564 D 2160 Welding Calculations

CLIENT: TECHNIP - NPCC CONSORTIUMVANDOR: DESCON ENGINEERING FZE. SHARJAH UNITED ARAB EMERITES.

DESIGN CALCULATION FOR THE WELD STRENGTH CALCULATION FOR

SADDLE BASEPLATE JOINT.EQUIPMENT NO: 564-D-2160

EQUIPMENT NAME: OPEN DRAIN DRUM

CALCULATION FOR THE WELDING OF BASEPLATE TO STRUCTURE

This calculation is based on the chapter 9 of "Shigley's Mechanical Engineering Design,9th Edition, andchapter 10 of Textbook of Machine Design R.S. Khurmi (14th Edition)CASES1. OPERATING CASE2. TRANSPORTATION FROM NPCC SITE TO INSTALLATION SITEFor Operating conditions data has been taken from compress calculations1.OPERATING CASE In Operating case only wind shall be considered values taken from compress calculationscompress.Input Data: Nomenclature

base plate length d= 2075 mmbase plate width b= 204 mmweld leg length h= 23 mm

H= 1475 mm

length of weld L= 2121 mmwidth of weld W= 250 mm

Aw= 77109.662 See Example 10.11 R.S Khurmi (14th Ed.)

Iu= Table 9-2 1928202396 Shigley 9th Edition

I= 3.1359E+10 topc: 9-4 Shigley 9th Edition

Wind Operating calculationthis calculation is based on the operating condition wind is acting only as

For Transverse Wind LoadsWind Pressure Pw 0.0031 bar(g)Multiplication factor 10000.00Gust factor G 0.85Shape Factor(Shell) Cf(sh) 1Shape Factor(Saddle) Cf(sa) 2Projected shell area A1 14.6Projected Saddle Area A2 0.1436Projected platorm Area A3 7.44Shape factor (Platform) Cf(p) 2Transverse Wind Shear, F (T) =

F (T) = 784.36572 Kg (f)= 7694.62771 N

For End Wind Condition Saddle Area Asad 1.18 meffective radius R 1.971 mShell shape factor CP(s) 0.5effective shell area As 6.10229397 including platformMultiplication factor 1.00E+04

F(e) = 222.981446 Kg (f)2187.44799 N

Frictional Load Operating Weight on One Saddle, W 16801 Kg

0.122016.12 Kg(f)

= 19778.1372 N

hight of saddle fromtrue center line

Area of weldAw=1.414 x h x (L+W)

mm²

Unit 2nd Moment of area

(d²/6)(3b+d)

mm³

2nd Moment of areaBASED ON WELDI=0.7071 x h x Iu

mm⁴

m²m²m²

Pw*G*(Cf(sh)*(Proj. shell areaA1) + Cf(sa)*(Proj. saddle areaA2)+Cf(p)*(platform Proj. AreaA3) Eqn#1

m²

End Wind Shear on Saddle F(e)= Pw*G*(Cf(shell)*p*Ro^2 + Cf(saddle)*(Proj. saddle area)) Eqn#2

Coefficiant of friction μ=Frictional Force F(f)=μ*W=





This Frictional Load will be added into the End Wind Load ConditionTotal End Force F(E) = F(e)+F(f) Eqn#3

2239.10145 Kg(f)21965.5852 N

1. PRIMARY SHEAR STRESS IN WELD DUE TO LONGITUDNAL AND TRANSVERSE LOADS

a. SHEAR STRESS IN WELD DUE TO TRANSVERSE LOADS

F(T) = 7694.62771 NAw = 77109.662 Weld area

τ(T) SHEAR STRESS IN TRANSVERSE CONDITION= F(T)/Aw Eqn#4

τ(T) = 0.0998 M Pa.

b. SHEAR STRESS IN WELD DUE TO END WIND LOADS

F(E) = 21965.5852 N TOTAL LONG. FORCEAw = 77109.662 Weld area

τ(L) = F(E)/Aw Eqn#50.28486165 M Pa.

c. COMBINED SHEAR STRESSES DUE TO TRANSVERSE AND LONGITUDNAL LOADS

τ'' = COMBINED LOADSτ'' = √(τ(T)²+τ(L)²) Eqn#6

= 0.3018341 M Pa.

