load check for shoe welded support.xls

7/28/2019 Load Check for shoe welded support.xls

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EPC FOR ASAB FFD PROJECT

JI-193

Page

STEEL MEMBER (or PLATE) - for Shoe / Line stop directly welded on bare pipe / pad

CS:

Member material = BS EN 10025 Grade S275 JR or equivalentYield strength of the structural member material,

Sy-member = N / mm2

LTCS:

Member material = BS EN 10113 Grade S355 NL/MLYield strength of the structural member material,

Sy-member = N / mm2

SS:

Member material = SS 316 PLATEYield strength of the structural member material,

Sy-member@250C = N / mm2

PIPE

Sy, Yield Strength of the pipe material, in N/mm

2

.from ASME Section II Part D, Subpart 1, Table Y-1.

Temperature, C -->

CS / LTCS

(for Structural member welded to pipe / pad of CS / LTCS)

SS

(for Structural member welded to pipe / pad of SS)

275.0

355.0

139.0

200 250 300 350 400 450 500

CS A106 Gr. B

- up to 24" NB & t 25.4 mm in all CS pipe classses), and

- 3" NB to 8" NB in CS-FBE lined pipe class 158028-X : lining

can be rapaired after shoe welding)

207 198 188 177 167 158 150

CS API 5L Gr. B

- 4" NB to 24" NB in CS-FBE lined pipe class 018028-X :

lining can be rapaired after shoe welding), and

- 3" NB to 8" NB in CS-Galvanised pipe class 018011-X - tobe galvanised after prefabrication).

Not available in ASME Section II Part D, Subpart 1, Table Y-1.

Hence considered same as that of A106 Gr. B / A333 Gr. 6 / A671

CC65, because Sh as per ASME B 31.3 of this material is also same

as these materials.

LTCS A333 Gr. 6- up to 24" NB in all LTCS pipe class), and

- up to 24" NB in CS pipe class, for t > 25.4 mm ), and

- in Alloy 825 pipe classes, for 8" NB to 24" NB LTCS

cladded with Alloy 625)

207 198 188 177 167 158 150

LTCS A671 CC65 Class-22

- above 24" NB, in all CS & LTCS pipe classes), and- in Alloy 825 pipe classes, for 24" NB & above, LTCS

cladded with Alloy 625)

207 198 188 177 167 158 150

A312 Gr.TP316/316L (Dual certified) - up to 8" NB

A358 Gr. TP316/316L Class 1 (Dual certified) - 10" NB &above

148 139 132 127 123 121 118

continued


2/313


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Page

Sy, Yield Strength of the pipe material, in N/mm2.

from ASME Section II Part D, Subpart 1, Table Y-1.

Temperature, C -->

SDSS (Super Duplex Stainless Steel)

(for Structural member welded to pipe / pad of SDSS)

ALLOY 825 Low Temp. (-50C to 250C)

(for Structural member welded to pad of Alloy 825)

200 250 300 350 400 450 500

continued from previous page

- up to 4" NB for all SDSS pipe classes, and

- up to 2" NB for all GRE pipe classes.420 404 396 393 393 - -

A928 S32750 Cl. 1- 6" NB & above for all SDSS pipe classes.

Not available in ASME Section II Part D, Subpart 1, Table Y-1.

Hence considered same as that of A790 UNS S32750, because this

material is also S32750.

B423 UNS N08825 - up to 6" NB. 201 193 186 181 177 175 174


3/313


JI-193

Pag

WELDING ELECTRODES, AND ALLOWABLE WELD STRESS

Pipe material Electrode

CS A106 Gr. B

CS - up to 24" NB,

for t 25.4 mm

(0C

to LTCS A333 Gr. 6

250C) - up to 24" NB, CS

for t > 25.4 mm

BS EN 10026 BS 5950

& LTCS A671 CC65 Grade S275 JR CLASS 42

- above 24" NB

[ Support codes :

CS - FBE CS API 5L Gr. B S1(CS), S3(CS), LS1(CS) ] 0.4 Sy-pipe @ 250C

LINED - up to 24" NB,- 3" NB [Only for or

& above CS-FBE Lined 150#,

and CS, Galvanized.] Sy-member

& or

Sy-pipe @ 250C Sy-member Su-weld

= N/mm2

= N/mm2

= N/mm2

0.3 Su-weld x 0.707

CS - Galv.

- 3" NB

& above Sy-pipe @ 250C Sy-member 0.3 Su-weld x 0.707

= N/mm2

= N/mm2

= N/mm2

= N/mm2

LTCS LTCS

LTCS A333 Gr. 6

(-50C - up to 24" NB, BS EN 10113

to Grade S355 NL/ML BS 5950 Sy-pipe @ 250C

250C) CLASS 42

LTCS A671 CC65 [ Support codes : or

- above 24" NB S1(LTCS), S3(LTCS),

& LS1(LTCS) ] Sy-member

or

ALLOY Sy-pipe @ 250C Sy-member Su-weld

825 = N/mm = N/mm = N/mm 0.3 Su-weld

x 0.707

Low Temp.

- 8" NB

& above, Sy-pipe @ 250C Sy-member 0.3 Su-weld x 0.707

(cladded) = N/mm2

= N/mm2

= N/mm2

= N/mm2

Pipe class

catagory

Structural material

(for Shoe / Line stop directlywelded on

bare pipe / pad)

Allowable weld stress

0.4

198 275 500

0.4 0.4

79.2 110 106.1 79.2

0.4

0.4

198 355 500

0.4 0.4

79.2 142 106.1 79.2

continued


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Pag


SS A312 Gr.SS TP316/316L LTCS

(Dual certified) (on SS pad)

(-20C - up to 8" NB,

to BS EN 10113 0.4 Sy-pipe @ 250C

250C) SS A358 Gr. Grade S355 NL/ML

TP316/316L or

(Dual certified) [ Support codes :

[Pipe class : - 10" NB & above S1(SS), S3(SS), LS1(SS) ] Sy-member

013411-X

033411-X or

063411-X Sy-pipe @ 250C Sy-member Su-weld

063421-X = N/mm2

= N/mm2

= N/mm2

0.3 Su-weld x 0.707

093411-X

153411-X]

Sy-pipe @ 250C Sy-member 0.3 Su-weld x 0.707

= N/mm2

= N/mm2

= N/mm2

= N/mm2

SS A312 Gr.

TP316/316L

(Dual certified) SS

SS - up to 8" NB,

SS 316 0.4 Sy-pipe @ 250C

(-73C SS A358 Gr.

to TP316/316L or

250C) (Dual certified) [ Support codes : S5]

- 10" NB & above Sy-member

[Pipe class :

013461-X or

093461-X Sy-pipe @ 250C Sy-member@250C Su-weld

153461-X] = N/mm2

= N/mm2

= N/mm2

0.3 Su-weld x 0.707


= N/mm2

= N/mm2

= N/mm2

= N/mm2

Pipe class

catagory

Structural material


bare pipe / pad)



?

0.4

139 355

0.4 0.4

55.6 142 0 55.6

?

0.4

139 139

0.4 0.4

55.6 55.6 0 55.6

continued


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JI-193

Page


SDSSSDSS A790

(-20C UNS S32750 LTCS

to - up to 4" NB, (on SDSS pad)

250C)

BS EN 10113 0.4 Sy-pipe @ 250C

& SDSS A928 S32750 Grade S355 NL/ML

- 6" NB & above or

GRE [ Support codes :

- up to S1(SS), S3(SS), LS1(SS) ] Sy-member

2" NB,

or

& Sy-pipe @ 250C Sy-member Su-weld

= N/mm2

= N/mm2

= N/mm2

0.3 Su-weld x 0.707CS - FBE

LINED

- up to Sy-pipe @ 250C Sy-member 0.3 Su-weld x 0.707

2" NB, = N/mm2

= N/mm2

= N/mm2

= N/mm2

LTCS

(on Alloy 825 pad)

ALLOY 825 BS EN 10113

B423 UNS N08825 Grade S355 NL/ML 0.4 Sy-pipe @ 250C

ALLOY - up to 6" NB,

825 [ Support codes : orLow Temp. S1(AS), S3(AS),

LS1(AS) ] Sy-member

(-50C

to or

250C) Sy-pipe @ 250C Sy-member Su-weld

= N/mm2

= N/mm2

= N/mm2

0.3 Su-weld x 0.707


= N/mm2

= N/mm2

= N/mm2

= N/mm2

ALLOY

400 /

Cu-Ni

(Nickel -

Copper

Alloy)

Pipe class

catagory

Structural material


bare pipe / pad)


0 142


?

0.4

404 355

0.4 0.4

0.4 0.4

161.6 142

77.2 142 0 77.2

No welded supports.

?

0.4

193 355


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P


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JI-193

Design of shoe Date : 14 Oct. 2009

Type S1 of support standard for pipe size 11/2" TO 4"

Fz

Hy Hx

Fy

FzHz

Fx

////////////////////////////////////////////////////

e

Max. stress

at this point

Weld analysis

done for this

weld at pipe

to shoe

interface.

x

'/////////////////////////////////////////////////////////////////////////////

b

'//////////////////////////////////////////////////

///////////////////////////

d

pipe

axis

sectional plan view of shoe at pipe andattachment interface

x

h

hy

Fy

y y

y

x

z (up)


8/313


JI-193



INPUT :

STEEL MEMBER


Sy-member = N / mm2

Hot Allowable Stress

Sh = N / mm2

WELDING ELECTRODE

Welding Electrode material = CLASS 42 BS 5950

Ultimate tensile strength of the Welding Electrode material,

Su-weld = N / mm2

(As per BS 5950-1 2000, Section 3.2.3 , Table 10 )

Line Pressure

P = bar (g)

P = N / mm2

Temperature, C -->

CS / LTCS


A333 Gr. 6 (up to 16" NB)

A671 CC60 EFW (18" NB & above)

ALLOWABLE STRESS ON WELD

Sweld-allow = Allowable weld stress for the welding

Sweld-allow = Taken from Sheet "Weld-Pipe to Shoe or Stop"

Sweld-allow = N / mm2

158

180 172 162 153

450 200

from ASME Section II Part D Table Y-1.

