MT

+= = 3.6t

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Splices & Connections