2. SECONDARY SHEAR STRESS IN WELD DUE BENDING DUE TO LONGITUDNAL AND TRANSVERSE LOADS

THIS TRANSVERSE AND LONGITUDNAL FORCES TENDS TO DEVELOP A MOMENT AND WILL TENDS TO BEND IT THEREFORE WITH THIS BENDING CONSIDERING SADDLE FIXED ONE END IS FREE (ATTACHED TO VESSEL). HENCE THIS BENDING STRESS WILL CREATE SHEAR STRESS (9-4)Shigley 9th Edition IT IS SECONDARY SHEAR STRESS

a. IN TRANSVERSE CASE

SADDLE HEIGHT H= 1475 mmCENTEROID x= 125 mmCENTEROID y= 1060.5 mmTransverse Load F(T)= 7694.62771 NMOMENT DUE TO TRANSVERSE LOAD

M(T)= F(T) x H Eqn#7= 1.13E+07 Nmm

τ'(T) = (M(T) x Y)/IWHERE I IS MOMENT OF INERTIA OF WELD

I= 3.1359E+10

τ'(T) = 0.38 M Pa.

b. IN LONGITUDNAL CASE

END LOAD F(E)= 21965.5852 NM(E)= 32399238 Nmm

τ'(E) = (M(E) x Y)/I Eqn#81.09568112 M Pa.

c. COMBINE SECONDARY STRESS

mm²

mm²

mm⁴





τ' = √(τ'(E)²+τ'(T)²) Eqn#9= 1.16096334 M Pa.





3. COMBINED PRIMERY AND SECONDARY SHEAR STRESSES

by combining the equation # 9 and equation # 6 will get combined shear acting on the weldτ = √(τ'²)+(τ''²)) Eqn#10

= 1.19955813 M Pa.

4. BENDING STRESSES IN WELDS


M(T) = 1.13E+07 N mmSECTION MODULUS Z TABLE 10.7 A TEXT BOOK OF MACHINE DESIGNWHERE t = THROAT THICKNESS BY R.S KHURMIt= 0.7071 x h= 16.2633 mm

σb(T) = M/Z Eqn#11FOR SECTION MODULUSL= 2121 mmW= 250 mmZ= 33011198σb(T) 0.34 M Pa.


M(E)= 32399238 NmmZ= 33011198

σb(E)= M(E)/Z Eqn#120.9814621 M Pa.

c. TOTAL BENDING STRESS

σb= √(σb(E)²)+(σb(T)²)) Eqn#131.039939 M Pa.

4. MAXIMUM NORMAL STRESS THEORY

σ t(max.) Eqn#14BENDING STRESS WILL BE TAKEN FROM EQUATION 13 AND SHEAR STRESS SHALL BE TAKEN FROM EQUATION 10σ t(max.)= 1.8273745 M Pa.

5. MAXIMUM SHEAR STRESS THEORY( VON MISES STRESS)

τ (MAX.)= Eqn#151.307405 M Pa.

6. VESSEL UP LIFT CHECK

VESSEL OPERATING WEIGHT= 33364 Kg327300.84 N

IF VESSEL OPERATING WEIGHT IS GREATER THAN THE END WIND LOAD THEN THIS VESSEL PRODUCE AN UPLIFT IF VESSEL OPERATING

IS LOWER THAN THE END WIND LOAD THEN THIS VESSEL WILL NOT CREATE AN UPLIFTEND WIND LOADF(E) 21965.5852 N

TOTAL LOADS THAT WILL CREATE AN UPWARD LIFT.

F(UP0)= F(E)21965.5852 N

t((W x L) + L²/3)

BENDING STRESS σb(T)

mm³

mm³

0.5 x σb +0.5((σb)²+4τ²)½

0.5((σb)²+4τ²)½





DESIGN CHECK NO UP LIFT WILL OCCUR





Hence No uplift shall be considered only shear stress due to longitudnal load shall be taken into account.Vessel weight in empty Condition 12000 KG.