207

144 189

196 188 177 167

182189

275.0

(As per BS 5950-1 Clause 6.8.5 Table 37, for S275 material, matching electrode is Class 35. However as per

contract specification ADCO DOC. No. 30.99.75.0704 Rev. E, for structural welding all electrodes shall be E70XX to

AWS D1.1. Therefore Class 42 electrode as per BS 5950 is specified in the Project WPS, which is equivalent to

E70XX to AWS D1.1.)

250

500.00

260 300 350 400

19.60

200


207

79.20

198

1.96

138.0


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INPUT FOR STRUCTURAL PROPERTIES

SHOE:

Shoe made up of: 1/2 UB 254 x 102 x 25

Plate thk. d = mm b = mm Hy = mm

Pipe

Pipe NB = " Sch = CA = mm

D = Pipe Outside Dia = mm

R = Pipe Radius = mm

tp = Pipe thk = mm

t = Pipe thk with CA = Pipe thk - CA mm

= - mm

= mm

hx

=

1.0

6

=6

mm

2

250

hx

hx 3

d=

250

60.32

2

mm

3.91

125

2

2

2

30.16

b

2.91

40

100

3.91 1.0

mm

mm

mm

mm

=

hy =

hy =

hy


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JI-193



Calculation of Weld Properties:

= mm3

( b x d )

( x )

= mm

=

=

= mm2

=

=

= mm2

Iwy-y

=

2 3

2

2

20833.3

Iwx-x

2

xd[2

125

]

x

hx

4500

)

= 4

=

2250

2

2

[ ]2

2604167

d

6

3

2604166.667

3

(

12

=

=

Zwy-y

Zwx-x

4500

250

Iwy-y

=

Iwy-y

Iwy-y

Iwy-y

Iwx-x

Iwx-x

=

Iwx-x

b

6

3

2

Iwx-xb

b

Zwx-x

Zwx-x

Zwy-y

Zwy-y

Iwx-x

b

6

1500

hy


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JI-193



= Length of Weld

= x

= mm

Jweld = Polar moment of inertia of weld joint = Iw z = Iw xx + Iw yy

= + = mm3

S = specified max weld size = mm

w = Fillet Weld LEG size

= Minimum of or or

= Minimum of or or

= mm

MOMENT ARM :

Hx =

= H + e

where,

e = possible additional moment arm due to

possibility of bearing plate / wear pad / steel packing at site

or thermal movement of pipe.

(considered as max. mm - all together.)

Therefore,

Hx = +

= mm

Hy =

=

= mm

Hz =

Hz = mm (considered max as 100 mm, for calculation purposes)

d

6.00

Moment arm, for calculating moments acting at pipe to shoe interface due to

loading along x axis (Fx)

2608667

6

S

6.00

Moment arm, for calculating moments acting at at pipe to shoe interface due to

loading along y axis

2.91

2604167 4500

25

Pipe thk

2.91

2

WL

WL

WL

Eccentricity for Torsion (Mz) at pipe to shoe interface due to loading along x

axis (Fx)

125

100 25

500

b

H

100

100


12/313


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P



CALCULATION BASED ON WELD STRENGTH:

MAXIMUM ALLOWABLE FORCE /mm BASED ON WELD

fw allowable = Sweld-allow x w

= x

= N/mm

fwresultant = Sz2 + Sx2 + Sy2 .(i)where;

fwresultant = Resultant weld loading

Sz = Weld loading along Z

Sx = Weld loading along X

Sy= Weld loading along Y

Sz = (Bending Stress) Lateral + (Bending Stress) Axial + (Tensile Stress) Vertical

Mx My

Zx Zy

Fy . Fx .

.

.

.

.

.Sy = Hz+()

79.20 2.91

)

)

230.5

+

)()(

( ]

Fx

(

+

+

Zwx-x

).

+

+)

+

Sx =

To calculate / consider maximum MzConsidering maximum possible eccentricity for

Fy (lateral load) = b/2, and

Considering for maximum possible eccentricity

Fx (axial load) = 1/2 of stopper width = Hz

Fz(

Fx .

+WL

[

Mz

2

Sy =

FyWL

Fy )

Sz =

(

Sx =

b

WL

hx

b

2

hy

)

.

Hx

(

( Fy

)

(Mz

( )

WL

+

Sz =

WL

Hy

)

hx

[ ] hy

Hz2

Hz

Fy)+

Jweld

Jweld

Jweld

Jweld

Zwy-y

Fx

( Fx ) (

+(Fx

WL

Fz

Mz = (Fy b )


13/313


JI-193

Page 13 of 313



Substituting in equation (i)

fwresultant = Fy . Fx .

.

.

Specifying FY & FX in terms of FZ

Let;FX = FX/Z * FZ

FY = FY/Z * FZ

. . . .

fwresultant = . . .

. . .

Solving for Fz:

fwresultant

. .

.

.

Let,

FX/Z =

FY/Z =

fwresultant = fw allowable

x x

x

x

Fx .

(+

2608666.667

+

100

hx)

}Fy ) [ Fy b )

)Fy

+Hx

Jweld

2

2

.

)

2

2

}

2. H

z

)]hx

2]+ Fz . hy. Hz

2Hz )+

}

}2

) 100

( 0.25 )

}+

0.5

++

+

0.5

( 0.5[

0.5

bFx .

}

+ .

Hx

+

(

(

(WL

Hz

Hz)

WL

Fx (

+

+WL 2

)

+

Fz

b

}2

3)

Jweld.

500

hy

1 )

}2

250

250

[x

2

2

2608666.667

0.25

( )( 0.5

FY/Z

FZ

1500

125

0.25

[

}

+ ) FX/Z

) FX/Zb

+

Zwy-y

0.25

20833.3

)FY/Z

FY/Z

+

WL

100

230.472

2)+

WL

Hy

500

500

b

FY/Z

Jweld

(

)(

Fz

Zwx-x+

Hy

1

FX/Z

))

(

)

FX/ZFz

Fz

FY/Z

b

2

WL+

)

( Fz )F

X/Z

}

+

2)+ [

FX/Z

FZ

( [( [

+

+ (FZ FY/Z

FX/Z

( ))+

[WL

+

+ +

+

}

FY/Z

2

FZ

2

WLHz )

2

FY/Z FX/Z )WL

(

Jweld

Hx )Zwy-yhx

Zwx-x)Hy

)0.25) ( x

2+

125)+

Fz =

=

Jweld

Zwx-xZwy-y

)Jweld

hy

{+

Fz-max weld

[ Filename : 154368447.xls.ms_off ice , Sheet : S-1 (2" to 4") calculation sht ]

. [Print Date : 6/27/2013]


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Fz-max weld = N

= KN

230.472

2

0.044866667

230.472

0.04509

5.11087Fz-max weld

{ {+}

230.472

2

0.001093439 0.004393272

1.93008E-05

{} }

0.002013018 1.19561E-06

Fz-max weld =

=Fz-max weld

=Fz-max weld

Fz-max weld =

Fz-max weld =

+

5110.87

230.472

+

2

++

+

[ +

}

}

2

0.0012 0.041666667 0.002+

2

+ +

0.0005+

}2

0.001 1.15001E-0631.25 50 ]

230.472

[( +2

4.79172E-0531.25 50+ ] }

+ 0.001 9.34385E-05 }+2

0.0005 + 0.003893272

}2

0.044866667


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CALCULATION BASED ON LOCAL STRESSES IN PIPE:

+ + Sh

L = Longitudinal

+ + Sh A = Axial

C = Circumferential

LP = Long. Press. Stress

CP = Circum. Press. Stress

. .

x . .

. .

.

.

.

Maximum local loading due to longitudinal moment (in N/mm length),

.

. .

. . .

SCP =

FL =

1.5

2.34 FC

IwY-Y

Sc

t

MY

FX Hx

RP

IwY-Y

1.5

This calculation is done with reference of eBMS trunnion load check calculation (Doc. No. SHJ-BMS-EN-VSS-L-3204-A).

FX/Z

IwY-Y

Hx

t1.5

hy

FZ

R0.5

R0.5

SCP

=

t1.5

1.17 FA

SA

SA

1.17 FL

LOCAL STRESSES (N/mm2)

t1.5

P R

=

=SLP

PRESSURE STRESSES (N/mm2)

hy

t

=

2

R0.5

hy

=

DUE TO EXTERNAL LOADING

LOCAL EXTERNAL LOADING (N/mm2)

The formulae used in this for trunion calculation for local stress are as per "Design of Piping Systems" by M.W.

Kellog Company based on the bending of a beam on an elastic foundation.

SL SLP

SL

SA

1.5SC

=

(Reference Doc. No. SHJ-

BMS-EN-VSS-L-3204-A))










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Maximum local loading due to circumferential moment (in N/mm length),

.

. .

. . . hx

Maximum local loading due to axial moment (in N/mm length),

Sh + +

. x . .

.

. . . . x .

. .

.

.