117720 NIN EMPTY CONDITION

UPLIFT FORCE IN EMPTY CASE,FUPE F(E)21965.59 N

DESIGN CHECK NO UPLIFT OCCURRED IN EMPTY CASE

FOLLOWING TABLE SHALL BE CONSIDERED FOR ALLOWABLE LOADS IN WELDING

REFERNCE : chapter 9 of "Shigley's Mechanical Engineering Design,9th Edition,WE WILL CONSIDER ELECTRODE AWS 7018 FOR OUR CALCULATIONS

HENCEFOR VON MISES STRESSESALLOWABLE STRESS 0.30Sut

36 M Pa.Sa ALLOWABLE IN BASE METAL=0.40Sy

ALLOWABLE IN BASE METAL 55.2 M Pa.

EQUATION 14 SHALL BE COMPARED ALLOWABLESDESIGN CHECK

7. ALLOWABLE STRENGTH CHECK





1DESIGN IS SAFE



SADDLE BASEPLATE JOINT.EQUPMENT NUMBER:564-D-2160

EQUIPMENT NAME:OPEN DRAIN DRUM

TORSION IN FIXED SADDLE DUE TO WIND FORCE

TORSION IN FIXED SADDLE DUE TO WIND FORCE IS CALCULATED IN FOLLOWING MODES:

DATA:WIND PRESSURE P = 0.0034 Bar(g) 340 N/m²EFFECTIVE LENGTH L= 5633 mmOuter Radius Shell R= 2152 mm

76166129.462

A 76.166129462 m² COMPLETE SURFACE AREAMOMENT ARM "L" IS THE DISTANCE OF THE CENTER OF FIXED SADDLE TO THE OUTER OF OPPOSITE HEAD

EFFECTIVE SURFACE AREA SHALL BE

Aeff= 38.08306 m²so the force, Fw exerted by wind shall be equal to wind pressure P x effective Surface areaF= P x Aeff Eqn#16

12948.24 NTHE MOMENT M ON FIXED SADDLE CREATED BY WIND SHALL BE EQUAL TO WIND FORCE INTO MOMENT ARMTORSIONAL MOMENT ON THE SADDLE (FIXED) IS Mt= F x L1 Eqn#16aL1 SHALL BE THE DISTANCE FROM THE CENTER OF FIXED SADDLE TO THE CENTER OF SHADDED AREA

L1= 2758 mm2.758 m

Mt= 35711251 N-mm35711.25 N-m

WELD STRENGTH CHECK IN TORSIONAL MOMENTTHIS CALCULATION IS BASED ON THE FOLLOWING SOURCES AS REFERENCES chapter 9 of "Shigley's Mechanical Engineering Design,9th Editionchapter 10 of Textbook of Machine Design R.S. Khurmi (14th Edition)

1. OPERATING CONDITION2. STAGE-2 TRANSPORTATION (NPCC TO INSTALLATION SITE)

1. OPERATING CONDITION

Methedology:As this vessel has one saddle fixed and one saddle sliding and not fixed with either boltings or welding therefore wind will create Torsional moment arround the fixed saddle. To calculate te The torsional moment following calculations are performed and this moment is then checked in welding as well. (See the attached Diagram for Illustration of Torsional Moment.

surface area A=2πRL= mm²As wind will be Exerted on the half of the surface area therefore surface area will be divided by 2 called as Effective. Surface Area





DATA:WELD LONG LENGTH d= 2121 mmWIDTH IN WELD b= 250 mmWELD LEG LENGTH h= 23 mm

1475 mm

2.22E+09 Eqn#17

3.61E+10 Eqn#18

CENTEROIDSx=b/2 125 mmy=D/2 1060.5 mm

1067.8413974 mm Eqn#19Now, the shear stress due to Torsion in welding

Eqn#20

τ(To)= 1.0555041.055504 M Pa.

COMBINE EFFECT OF THE FOLLOWING STRESSES SHALL BE CHECK AND WILL BE COMPARED WITH THEALLOWABLES 1. BENDING STRESS (DUE TO TRANSVERSE LOAD BENDING MOMENT(EQN# 11)2. SHEAR STRESS (DUE TO TRANSVERSE LOAD (EQN# 6)3. SHEAR STRESS DUE TO TORSION (EQUATION # 20)

SHEAR STRESS DUE TO TRANSVERSE LOAD IS COMBINATION OF PRIMARY AND SECONDARY SHEAR STRESSES DUE TO TRANSVERSE LOADINGS

FROM EQUATION # 4 AND 7

τ(T)= 0.099788 M Pa. PRIMARY SHEAR STRESS DUE TO TRANSVERSE LOADSτ'(T) 0.38 M Pa.TOTAL SHEAR STRESS DUE TO TRANSVERSE LOADS

τ(T)(Tot.)= √(τ(T)²+τ'(T)²) Eqn#21τ(T)(Tot.)= 0.396581 M Pa. Eqn#22

BENDING STRESS DUE TO TRANSVERSE LOADSσb(T) = 0.34 M Pa. FROM EQUATION 11

STRESSES IN WELDED JOINTS IN TORSION CALCULATION IS BASED ON SHIGLEY'S MECHANICAL ENGINEERING DESIGN 9TH ED.

HEIGHT OF THE SADDLEFROM TRUE CL H=

Unit Polar Moment of Inertia of weld Ju = (b+d)³/6

mm³

Polar Moment of Inertia of weld J = 0.7071 x h x Ju mm⁴

r=√x2+y2

τ(To)=Mr/J

N/mm²





COMBINED SHEAR STRESS ACTING DUE TRANSVERSE LOAD AND TORSIONAL MOMENT

τ= TOTAL SHEAR STRESS DUE TO TRANSVERSE AND TORSIONAL MOMENTτ= = √(τ(TO)²+τ'(T)(Tot.)²) Eqn#23

= 1.1275479448 M Pa.


σ t(max.)= 0.5 x σb +0.5((σb)²+4τ²)½ SHEAR STRESS SHALL BE TAKEN FROM EQUATION 23 AND BENDING STRESS SHALL BE TAKEN FROM EQUATION 11

σ t(max.)= 1.312482 M Pa. Eqn#24


τ (MAX.)= Eqn#25τ (MAX.)= 1.140577 M Pa.

COMPARING EQN# 24 AND 25 WITH ALLOWABLES

ALLOWABLE STRESS IN WELD S= 0.30Sut36 M Pa.

Sa ALLOWABLE IN BASE METAL=0.40SySa 55.2 M Pa.

DESIGN CHECK

1. FOR WELDING

DESIGN IS SAFE

2. FOR BASE METAL

DESIGN IS SAFE

0.5((σb)²+4τ²)½





CALCULATION FOR THE WELDING OF BASEPLATE TO STRUCTURE

This calculation is based on the chapter 9 of "Shigley's Mechanical Engineering Design,9th Edition, andchapter 10 of Textbook of Machine Design R.S. Khurmi (14th Edition)CASES1. OPERATING CASE2. TRANSPORTATION FROM NPCC SITE TO INSTALLATION SITEFOR TRANSPORTATION CASE VALUES TAKEN FROM TRANSPORTATION CALCULATIONS2.Stage 2 Transportation FOR TRANSPORTATION CASE VALUES TAKEN FROM TRANSPORTATION CALCULATIONS DONE Icompress.Input Data: Nomenclature

base plate length d= 2075 mmbase plate width b= 204 mmweld leg length h= 23 mm

H= 1475 mm

length of weld L= 2121 mmwidth of weld W= 250 mm

Aw= 77109.662 See ExamplR.S Khurmi (EQN#26

Iu= Table 9-2 EQN#271928202396 Shigley 9th Edition

I= 3.1359E+10 topc: 9-4Shigley 9th Edition

Wind Operating calculationthis calculation is based on the operating condition wind is acting only as