Hy

MX hx

IwX-X

IwX-X

SLP

IwX-X

FZ=FA

FZ

WL

FY Hy hx

SA

R

Hx

R

t

IwY-Y

t

1.17

hy +1.5

2

FA

P

2+

FX/ZFZ

SL

FL

t1.5

R0.5

t1.5

1.5

=

Sh1.17 R

0.5

t1.5

P

R0.5

+ +

t1.5

R0.5

=

=

FC

1.17

1.17

1.5

1.5

1.5 Sh

FZ FY/Z

WL


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Solving for Fz,

.

.

. . . x

. . .

x

x

x x x x x x

x x

x x x x x x

x x

+

= N

= KN

Sh + +

. x . .

. . . . hx x .

. .

.

.

. . . . x x

. x

x

0.5

2 2.91

1.5 1.17 R0.5

125 30.2

+

125

2

IwY-Y

=FZ-max (SL)

FZ-max (SL)

hy

-1381.96

FZ-max (SL)

1.5

1.5 ShxP R

-

FX/Z Hx

FZ-max (SL)

1.17

2.911.5

2.911.5

0.5

500

0.51.17

1.5 SC SA SCP

R0.5

1.5+

R

t1.5

FZ FY/Z Hy 1.5 1.17 R0.5

FZ

t

R-

+

t

1.5 Sh.

138

t1.5

P

WL

P R

P

R0.5

t1.5

WL

1 96

R0.5

FY/Z hx

IwX-X

+

1 5

1.17

t1.5

Hy 1.5

t1.5

IwX-X

t1.5

25346

t1.5

+

+

1.17 FA R0.5

2.34

FC

t

R0.5

25.35

1.17 R0.5

0.00777

FZ-max (SL)

=

=

FZ-max (SL) =

=

1.5 Sh

1.5 Sh

2.34

2.34

2604167+

FZ-max (SC)

30 16

x

30.16

1.530.16

t

t1.5

WL

0.5 125 125+

-

FZ-max (SL) =1.17

500

207 10.16

5.491812

4.96

5.49181

4.96

1.5 1.17

2604167

196.843

196.843

0.00388 0.00388


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x x x x x xx x

+

= N

= KN

= Minimum of

= KN , = KN , = KN

= KN

Setting a standard for FY & FX

As,

FX = FX/Z * Fz min

FY = FY/Z * FZ min

FX/Z =

FY/Z =

FX = x

Axial Load, FX = KN

Fy = x

Lateral Load, Fy = KN

5.4918124.96409 4.964094500 500

0.25 3 + 1.5 1.17

-

FZ-max (SC) =

20.31395

5.491812

FZ-max (SC) =

2.34

207

186.686

0.04315 0.00388

100

Fzmin

3.970.25

Fz-max weld

1.9847901

0.992395

3.9725.35

3970

3.97

0.04703

FZ-max (SC)

FZ-max (SC)

=FZ-max (SC)

Fz-max 3.97

0.5 3.97

5.111 FZ-max (SC)FZ-max (SL)

0.5

0.25

186.686


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MISCELLANEOUS CALCULATIONS

To Calculate,Fz-(x,y=0) = Fz maximum allowed, if Fx & Fy =0

Fz-(x,y=0) max based on weld strength calculations:

Fy . Fx .

.

.

Substituting Fx = 0 & Fy = 0

Solving for Fz, and substituting fw resultant with fw allowable

Fz-(x,y=0) max weld .

Fz-(x,y=0) max weld x

Fz-(x,y=0) max weld KN

Fz-(x,y=0) max based on local stress calculation due to longitudinal loading:

Sh + +

. x . . .

. t

FA R0.5

t1.5

t1.5

2

SLP

R0.5

+

+P R

]

1.5 Sh 1.17 FL 1.17

( Hy +Fy

Zwy-y Zwx-x WL+

Hx(Hz+ ( Fx )+ (Fx( )]b ) .

Fy b ) (

+hx }

2

WL 2

[( )Fx

[

}2

WL 2

2

= 230.472 500

+{

= 115.236

1.5 SL SA

fwresultant =

fwresultant =

fwallowable

x

( FzWL

+

+1.5

Jweld

Jweld

hy

=

( Fz

WL

}2

Hz(Fy )


20/313


JI-193



But,

. .

Therefore,= (since, Fx=0)

Therefore equation reduces to,

x . .

.

Substituting

x . .

. .

Solving for Fz,

.

.

. .

.

x

x

x x

x

-

x x

x4.96

207 10.16

-

30.16

WL

1.5

R.

t

P R

2

500

1.96

2138.0

=

0.5

2.911.5

500

2.91

t

R0.5

+

1.17

t1.5

WL

1.5 ShR

1.17 FA

FZ

+

P1.5

hy

2

P

1.5 1.17 R0.5

FL =FX Hx

R

2

x

t1.5

Sh -1.5

WL

t1.5

FL 0

FzFa =

1.5

1.5

=Fz-(x,y=0) max SL

Fz-(x,y=0) max SL

Fz-(x,y=0) max SL =

1.5 1.17 5.492

Sh

1.5

t

1.17

30.16

IwY-Y


21/313


JI-193



= N

Fz-(x,y=0) max SL = KN

Fz-(x,y=0) max based on local stress calculation due to circumferential loading:

Sh + +

. x . .

But, . .

Therefore, Fc =0 (since Fy=0)

Therefore, equation reduces to,

x . .

Substituting

x . .

.

Solving for Fz

.

x

.

=

Fz

WLFA =

1.17Sh

0.00388314

196.843024

Fy Hy hx

IwX-X

Fz-(x,y=0) max SC

Fc

1.5 1.17 R0.5

t

FZ R

-

+P

t1.5

WL

R

R

t1.5

t

t1.5

WL

R0.5

t

1.17

Sh

R0.5

P1.5

FA R0.5

+P

1.5 Sh 1.5

1.5

+ P R

t1.5

t1.5

t

R0.5

Fz-(x,y=0) max SL

Fz-(x,y=0) max SL

SC SA

1.17+ 1.52.34 FC

50.6917

FA

SCP

=

1.5

=

50691.7

1.5

1.5 Sh


22/313


JI-193



x

x x

x

-

x x

x

= N

= KN

Thus,



= KN

Therefore,

= KN

50.6917

= 115.236

Fz-(x,y=0) max 48.076

=Fz-(x,y=0) max SC

Fz-(x,y=0) max SC

30.16

30.16

0.00388314

186.686

1.5500

500

Fz-(x,y=0) max SC

48076

48.076

1.5 1.170.5

2.91=

1.5 x1.96

2.91

Fz-(x,y=0) max SC

Fz-(x,y=0) max SC

Fz-(x,y=0) max SC

138.0

48.076

=

20.31

5.492

4.964

1.5 1.17

-

207.0


23/313


JI-193

Page 23 o



To Calculate,

Fx-(y,z=0) = Fx maximum allowed, if Fy & Fz =0

Fx-(y,z=0) max based on weld strength calculations:

Fy . Fx .

.

.

Substituting Fy = 0 & Fz = 0

Fx .

Solving for Fx, and substituting fwresultant with fwallowable

Fx (y,z=0)-max weld =

hy

+

hx

2608666.667

Jweld

Jweld

2608666.667

Jweld

Jweld

hy

Jweld

Jweld


hy

Fx )+(

3( )2

125

500

100+{

2

1500100

125+

+

{ 1

Fy +

230.472

{ }2

)

+

2

WL 2

bFx

{

}2

WLZwx-x

2

Hz)+ (

2

WL 2(

Fy b[Fy )+

fwresultant =

+{(Fx

[(

+

]

) hx

hx

]Hz

)

2

{( )+(Hx

WL

)

(Fx x

2

Fx

(

Fz

(Hz

(. Hz

Zwx-x)

2

Zwx-x WL

+{ )

Hx

{

)}2

+{(2

fwallowable

x Hz

1 )+

(Hz

Fx

{(

Zwy-y

+

)

(

Hy

fwresultant = +{

.

) )

{(Hx )


24/313


JI-193



Fx (y,z=0)-max weld = N

Fx (y,z=0)-max weld = KN

Calculating Fx-(y,z=0)max based on stress calculation due to longitudinal loading

Sh + +

. x . .

.

But,

Therefore,

= (since Fz = 0)


2756.52164

Fz

WLFA =

FA 0

0.006991





+ 1.32253E-084.61275E-05

2

0.08333 0.002

0.000115001+}

2

0.00479172

230.472

0.00694

+

+

230.472

R

t1.5

t1.5

2 t

1.17 FL R0.5

1.5 SL SA SLP

1.5 Sh

2.757

++1.5 1.17 FA R

0.5P

230.472

0.083609719

230.472

2

+

{


25/313


JI-193



. .

.

Substituting,

. .

. . . .

. .

Solving for Fx,

.

.

. .

.

x

x

x x x

x

-

x x x

x

Fx-(y,z=0)max (SL) = N

-138.00

Fx-(y,z=0)max (SL)

125

2604166.67

207 10.157

5.49181209

4.96

125

1.5 Sh 1.17 FL R

0.5

+t1.5

P R

2 t

FL =FX Hx hy

IwY-Y

Fx-(y,z=0)max (SL)

Fx-(y,z=0)max (SL)

Fx-(y,z=0)max (SL)

Fx-(y,z=0)max (SL) =

1.17

=

30.16

25345.84

Hx

KN

125

IwY-Y

2604166.67

1.96

2 t

1.17 R0.5

t1.5

1.5

196.843

0.007766

2.91

hy

t1.5

IwY-Y

hy

0.5

1.17 125

1.5

2

x

1.5 Sh 1.17 Fx

30.16

=

2.91

=1.5

Hx

=

R0.5

P R

2

25.35

t

Sh -

P R

+


26/313


JI-193



Calculating Fx-(y,z=0)max based on stress calculation due to circumferential loading

Sh + +

. x . .