For Transverse LoadsMax. Transverse loads shall be taken from the calculations as attached

max. Trans. Load F(T) 30 KN30000 N

these values has been taken from the calculations done for Stage-2 Transportation in PV Elite based

on G Values provided by client

F (T) = 3059.13 Kg (f)= 30000.0008 N

For End Wind Condition Max. Longitudnal loads shall be taken from the calculations as attached

Max. Long. Load F(e)= 64 KN64000 N

these values has been taken from the calculations done for Stage-2 Transportation in PV Elite based

on G Values provided by client

F(e) = 6526.144 Kg (f)64000.0016 N

Frictional Load Operating Weight on One Saddle, W 13000 Kg

0.121560 Kg(f)

= 15303.6 N

hight of saddle fromtrue center line

Area of weldAw=1.414 x h x (L+W)

mm²

Unit 2nd Moment of area

(d²/6)(3b+d)

mm³

2nd Moment of areaBASED ON WELDI=0.7071 x h x Iu

mm⁴

End Wind Shear on Saddle F(e)= Pw*G*(Cf(shell)*p*Ro^2 + Cf(saddle)*(Proj. saddle area)) Eqn#2

Coefficiant of friction μ=Frictional Force F(f)=μ*W=





This Frictional Load will be added into the End Wind Load ConditionTotal End Force F(E) = F(e)+F(f) EQN#28

8086.144 Kg(f)79325.0726 N

1. PRIMARY SHEAR STRESS IN WELD DUE TO LONGITUDNAL AND TRANSVERSE LOADS

a. SHEAR STRESS IN WELD DUE TO TRANSVERSE LOADS

F(T) = 30000.0008 NAw = 77109.662 Weld area

τ(T) SHEAR STRESS IN TRANSVERSE CONDITION= F(T)/Aw EQN#29

τ(T) = 0.3891 M Pa.

b. SHEAR STRESS IN WELD DUE TO END WIND LOADS

F(E) = 79325.0726 N TOTAL LONG. FORCEAw = 77109.662 Weld area

τ(L) = F(E)/Aw EQN#301.02873065 M Pa.

c. COMBINED SHEAR STRESSES DUE TO TRANSVERSE AND LONGITUDNAL LO(COMBINING EQUATION 29 AND 30)τ'' = COMBINED LOADSτ'' = √(τ(T)²+τ(L)²) EQN#31

= 1.09984161 M Pa.

2. SECONDARY SHEAR STRESS IN WELD DUE BENDING DUE TO LONGITUDNAL AND TRANSVERSE LOADS

THIS TRANSVERSE AND LONGITUDNAL FORCES TENDS TO DEVELOP A MOMENT AND WILL TENDS TO BEND IT THEREFORE WITH THIS BENDING CONSIDERING SADDLE FIXED ONE END IS FREE (ATTACHED TO VESSEL). HENCE THIS BENDING STRESS WILL CRESHEAR STRESS (9-4)Shigley 9th Edition IT IS SECONDARY SHEAR STRESS


SADDLE HEIGHT H= 1475 mmCENTEROID x= 125 mmCENTEROID y= 1060.5 mmTransverse Load F(T)= 30000.0008 NMOMENT DUE TO TRANSVERSE LOAD

M(T)= F(T) x H EQN#32= 4.43E+07 Nmm

τ'(T) = (M(T) x Y)/IWHERE I IS MOMENT OF INERTIA OF WELD

I= 3.1359E+10

τ'(T) = 1.50 M Pa.


END LOAD F(E)= 79325.0726 NM(E)= 117004482 Nmm

τ'(E) = (M(E) x Y)/I EQN#333.95687089 M Pa.

c. COMBINE SECONDARY STRESS

mm²

mm²

mm⁴





τ' = √(τ'(E)²+τ'(T)²) EQN#34= 4.23038941 M Pa.





3. COMBINED PRIMERY AND SECONDARY SHEAR STRESSES

by combining the equation # 31 and equation # 34 will get combined shear acting on the welτ = √(τ'²)+(τ''²)) EQN#35

= 4.37102347 M Pa.

4. BENDING STRESSES IN WELDS


M(T) = 4.43E+07 N mmSECTION MODULUS Z TABLE 10.7 A TEXT BOOK OF MACHINE DESWHERE t = THROAT THICKNESS BY R.S KHURMIt= 0.7071 x h= 16.2633 mm

σb(T) = M/Z EQN#37FOR SECTION MODULUSL= 2121 mmW= 250 mmZ= 33011198σb(T) 1.34 M Pa.


M(E)= 117004482 NmmZ= 33011198

σb(E)= M(E)/Z EQN#383.5443877 M Pa.

C. TOTAL BENDING STRESS

σb= √(σb(E)²)+(σb(T)²)) EQN#393.7893933 M Pa.


σ t(max.) EQN#40BENDING STRESS WILL BE TAKEN FROM EQUATION 39 AND SHEAR STRESS SHALL BTAKEN FROM EQUATION 35σ t(max.)= 6.6586993 M Pa.


τ (MAX.)= EQN#414.7640027 M Pa.

6. VESSEL UP LIFT CHECK

VESSEL OPERATING WEIGHT= 33364 Kg327300.84 N

IF VESSEL OPERATING WEIGHT IS GREATER THAN THE END WIND LOAD THEN THIS PRODUCE AN UPLIFT IF VESSEL OPERATING IS LOWER THAN THE END WIND LOAD THEN THIS VESSEL WILL NOT CREATE AN UPLIFTEND WIND LOADF(E) 79325.0726 N

TOTAL LOADS THAT WILL CREATE AN UPWARD LIFT.

F(UP0)= F(E)79325.0726 N

t((W x L) + L²/3)

BENDING STRESS σb(T)

mm³

mm³

0.5 x σb +0.5((σb)²+4τ²)½

0.5((σb)²+4τ²)½





DESIGN CHECK NO UP LIFT WILL OCCURHence No uplift shall be considered only shear stress due to longitudnal load shall be taken into account.

Vessel weight in empty Condition 13000 KG.127530 N

IN EMPTY CONDITION

UPLIFT FORCE IN EMPTY CASE,FUPE F(E)79325.07 N

DESIGN CHECK NO UPLIFT OCCURRED IN EMPTY CASE

FOLLOWING TABLE SHALL BE CONSIDERED FOR ALLOWABLE LOADS IN WELDING

REFERNCE chapter 9 of "Shigley's Mechanical Engineering Design,9th Edition,WE WILL CONSIDER ELECTRODE AWS 7018 FOR OUR CALCULATIONS

HENCEFOR VON MISES STRESSESALLOWABLE STRESS 0.30Sut EQN#42

36 M Pa.Sa ALLOWABLE IN BASE METAL=0.40Sy EQN#43

ALLOWABLE IN BASE METAL 55.2 M Pa.

EQUATION 14 SHALL BE COMPARED ALLOWABLES

7. ALLOWABLE STRENGTH CHECK





DESIGN CHECKTRUE

DESIGN IS SAFE





TORSION IN FIXED SADDLE DUE TO WIND FORCE

DATA:WIND PRESSURE P = 0.0034 Bar(g) 340 N/m²EFFECTIVE LENGTH L= 5633 mm TOTAL LENGTH OF VESSEL FROM FIXED SADDLE CL.Outer Radius Shell R= 2152 mm

76166129.462

A 76.166129462 m² COMPLETE SURFACE AREAMOMENT ARM "L" IS THE DISTANCE OF THE CENTER OF FIXED SADDLE TO THE OUTER OF OPPOSITE HEADEFFECTIVE SURFACE AREA SHALL BE Aeff= 38.08306473 m²so the force, Fw exerted by wind shall be equal to wind pressure P x effective Surface areaF(WIND)= P x Aeff= 12948.24201 N Eqn#44