But,

. . . hx

and

Therefore, Fc =0 & FA =0 (since Fy=0 & Fz= 0)


.

Therefore this equation is independent of Fx, and Fx does not have any effect on this stress check.

Fx-(y,z=0)max (SC) =

Thus,




Therefore,

Fx-(y,z=0)max = 2.76

2.76

25.35

Not applicable

Not applicable

KN

KN

R

t

KN

FA = FZ

WL

1.5 Sh P

=FZ FY/Z Hy

IwX-X

SA SCP

1.17 FAR0.5

FC

FC

P R

t1.5

t1.5

t

+1.5 R

0.5

+1.5 Sh 2.34

1.5 SC


27/313


JI-193

Page 27 of



To Calculate,

Fy-(x,z=0) = Fy maximum allowed, if Fx & Fz =0

Fy-(x,z=0) max based on weld strength calculations:

Calculating Fy-(x,z=0)max based on weld strength calculation

Fy . Fx .

.

.

Substituting Fx=0 & Fz=0

. .

.

Solving for Fy , and substituting fwresultant with fwallowable

0.000144 +

( 6.38E-05 )++( 2.07E-08 )

0.00599

Fy(x,z=0)-max weld =2

( 0.00002304 )

0.0048 0.002+()2

)22

+(

2608666.667

2608666.667

Jweld

Jweld

Jweld

Jweld

Jweld

Jweld

hyHz )]

2

hy+

HyFy

Fy

fwresultant =

Fy(x,z=0)-max weld =



(

hy

b

b

WL

)}

230.472

100

230.472

{(

({

+{250

+{(1 )+ 125 }

2

500 2( )250

}3 2

20833.3333 2))}

2

(2

+{Hy

Zwy-y

( ) }

2

2+

{(1

)+

2

2) hx

fwallowable

WL 2

2

}b )](Fy

}+[

2

b

+{(Fy )

] hx })2

( )

}2

WL 2

b

Fy

2

WL 2

Fx .(

+{[({( )Zwy-y

Fz() )( )}2

Zwy-y Zwx-x WL

Hy + +

fwresultant = )]

Hx

hx }+ + ){(Fx b +

+ )

. Hz) [(Fy Fx

+ )

{

{(Fy

[ Filename : 154368447 xls ms office Sheet : S-1 (2" to 4") calculation sht ]


28/313


JI-193



Fy(x,z=0)-max weld = KN


Sh + +

. x . .

.

But,

. .

and

Therefore, FL =0 & FA =0 (since Fx=0 & Fz= 0)


.

Therefore this equation is independent of Fy, and Fy does not have any effect on this stress check.

Fy-(x,z=0)max (SC) =

8.6895E-05

R

FL

Not applicable

24.72

Fy(x,z=0)-max weld = 24724.09209 N

FL

FA

1.5 Sh

=FX Hx hy

t

IwY-Y

=FZ

WL

P

0.009321758


R0.5

230.472

1.17 1.5 1.17 FA

1.5 SL SA SLP


R

t+1.5 Sh +

P

t1.5

t1.5

2

R0.5


29/313


JI-193



Calculating Fy(x,z=0)max based on local stress calculation due to circumferential loading:

Sh + +

. x . .

But,

Therefore FA=0 (since Fz=0),


. .

Substituting,

. . . hx

. . . .

.

Solving for Fy,

.

. .

.

x

x x x

x

x x x

x

Fy (x,z=0)-max (SC) = N1081.71

Fy (x,z=0)-max (SC) = 2.34 100

4500

207 20.31

5.491812

Fy (x,z=0)-max (SC)

2.34 R

Fy (x,z=0)-max (SC)

Fy (x,z=0)-max (SC)

Fy (x,z=0)-max (SC)

Fz

WL

=

1.5 Sh

FA

=Fc

FC R0.5

t1.5

+P

t

t1.5

FZ FY/Z Hy

-

= 1.08171 KN

IwX-X

1000.5

30.16

1.52.91

3

4500

=0.17258

4.96

-

3

186.69

x

1.5

2.34 Hy

t1.5

IwX-X

hx

2.91

x -1.96 30.2

1.5 SC

=

2.34

=

1.5 Sht1.5

SCP

2.34 R

SA

t

1.5 Sh 2.34 R

0.5

+

R

R0.5

P

Fy Hy hx

t1.5

IwX-X

R0.5

+FC

138.0

R0.5

1.5P R

t

1.5 1.17

Sh

+P

t

FA


30/313


31/313


JI-193


Type S1 of support standard for pipe size 6" TO 24"


Fy

x

hy

x

hx

y y

'//////////

b-100

'

'/////////

'////////////

'///////////

d

i

e

axis

'////////////////////////////

' '///

/////

ds '///////////// '////////////

'//////////// '////////////

'///////////////

'///////////////

y

x

z (up)

l

Hy

'///////////'/////////////////////////////////////////////////////////////////////

Hx

Fy

FzHz

Fx

e

Max. stress at

this point

Weld analysis

done for this

weld at pipe to

shoe interface.

Fz


32/313


JI-193



INPUT :

STEEL MEMBER


Sy-member = N / mm2


Sh = N / mm2

WELDING ELECTRODE



Su-weld = N / mm2


Line Pressure

P = bar (g)

P = N / mm2

Temperature, C -->

CS / LTCS


A333 Gr. 6 (up to 16" NB)

A671 CC60 EFW (18" NB & above)




Sweld-allow = N / mm279.20

158 207

189 182 180 172 162 153 144 189

207 198 196 188 177 167

from ASME Section II Part D Table Y-1.

200 250 260 300 350 400 450 200

138.0

(As per BS 5950-1 Clause 6.8.5 Table 37, for S275 material, matching electrode is Class 35. However as per

contract specification ADCO DOC. No. 30.99.75.0704 Rev. E, for s tructural welding all electrodes shall be E70XX to

AWS D1.1. Therefore Class 42 electrode as per BS 5950 is specified in the Project WPS, which is equivalent to

E70XX to AWS D1.1.)

500.00

19.60

1.96


275.0


33/313


JI-193




SHOE:


Plate thk. d = mm b = mm H = mm l = mm

ds = mm

Pipe




tp = Pipe thk = mm

wp = Wear Pad = mm

t = Pipe thk with CA = Pipe thk - CA mm

= - mm

= mm

80

0.00

hx =80

mm2

hx = 40 mm

hy = 150 mm

hx =l

mm2

hy = b mm2

hy =300

mm2

168.3

84.14

7.11

7.11 1.0

6.11

6 300 100

6 40 1.0

10


34/313


JI-193




b ( - ) . ( - )2

( - ) . ( - )2

( - ) . ( - )2

( - ) . ( - )2

( - ) x ( - )2

( - ) x ( - )2

( ) x ( )2

( ) x ( )2

( ) x ( ) ( ) x ( )

= mm3

( . ) x

( . ) x

( b . d ) ( . ) x

( x ) ( - ) x

( x ) ( - ) x

( ) x

= mm

=

=

= mm2

=

=

= mm2Zwx-x 4399.87

5117845400

4

12+

Iwx-x + 170594.667

4

2 12

36 512000 216+

=

Iwx-x

Iwx-x

Iwx-x

Iwx-x =

l 3 d 3 4

+ 803

34

63

4

12

12

d

+l

3

12

l3

d

+

d3

4

12

2

Iwy-y = +

4500000

4500000 + 1480000

=

74 200

+ 74 48400

2

Iwy-y

Iwy-y

Iwy-y

Iwy-y

Iwy-y

+

+ 80

+ 74 40000

2

100

2 2

6 300 80

2

220

100

12 2 2

+ l d

d b 80

+

+ 80

l d

6 300 100

2

bb 80

b

+ l d

2

74

1790800

+

l

Zwx-x Iwx-x

hx

Zwx-x 17599540

hy

Zwy-y 7770800

150

Zwy-y 51805.3

Iwx-x 175994.6667

Zwy-y Iwy-y

300 6

Iwx-x =

300

5400

2

2

=

=

2

2

2

] += 2 [ b .4

(d

)

2

]2

= 2 [ b .Iwx-x

= 3003

6

Iwy-y 7770800

6

2.7E+07= +

= b3

6

=

23


35/313


JI-193



= Length of Weld

= x ( - x ) + x ( - )

= x ( - x ) + x ( - )

= x ( - ) + x ( )

= x ( ) +

= +

= mm

Jweld = Polar moment of inertia of weld joint = Iw z = Iw xx + Iw yy

= + = mm3



= Minimum of or or

= Minimum of or or

= mm

MOMENT ARM :

Hx =

= H + e

where,





Therefore,

Hx = +

= mm

Hy =

=

= mm

Hz =

Hz = mm (considered max as 100 mm, for calculation purposes)

280 296

WL 560 296

WL 2

80

2 ds

6

WL 2 300 20 4 74

WL 2 300 2 10 4

4 l d

Moment arm, for calculating moments acting at at pipe to shoe interface due to

loading along y axis

H

Eccentricity for Torsion (Mz) at pipe to shoe interface due to loading along x

axis (Fx)

100

110.117

6.00

Moment arm, for calculating moments acting at pipe to shoe interface due to

loading along x axis (Fx)

25

110.1 25

135.117

Pipe thk d S

6.11 6.00 6.00

WL 856

7770800 175995 7946795

6

WL

WL 2 b


36/313


JI-193

Pag



CALCULATION BASED ON WELD STRENGTH:



= x

= N/mm


fwresultant = Resultant weld loading



Sy = Weld loading along Y


Mx My

Zx Zy

Fy . Fx .

.

.

.

.

.

WL 2 JweldFx . Hz )] hy(Fy b )+ (

hy )WL Jweld

Sy = ( Fy )+ [

WL 2 Jweld

Sy = ( Fy )+ (Mz

Fx . Hz )] hx(Fy b )+ (

hx )WL Jweld

Sx = ( Fx )+ [

Hz )2

Sx = ( Fx )+ (Mz

To calculate / consider maximum MzConsidering maximum possible eccentricity for

Fy (lateral load) = b/2, and

Considering for maximum possible eccentricity

Fx (axial load) = 1/2 of stopper width = Hz

Mz = (Fy b )+ (Fx .

Hx )+ ( Fz )Zwx-x WL

Sz = ( Hy )+ (Zwy-y

475.2

Sz = + +Fz

WL

79.20 6.00


37/313


JI-193

Page 37 of 313





.

.


Let;FX = FX/Z * FZ

FY = FY/Z * FZ

. . . .

fwresultant = . . .

. . .

Solving for Fz:

fwresultant

. .

.

.

Let,

FX/Z =

FY/Z =


x x

x

x

150

}

2

856 2 7946794.6670.5 x 100

)]0.25 300

)++

{

0.25

)+

40 }2

856 2 7946794.6670.5 x 100 )]0.25 300 )++{ 0.5 )+

)2

51805.3 4400 856+ ]+

0.5

0.25

1

Fz-max weld =475.2

{ 0.25 110 ) 0.5 135.117

hy }(FY/Z b 2

WL 2 JweldFX/Z . Hz )])++{

FY/Z )+

hx }2

WL 2 JweldFX/Z . Hz )]( FY/Z b )++{

FX/Z )+

1 )}2

Zwy-y Zwx-x WL

FX/Z Hx )+ ({FY/Z Hy )+

2

WL 2 Jweld

Fz =

. Hz ) hy)+ Fz . FX/Z+ (FZ FY/Z b

2

WL 2 Jweld

+{FZ FY/Z )

. Hz

)hx

)+ Fz . FX/Z+

(FZ FY/Z b

+

{FZ FX/Z

)

+ ( Fz )}2

WL+

Fz FX/Z Hx )Zwx-x{

Fz FY/Z Hy )Zwy-y

hy

}2

WL 2 Jweld(Fx . Hz )[ Fy b )++{

Fy )+

hx

}2

WL 2 Jweld+ Fx . Hz )+ (Fy b )

2

Zwy-y Zwx-x WL

+{Fx )

)+ Fz )}{(Hy )+ Hx

[ Filename : 154368447.xls.ms_office , Sheet : S-1 (2) ]

. [Print Date : 6/27/2013]


38/313


JI-193

P



Fz-max weld = N

= KN

CALCULATION BASED ON LOCAL STRESSES IN PIPE:

L = Longitudinal

+ + Sh A = Axial

C = Circumferential

+ + Sh LP = Long. Press. Stress


LOCAL STRESSES DUE TO LOCAL EXTERNAL LOAD (N/mm )

. . (Reference Doc. No. SHJ-FL

SC SA SCP 1.5

=1.17 R

0.5

This calculation is done with reference of eBMS trunnion load check calculation (Doc. No. SHJ-BMS-EN-VSS-

L-3204-A).

The formulae used in this for trunion calculation for local stress are as per "Design of Piping Systems" by M.W.

Kellog Company based on the bending of a beam on an elastic foundation.

DUE TO EXTERNAL LOADING

SL SA SLP 1.5

Fz-max weld =475.2

0.0172

27635.6

Fz-max weld 27.6356

1.04968E-06 + 3.77784E-06

}

2

Fz-max weld =475.2

0.000290848 +

0.001024541 }2

+ { 0.001943665

2

Fz-max weld =475.2

{ 0.017054265 }2

+ {

2

+

{0.000292056 + 0.001651609

}

0.000584112 + 0.000440429 }{ 0.017054265 }2

+{

] 1.88755E-05 }2

Fz-max weld =475.2

37.5 + 50

5.03348E-06 }50 ]2

+{(0.00029 +[

+

2

+{ 0.00058 +[ 37.5+ 0.00117

}

Fz-max weld =475.2

{0.0005314 + 0.015354644


39/313


40/313


JI-193



Maximum local loading due to circumferential moment (in N/mm length),

.

. .

. . .

Maximum local loading due to axial (radial) load (in N/mm length),

Sh + +

. . x . . .

.

. . . . . x . .

. .

.

.

FZ

WL

FL FA

1.17 R0.5

t1.5

IwY-Y

P

2+

FY Hy hx

FZ Hy hx

t1.5

FX/Z hy 1.51.5 Sh

1.17 R0.5

FZ Hx +

t1.5

t1.5

+1.5 1.17

1.5 SL

RR0.5

SLP

1.5 Sh1.17 R

0.5

SA

hx

FA =FZ

WL

MX

=IwX-X

=FY/Z

IwX-X

FC =

IwX-X

t

+P R

2 t


41/313


JI-193



Solving for Fz,

.

.

. . . x

. . .

x

x

x x x x x x

x x

x x x x x x

x x

+

= N

= KN

Sh + +

. x . .

. . . . hx x .

. .

.

.

. . . . x x

. x

x

1.5

1.5 x 138 - 1.96 84.14

hx+

1.5 1.17 R0.5

t1.5

IwX-X t1.5

WL

P R

FZ-max (SC) =t

2.34 R0.5

FY/Z Hy

+P R

t

1.5 . Sh -

FZ

t1.5

IwX-X t1.5

WL

FZ FY/Z Hy 1.5 1.17

P R

t1.5

t1.5

t

1.5 1.17

R0.5

Sh 2.34 R

0.5

+FA

+

SCP

R0.5

+1.5 Sh 2.34 FC R

0.5

FZ-max (SL) 89098

FZ-max (SL) 89.1

1.5 SC SA

FZ-max (SL) =193.505

0.00093 0.00125

FZ-max (SL) =193.505

0.00217

+1.5 1.17 9.17251

15.10 7770800 15.10 856

207 - 13.49FZ-max (SL) =

1.17 9.172513 0.5 135 150

+1.5 1.17

0.5

6.111.5

7770800 6.111.5

856

FZ-max (SL) =2 6.11

1.17 84.140.5

0.5 135 150

WL

1.5 x 138 -1.96 84.14

FZ-max (SL)2 t

1.17 R0.5

FX/Z Hx hy+

1.5=

1.5 x Sh - P

t1.5

IwY-Y

84.14

R

1.17 R0.5

t1.5


42/313


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x x x x x xx x

+

= N

= KN

= Minimum of

= KN , = KN , = KN

= KN

Calculating FX & FY :

As,

FX = FX/Z * FZ

FY = FY/Z * FZ

, and

FX/Z =

FY/Z =

FX-max = x

Axial Load, FX-max = KN

FY-max = x

Lateral Load, FY-max = KN

FZ-max (SC) =

207 - 26.9893FZ-max (SC) =

2.34 9.172513 0.25 110 40 +15.103 2E+05 9.17251315.103 8561.5 1.17

4.439399

0.5

0.25

0.5 17.76

8.8787979

0.25 17.76

FZ-max (SL) 89.1 FZ-max (SC) 17.76

FZ-max 17.76

FZ-max (SC) 17758

FZ-max (SC) 17.76

Fz-max

Fz-max weld 27.64

FZ-max (SC) =180.011

0.01014

180.011

0.00889 0.00125


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MISCELLANEOUS CALCULATIONS

To Calculate,Fz-(x,y=0) = Fz maximum allowed, if Fx & Fy =0

Fz-(x,y=0) m ax based on weld strength calculations:

Fy . Fx .

.

.

Substituting Fx = 0 & Fy = 0

Solving for Fz, and substituting fw resultant with fw allowable

Fz-(x,y=0) max weld .

Fz-(x,y=0) max weld x


Fz-(x,y=0) m ax based on local stress calculation due to longitudinal loading:

Sh + +

. x . . .

.

P R

t1.5

t1.5

2 t+

1.5 1.17 FA R0.5

+1.5 Sh 1.17 FL R

0.5

= 406.7712

1.5 SL SA SLP

= fwallowable WL

= 475.2 856

fwresultant = ( Fz )2

WL

]hy

}2

WL 2 Jweld

Fx x Hz[(Fyb

)++ (Fy

+

] hx }2

WL 2 Jweld+(Fx . Hz )+ [(Fy b )fwresultant = + Fx

}2

Zwy-y Zwx-x WL

Hx+

Fz{Hy )+


44/313


JI-193



But,

. .

Therefore,= (since, Fx=0)


x . .

.

Substituting

x . .

. .

Solving for Fz,

.

.

. .

.

x

x

x x

x

-

x x

x 85615.10

6.111.5

856

Fz-(x,y=0) max SL =207 13.49

1.5 1.17 9.173

84.14

Fz-(x,y=0) max SL =2 6.11

1.5 1.17 84.140.5

t1.5

WL

1.5 x 138.0 -1.96

Fz-(x,y=0) max SL =2 t

1.5 1.17 R0.5

1.5 Sh -P R

R0.5

FZ

+

P R

WL 2 t

Fa =Fz

WL

1.5 Sh

1.5 1.17

1.17

t1.5

FA R0.5

+P R

t1.5

2 t

FL 0

1.5 Sh 1.5

FL =FX Hx hy

IwY-Y


45/313


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= N


Fz-(x,y=0) max based on local stress calculation due to circumferential loading:

Sh + +

. x . .

But, . .

Therefore, Fc =0 (since Fy=0)

Therefore, equation reduces to,

x . .

Substituting

x . .

.

Solving for Fz

.

x

.

R

Fz-(x,y=0) max SCt

1.5 1.17 R0.5

=

1.5 Sh -P

t1.5

WL

1.17 R0.5

FZ+

P R

t1.5

WL t

FA =Fz

WL

1.5 Sh 1.5

FA R0.5

1.5 Sh 1.5

+P R

t1.5

t

IwX-X

1.17

P R

t1.5

t1.5

t

Fc =Fy Hy hx

+ 1.5 1.17 FA R

0.5

+1.5 Sh 2.34 FC R

0.5

Fz-(x,y=0) max SL 155404

155.404

1.5 SC SA SCP

Fz-(x,y=0) max SL =193.505352

0.00124517


46/313


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x

x x

x

-

x x

x

= N

= KN

Thus,



= KN

Therefore,

= KNFz-(x,y=0) max 144.567

Fz-(x,y=0) max SC 144.567

= 406.771

155.404

Fz-(x,y=0) max SC 144.567

Fz-(x,y=0) max SC =

180.011

0.00124517

Fz-(x,y=0) max SC 144567

=Fz-(x,y=0) max SC207.0 26.99

1.5 1.17 9.173

15.103 856

84.14

Fz-(x,y=0) max SC 6.11

1.5 1.17 84.140.5

=

1.5 x 138.0 -1.96

6.111.5

856


47/313


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Page 47 of 31



To Calculate,

Fx-(y,z=0) = Fx maximum allowed, if Fy & Fz =0

Fx-(y,z=0) max based on weld strength calculations:

Fy . Fx .

.

.

Substituting Fy = 0 & Fz = 0

Fx .

Solving for Fx, and substituting fwresultant with fwallowable

+{ ( 100 )

100 )856

40 2

7946794.667

135.1168 }2

7946794.667+{( 1 )+{ 135.117 )}

2

4399.87

Fx (y,z=0)-max weld =475.2

+{ (Hz )

Hz )WL

hy2

Jweld

hx }2

Jweld

+{(1 )+{

Hx )}2

Zwx-x

hy

}2

Jweld

Fx (y,z=0)-max weld =fwallowable

Fx x Hz

2

Zwx-x WL JweldFx . Hz

fwresultant = +{

{Hx ) ) hx+{(

Fx )+}2

]hy

2

WL 2 Jweld

Fx x Hz )[(Fy b )++{

Fy )+

] hx }2

WL 2 Jweld+ Fx . Hz )+ [(Fy b )fwresultant = +{

Fx )

Fz )}2

Zwy-y Zwx-x WL+ ( Hx )+({

Hy )


. [Print Date : 6/27/2013]


48/313


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Fx (y,z=0)-max weld = N

Fx (y,z=0)-max weld = KN


Sh + +

. x . .

.

But,

Therefore,

= (since Fz = 0)


FA =Fz

WL

FA 0

P R

t1.5

t1.5

2 t+

1.5 1.17 FA R0.5

+1.5 Sh 1.17 FL R

0.5

15405.02694

15.41

1.5 SL SA SLP


0.000952


0.030847074

8.22825E-06 + 2.53359E-07

2


0.00094 +

0.001700268

}

2

+ ( 0.000503348

+

{0.00117 +0.03071

2



49/313


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. .

.

Substituting,

. .

. . . .

. .

Solving for Fx,

.

.

. .

.

x

x

x x x

x

-

x x x

x

Fx-(y,z=0)max (SL) = N104410.9

Fx-(y,z=0)max (SL) = 104.41 KN

135.1 150

15.10 7770800

Fx-(y,z=0)max (SL) =193.5054

0.001853

150

6.111.5

7770800

Fx-(y,z=0)max (SL) =207 13.4946

1.17 9.17251329

Fx-(y,z=0)max (SL) =2 6.11

1.17 84.1350.5

135.1

1.5 x 138.00 -1.96 84.14

R

Fx-(y,z=0)max (SL)2 t

1.17 R0.5

Hx hy

=

1.5 Sh -P

t1.5

IwY-Y

Hx hy+

P R

t1.5

IwY-Y 2 t1.5 Sh

1.17 R0.5

Fx

FL =FX Hx hy

IwY-Y

+P R

t1.5

2 t1.5 Sh

1.17 FL R0.5


50/313


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Calculating Fx-(y,z=0)max based on stress calculation due to circumferential loading

Sh + +

. x . .

But,

. . . hx

and

Therefore, Fc =0 & FA =0 (since Fy=0 & Fz= 0)


.

Therefore this equation is independent of Fx, and Fx does not have any effect on this stress check.


Thus,




Therefore,

Fx-(y,z=0)max =

Not applicable

15.41 KN

t

Not applicable

15.41 KN

104.41 KN

IwX-X

FA =

FZ

WL

1.5 Sh P R

P R

t1.5

t1.5

t

FC =FZ FY/Z Hy

+1.5 1.17 FA R

0.5

+1.5 Sh 2.34 FC R

0.5

1.5 SC SA SCP


51/313


JI-193

Page 51 of 313



To Calculate,

Fy-(x,z=0) = Fy maximum allowed, if Fx & Fz =0

Fy-(x,z=0) m ax based on weld strength calculations:

Calculating Fy-(x,z=0)max based on weld strength calculation

Fy . Fx .

.

.

Substituting Fx=0 & Fz=0

. .

.

Solving for Fy , and substituting fwresultant with fwallowable

1.6E-05 )( 5.7E-07 ) +( 4.5181E-06 ) +

0.002831 )2


)2

+ 0.001168 +

Fy(x,z=0)-max weld =475.2

( 0.00212559 )2

+ ( 0.000755

150

}2

856 2 7946794.667+ ( 300 )

}2300 ) 4051805.3333 2 7946794.667

+{(1 )


{(110.116771 )}

2

+{

hy

}

2

WL 2 Jweld+

(

b

)

2

Zwy-y 2 Jweld

b )+

{(

1

)

+{

Fy(x,z=0)-max weld =fwallowable

{(Hy )}

2

Fy b)+

hx

{(Fy hy

}WL 2 Jweld[(

] hx

)]2

fwresultant =Zwy-y 2 Jweld

+

}22

+{ (Fy b ){(

Fy Hy )}

]hy

}2

WL 2 Jweld(Fx . Hz )[(

Fy b )++{(Fy )+

] hx }2

WL 2 Jweld+ (Fx . Hz )+ [(Fy b )fwresultant = +{(

Fx )

)}2

Zwy-y Zwx-x WL

Hx )+ Fz{(Hy )+


. [Print Date : 6/27/2013]


52/313


JI-193



Fy(x,z=0)-max weld = KN


Sh + +

. x . .

.

But,

. .

and

Therefore, FL =0 & FA =0 (since Fx=0 & Fz= 0)


.

Therefore this equation is independent of Fy, and Fy does not have any effect on this stress check.


R

t

Not applicable

FA =FZ

WL

1.5 Sh P

FL =FX Hx hy

IwY-Y

P R

t1.5

t1.5

2 t+

1.5 1.17 FA R0.5

+1.5 Sh 1.17 FL R

0.5

103.49

1.5 SL SA SLP


0.004591799

Fy(x,z=0)-max weld = 103488.8551 N

2.1085E-05Fy(x,z=0)-max weld =

0


53/313


JI-193



Calculating Fy(x,z=0)max based on local stress calculation due to circumferential loading:

Sh + +

. x . .

But,

Therefore FA=0 (since Fz=0),


. .

Substituting,

. . . hx

. . . .

.

Solving for Fy,

.

. .

.

x

x x x

x

x x x

x

Fy (x,z=0)-max (SC) = N5061.07

Fy (x,z=0)-max (SC) = 5.06107 KN

15.10 2E+05

Fy (x,z=0)-max (SC) =180.01

0.03557

9.172513 40

6.111.5

2E+05

207 - 26.99

110.1

84.1

Fy (x,z=0)-max (SC) =6.11

2.34 84.140.5

40110.1

Fy (x,z=0)-max (SC)

t1.5

IwX-X

1.5 x 138.0 -1.96

=2.34

Fy (x,z=0)-max (SC) =t

2.34 R0.5

Hy hx

1.5 x Sh -P R

Fy Hy hx+

P R

IwX-X t

IwX-X

1.5 Sh 2.34 R

0.5

t1.5

+P R

t1.5

t

R0.5

Fc =FZ FY/Z Hy

1.5 Sh 2.34 FC

P R

t1.5

t1.5

t

FA =Fz

WL

+1.5 1.17 FA R

0.5

+1.5 Sh 2.34 FC R

0.5

1.5 SC SA SCP


54/313


JI-193



Thus,

Fy (x,z=0)-max weld =

Fy-(x,z=0)max (SL) =


Therefore,

Fy-(x,z=0)max =

Fx-(y,z=0) 15.41 104.41 Not Applicable 15.41

144.57

Fy-(z,x=0) 103.49 Not Applicable 5.06 5.06

F-max (SC) (KN) F-max (KN)

Fz-(x,y=0) 406.77 155.40 144.57

Not applicable

5.06 KN

103.49 KN

F-max weld (KN) F-max (SL) (KN)

103.49 KN


55/313


JI-193



Hy

'///////////'////////////////////////////////////////////////////////////////////

Hx

Fy

FzHz

Fx

e

Max. stress at

this point

Weld analysisdone for thisweld at pipe toshoe interface.

Fz


Fy

x

hy

x

hx

y y

'

b

b-100

'

'/////////

'///////////

'//////////

d

i

e

axis

'//////////////////////////

///

' '

ds'///////////// '///////////

'////////////'///////////

'///////////////

'///////////////

y

x

z (up)

l


56/313


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INPUT :

STEEL MEMBER


Sy-member = N / mm2


Sh = N / mm2

WELDING ELECTRODE



Su-weld = N / mm2


Line Pressure

P = bar (g)

P = N / mm2





275.0

138.0

(As per BS 5950-1 Clause 6.8.5 Table 37, for S275 material, matching electrode is Class 35. However as per contract

specification ADCO DOC. No. 30.99.75.0704 Rev. E, for structural welding all electrodes shall be E70XX to AWS

D1.1. Therefore Class 42 electrode as per BS 5950 is specified in the Project WPS, which is equivalent to E70XX to

AWS D1.1.)

500.00

19.60

1.96

79.20


57/313


JI-193

Pa




SHOE:


Plate thk. d = mm b = mm H = mm l = mm

ds = mm

Pipe




tp = Pipe thk = mm

wp = Wear Pad = mm

t = Pipe thk with CA = Wear Pad + Pipe thk - CA mm

t = Pipe thk with CA = mm

6 300 100 80

10

6 40 1.0

168.3

84.1

7.11

0.00

6.11

hy =b

mm2

2

hy =300

mm2

hy = 150 mm

hx =l

mm2

hx =80

mm

hx = 40 mm


58/313


JI-193




b ( - ) . ( - ) 2 ( - ) . ( - ) 2

( - ) . ( - ) 2 ( - ) . ( - ) 2

( - ) . ( - ) 2 ( - ) . ( - ) 2

= mm3

( . ) x

( . ) x

( b . d ) ( . ) x

( x ) ( - ) x

= mm3

=

=

= mm2

=

=

= mm2

+ +

lIwy-y=

2 3 d b 100

12 2 2

l d b 80

l dIwy-y =b 3

+l

Iwy-y =300 3

+

d

+80

b 100

6 2 2

b 80+

6 300 100

6 2 2

80 6 300 80

Iwy-y 7770800

Iwx-x = 2 [ b . (d

)

2

]2

+ l3

d3

4

12

Iwx-x = 2 [ b .d 2

] + l3

d3

4

4 12

Iwx-x =2

Iwx-x =2

+

2

34

d 3 4

2 12

l 3

6

12

Iwx-x 175994.7

Zwy-y Iwy-y

+ 803

6300

hy

Zwy-y 7770800

150

Zwy-y 51805.3

Zwx-x Iwx-x

hx

Zwx-x 17599540

Zwx-x 4399.87


59/313


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Pag



= Length of Weld

= x ( - x ) + x ( - )

= x ( - x ) + x ( - )

= mm

Jweld = Polar moment of inertia of weld joint = =

= + = mm3



= Minimum of or or

= Minimum of or or

= mm

MOMENT ARM :

Hx =

= Hy + e

where,





Hx = mm

Hy =

Hy = mm

Hz =

WL

WL 2 b 2 ds 4 l d

WL 2 300 2 10 4 80 6

WL 856

Izz Ixx + Iyy

7770800 175995 7946795

6

Pipe thk d S

6.11 6.00 6.00

6.00

Moment arm, for calculating moments acting at base of Member / at weld due

to loading along x axis

25

135.12


to loading along y axis

110.12


to loading along y axis


60/313


JI-193

Page 60 of 31





= x

= N/mm


fwresultant = Resultant welding loaing



Sy = Weld loading along Y


Mx My Fz

Zx Zy 2*b

Fy . Fx .

Sx = .

Sx = .

x Jweld

Sx = .

Sy = . +

x Jweld

79.20 6.00

475.2

Sz = + +

Sz = ( Hy )+(Zwy-y

+ ( Fz )Zwx-x WL

)+(Mz

Hx )

WL WL

( Fx )+ (

( Fx

)+ (Fx

hx )

. Hz ) lWL 2 2

Fy b

( Fx )+(Mz hx )WL WL

( Fx )+ ( Hz ) bWL 2 2

Fy b ) (Fx x

To calculate / consider maximum Mz, considering maximum possible eccentricity for Fy

(lateral load) = b/2, and considering for

Fx (axial load) = 1/2 of stopper width

Fx (axial load) = Hz

Hz = 100 mm

[ Filename : 154368447.xls.ms_office , Sheet : S-1 (6" to 24") calculation sht ]

. [Print Date : 6/27/2013]


61/313


JI-193

Page 61 of 313



.



.

x

. b

x


Let;FX = FX/Z * FZ

FY = FY/Z * FZ

. . . .

fwresultant = . . .

x

. . .

xSolving for Fz:

fwresultant

. .

Fz =

.

x

.

xLet,

Fx/z =

Fy/z =


Fz-max weld =

x x ( ) +

x +

x

x +

x

Fz-max weld =

Fx . HzMz = Fy b

+Fz

2

( Hy )+

+

) 2Zwy-y Zwx-x WL(Hx )

+ ( Fx )+ Hz )[(Fy b )+ l 2WL 2 2 Jweld

(Fx .+

Fy )+ Hz )[(Fy b )+ 2WL 2 2 Jweld

(Fx x

(Fz FY/Z Hy )Zwy-y + (Fz FX/Z Hx )Zwx-x + (

Fz ) 2WL

+ [(FZ FX/Z )+ [(FZ FY/Z b )+ (Fz . FX/Z . Hz )] l 2WL 2 2 Jweld

+ [(FZ FY/Z )+ [(FZ FY/Z ] bb )+ (Fz . 2WL 2 2 Jweld

FX/Z . Hz )

)+ (( FY/Z Hy )+ 1 ) 2Zwy-y Zwx-x WL(FX/Z Hx

+ [( FX/Z )+ Hz )[( FY/Z b )+ l 2WL 2 2 Jweld

( FX/Z .+ [( FY/Z )+ Hz )[( FY/Z b )+ b 2WL 2 2 Jweld( FX/Z .

0.5

0.25

475.2

{ ( 0.25 110.12 ) + [0.5 135.117 ] ( 1 ) }

2

51805.3 4400 856

+ { (0.5 ) 100 ) ]+ [ (

0.25 300 ) 80 }2

856 2 2 7946795( 0.5 x

+ { (0.25 ) + ]

300[ ( 0.25 300 ) ( }2

856 2 2 79467950.5 x 100 )

475.2

{ 0.017054265 }2

+ { 0.001024541 }2

+ { 0.001943665 }2


. [Print Date : 6/27/2013]


62/313


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Pag



Fz-max weld =

Fz-max weld =

Fz-max weld = N

= KN

+ + x Sh L = Longitudinal

A = Axial

+ + x Sh C = Circumferential

LP = Long. Press. Stress


. .

x . .

. .

.

.

.

.

. .

. . .

.

475.2

{ 0.000290848 }+ { 1.04968E-06 }+ { 3.77784E-06 }

475.2

0.0172

27635.6

Fz-max weld 27.6356

This calculation is done with reference of Ebms trunnion load check calculation (Doc. No. SHJ-BMS-

EN-VSS-L-3204-A).

The formulae used for trunion calculation are as per "Design of Piping Systems" by M.W. Kellog

Company based on the bending of a beam on an elastic foundation.

SL SA SLP 1.5

SC SA SCP 1.5

SL =1.17 FL R

0.5

t1.5

SA =1.5 1.17 FA R0.5

t1.5

Sc =2.34 FC R

0.5

t1.5

SLP =P R

2 t

SCP =P R

t

FL =MY hy

IwY-Y

FL =FX Hx hy

IwY-Y

FL =FZ FX/Z Hx hy

IwY-Y

FC =MX hx

IwX-X


63/313


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. .

. . . hx

. Sh + +

. . x . . .

.

. . . . . x . .

. .

.

.

.

.

. . . . . .

. . .

x - x

x

x x x x x x

x x

= N

= KN

FC =FY Hy hx

IwX-X

FC =FZ FY/Z Hy

IwX-X

FA =FZ

WL

1.5 SL SA

R0.5

+

SLP

1.5 . Sh 1.17 F

LR

0.5P R

t1.5

t1.5

2 t+

1.5 1.17 FA

hy+

1.51.5 . Sh

1.17 R0.5

1.17 R0.5

FZ

t1.5

IwY-Y t1.5

WL

FZ FX/Z H

+P R

2 t

=

1.5 x Sh -P R

FZ-max (SL)

2 t

1.17 R0.5 FX/Z Hx hy+

1.5 1.17 R0.5

t1.5

IwY-Y t1.5

WL

1.5 138 1.96 84.1

FZ-max (SL) =2 6.11

1.17 84.1 0.5 0.5 135 150+

1.5 1.17 84.1 0.5

6.111.5

7770800 6.111.5

856

FZ-max (SL) =194

0.00217

FZ-max (SL) 89098

FZ-max (SL) 89.1


64/313


JI-193



. Sh + +

. . x . . .

. . . . . hx x . .

. .

.

.

. . . . x x

. x

x

x x x x x x

x x

= N

= KN

= Minimum of

= KN , = KN , = K

= KN

1.5 SC SA SCP

FA R0.5

1.5 . Sh 2.34 FC

+P R

t1.5

t1.5

t

R0.5

+1.5 1.17

+1.5 1.17

1.5 . Sh 2.34 R

0.5R

0.5FZ

t1.5

IwX-X t1.5

WL

FZ FY/Z Hy

+P R

t

1.5 . Sh -P R

FZ-max (SC) =t

2.34 R0.5

FY/Z Hy hx+

1.5 1.17 R0.5

t1.5

IwX-X t1.5

WL

1.5 x 138 -1.96 84.1

FZ-max (SC) =6.11

2.34 84.1 0.5 0.25 110.12 40+

1.5 1.17 84.1 0.5

6.111.5

2E+05 6.111.5

856

FZ-max (SC) =180

0.01014

FZ-max (SC) 17758

FZ-max (SC) 17.76

Fzmin

Fz-max weld 27.64 FZ-max (SL) 89.1 FZ-max (SC) 17.76

Fz-max 17.76


65/313


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Page 65



Setting a standard for FY & FX

As,

FX = FX/Z * Fz min

FY = FY/Z * FZ min

Fx/z =

Fy/z =

FX = x

Axial Load, FX = KN

Fy = x

Lateral Load, Fy = KN

Miscellaneous Calculations

Fz-(x,y=0)

Fy . Fx .

.

x

. b

x

.

.

KN

0.5

0.25

0.5 17.76

8.8787979

0.25 17.76

+ ( Fz

4.439399

( Hy )+ ) 2Zwy-y Zwx-x WL

( Hx )fwresultant =

+ ( Fx ) Fx . Hz )+ [(Fy b ) ] l 2WL 2 2 Jweld

+ (

+ (Fy

)+ Hz )[(Fy b

)+ ]2

WL 2 2 Jweld(Fx x

fwresultant = ( Fz )2

WL

Fz-max weld = fwallowable WL

Fz-max weld = 475.2 856

Fz-max weld = 406.7712


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. Sh + +

. . x . . .

.

x . . .

.

x . . .

. .

.

.

. ..

x

x

x x

x

FZ-max (SL) = KN

. Sh + +

. . x . . .

x . . .

x . . .

.

.

. .

1.5 SL SA

1.5 . Sh 1.17 FL 1.5 1.17 FA R

0.5

+

SLP

R0.5

FA R0.5

+

P R

t1.5

t1.5

2 t+

1.5 . Sh 1.5 1.17 P R

t1.5

2 t

t1.5

WL 2 t

R1.5 . Sh

1.5 1.17

Sh -P

t1.5

WL

R0.5

FZ+

P

R

FZ-max (SL)2 t

1.5 1.17 R0.5=

1.5 x

x 138.0 -1.96

6.111.5

856

84.1

FZ-max (SL)2 6.11

1.5 1.17 84.10.5=

1.5

FZ-max (SL) =194

0.00125

155.404

1.5 SC SA SCP

1.5 . Sh 2.34 FC R

0.5

+1.5 1.17 FA R

0.5

+P R

t1.5

t1.5

t

1.5 . Sh 1.5 1.17 FA R

0.5

+P R

t1.5

t

R

t1.5

WL t1.5 . Sh

1.5 1.17

- P

R0.5

FZ+

P

R

FZ-max (SC)t

1.5 1.17 R0.5=

1.5 . Sh


67/313


JI-193

Page 67 of 313



x

x x

x

= KN

Fx-(z,y=0)

Fy

. Fx

.

.

x

. b

x

Fx .

x

b

x

x

b

x

138.0 -1.96

6.11 1.5 856

84.1

FZ-max (SC)6.11

1.5 1.17 84.1 0.5=

1.5 x

FZ-max (SC) =180

0.00125

FZ-max (SC) 144.567

Fz

)(H

y

)+

(2

Zwy-y Zwx-x WL

Hx

)+

(fwresultant = + ( Fx ) Fx . Hz )+ [(Fy b ) ] l 2

WL 2 2 Jweld+ (

+ (Fy )+ Hz )[(

Fy b )+ ]2

WL 2 2 Jweld(Fx x

( Hx ) 2 ) l+ ( Fx )+ 2Zwx-x WL 2 Jweld

(Fx . Hzfwresultant =

+ Fx x Hz2

2 Jweld

fwresultant

Fx-max weld = ( Hx ) Hz ) l2

+ ( 1 )+ (2

2 Jweld

2

Zwx-x WL 2 Jweld

+ [(Hz )]


. [Print Date : 6/27/2013]


68/313


JI-193

P



x

x

KN

. Sh + +

. . x . . .

.

. . . . .

. .

.

.

. . .

.

x

x

x x x

x

2+ + 100

475.2

Fx-max weld = 135.117 80 2

4399.87 856 2 7946795

1

+ [ 100 )] 3002

2 7946795

Fx-max weld =475.2

0.030812617

Fx-max weld = 15.42

1.5 SL SA

R0.5

+

SLP

1.5 . Sh 1.17 FL R

0.5P R

t1.5

t1.5

2 t+

1.5 1.17 FA

1.5 . Sh 1.17 R

0.5Fx H hy

+P R

t1.5

IwY-Y 2 t

x Sh -P

t1.5

IwY-Y

R

Fx-max (SL)2 t

1.17 R0.5 Hx hy=

1.5

1.5 x 138.00 -1.96 84.1

2 6.11

1.17 84.135 0.5 135.1 150

6.11 1.5 7770800

Fx-max (SL) =193.5054

0.001853

Fx-max (SL) =

Fx-max (SL) = 104.4109 KN


69/313


JI-193

Page 69 of 313



. Sh + +

. . x . . .

Fy-(z,x=0)

Fy . Fx .

.

.

. b

.

. .

.

. b

.

.

.

1.5 SC SA SCP

FA R0.5

1.5 . Sh 2.34 FC +

P R

t1.5

t1.5

t

R0.5

+1.5 1.17

Fx-max (SC) = 0 KN

( Hy )+ ( Hx )+ ( Fz ) 2Zwy-y Zwx-x WL

fwresultant =+

Fx ) Fx . Hz )+ [(Fy b ) ] l 2WL 2 2 Jweld

+ (

+

Fy

)+ Hz

)[

Fy b

)+

]

2

WL 2 2 Jweld(Fx x

( Fy Hy )Zwy-y

2

2 2 Jweld

2+ [( Fy b )] l

[(Fy b )fwresultant = + Fy )

+

] 2WL 2 2 Jweld

+

l

fwallowable

Fy-max weld = Hy 2 2

Zwy-y 2 2 Jweld

b

b )] b+ 1 )+ 2WL 2 2 Jweld

[(


. [Print Date : 6/27/2013]


70/313


JI-193

P



x

x

KN

. Sh + +

. . x . . .

.

. Sh + +

. . x . . .

. . . . .

.

.

. . .

.

+

80

475.2

Fy-max weld = 110.12 2 2

51805.3333 2 2 7946795

300

] 300+ ( 1 +2

856 2 2 7946795[( 300

Fy-max weld =475.2

0.004591799

Fy-max weld = 103.5

1.5 SL SA

R0.5

+

SLP

1.5 . Sh 1.17 FL R

0.5P R

t1.5

t1.5

2 t+

1.5 1.17 FA

Fy-max (SL) = 0 KN

1.5 SC SA SCP

1.5 . Sh 2.34 FC R

0.5

+1.5 1.17 FA R

0.5

+P R

t1.5

t1.5

t

1.5 . Sh 2.34 R

0.5Fy Hy hx

+P R

t1.5

IwX-X t

1.5 x Sh -P R

Fy-max (SC) =t

2.34 R0.5

Hy hx

t1.5

IwX-X


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JI-193



x

x x x

x

1.5 x 138.0 -1.96 84.1

Fy-max (SC) = 6.112.34 84.14 0.5 110.12 40

6.11 1.5 175994.7

Fy-max (SC) =180

0.03557

15.422 104.411 0.000

Fy-max (SC) = 5.06107 KN

F-max weld (KN) F-max (SL) (KN)

Fy-(z,x=0) 103.489 0.000 5.061

F-max (SC) (KN)

Fz-(x,y=0) 406.771 155.404 144.567

Fx-(y,z=0)


72/313


JI-193



'///////////'///////////

e

HyHx

Fy

FzHz

Fx

Max. stress at

this point

Weld analysis

done for this

weld at pipe to

shoe interface.

Fz

'/////////////////////////////////////////////////////////////////////////////////////////

pipe

axis

b

Fy

hy

'///////////////////////////

'///////////

'/////////////

d

'///////////////////////////

'/////

/////////////////////

d '

'/////////////////////////////////////

//////////////////////

'

'

'

'/////////////////////////////////////

//////////////////////

a

l

A

'/////////////

'/////////////

x

x

hx

yy y

x

z (up)



73/313


JI-193



INPUT :

STEEL MEMBER


Sy-member = N / mm2


Sh = N / mm2

WELDING ELECTRODE



Su-weld = N / mm2


Line Pressure

P = bar (g)

P = N / mm2





275.0

138.0

(As per BS 5950-1 Clause 6.8.5 Table 37, for S275 material, matching electrode is Class 35. However as per contract

specification ADCO DOC. No. 30.99.75.0704 Rev. E, for structural welding all electrodes shall be E70XX to AWS

D1.1. Therefore Class 42 electrode as per BS 5950 is specified in the Project WPS, which is equivalent to E70XX to

AWS D1.1.)

500.00

19.60

1.96

79.20


74/313


JI-193

Pa




SHOE:


Plate thk. d = mm b = mm H = mm A = mm

Pipe




tp = Pipe thk = mm

wp = Wear Pad = mm

t = Pipe thk with CA = Wear Pad + Pipe thk - CA mm

t = Pipe thk

load check for shoe welded support.xls

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