SIMULTANIOUSLY TRANSVERSE FORCE ACTING DUE TO G LOADINGS WILL CREATE A TORSIONAL FORCE THEREFORE THE TRANSVERSE FORCE FROM THE CALCULATIONS WILL BE ADDED TO THIS WIND FORCE AND THUS RESULTS IN THE TORSIONAL MOMENT ON FIXED SADDLE AND THUS SHEAR STRESS WILL BE CALCULATED FROM THIS TOTAL FORCEF(TRANS.) 30000.00076 N Eqn#44(a)

HENCE THE TOTAL FORCE WILL BE THE SUMM OF THESE TWO FORCES F(TOR.)= √(F(WIND)²)+(F(TRANS)²)) Eqn#45F(TOR.)= 3.27E+04 N

THE MOMENT M ON FIXED SADDLE CREATED BY TOTAL FORCE SHALL BE EQUAL TO WIND FORCE INTO MOMENT TORSIONAL MOMENT ON THE SADDLE (FIXED) IS Mt= F x L1L1 SHALL BE THE DISTANCE FROM THE CENTER OF FIXED SADDLE TO THE COG VESSEL (FULLY DRESSED)IN TRANSPORTATION CASE.(FROM PV ELITE CALCULATIONS)L1= 1838.832 mm COG IN TRANSPORTATION CONDITION(FROM DATUM) 2638.832 mm

1.838832 m DISTANCE OF FIXED SADDLE FROM DATUM 800 mmMt= 6.01E+07 N-mm EFFECTIVE LENGTH L1 1838.832 mm

6.01E+04 N-m

WELD STRENGTH CHECK IN TORSIONAL MOMENTTHIS CALCULATION IS BASED ON THE FOLLOWING SOURCES AS REFERENCES chapter 9 of "Shigley's Mechanical Engineering Design,9th Edition

2. STAGE-2 TRANSPORTATION (NPCC TO INSTALLATION SITE)

Methedology:As this vessel has one saddle fixed and one saddle sliding and not fixed with either boltings or welding therefore wind will create Torsional moment arround the fixed saddle. To calculate te The torsional moment following calculations are performed and this moment is then checked in welding as well. (See the attached Diagram for Illustration of Torsional Moment).

surface area A=2πRL= mm²As wind will be Exerted on the half of the surface area therefore surface area will be divided by 2 called as Effective. Surface Area





chapter 10 of Textbook of Machine Design R.S. Khurmi (14th Edition)

DATA:WELD LONG LENGTH d= 2121 mmWIDTH IN WELD b= 250 mmWELD LEG LENGTH h= 23 mm

1475 mm

2.22E+09 Eqn#46

3.61E+10 Eqn#47

CENTEROIDSx=b/2 125 mmy=D/2 1060.5 mm

1067.8413974 mm Eqn#48Now, the shear stress due to Torsion in welding

Eqn#49

τ(To)= 1.7758758471.775875847 M Pa.

COMBINE EFFECT OF THE FOLLOWING STRESSES SHALL BE CHECK AND WILL BE COMPARED WITH THEALLOWABLES 1. BENDING STRESS (DUE TO TRANSVERSE LOAD BENDING MOMENT(EQN# 11)2. SHEAR STRESS (DUE TO TRANSVERSE LOAD (EQN# 6)3. SHEAR STRESS DUE TO TORSION (EQUATION # 20)

SHEAR STRESS DUE TO TRANSVERSE LOAD IS COMBINATION OF PRIMARY AND SECONDARY SHEAR STRESSES DUE TO TRANSVERSE LOADINGS

FROM EQUATION # 29 AND 32

τ(T)= 0.389056313 M Pa. PRIMARY SHEAR STRESS DUE TO TRANSVERSE LOADSτ'(T) 1.50 M Pa.TOTAL SHEAR STRESS DUE TO TRANSVERSE LOADS

τ(T)(Tot.)= √(τ(T)²+τ'(T)²) Eqn#50τ(T)(Tot.)= 1.546199253 M Pa. Eqn#51

STRESSES IN WELDED JOINTS IN TORSION CALCULATION IS BASED ON SHIGLEY'S MECHANICAL ENGINEERING DESIGN 9TH ED.

HEIGHT OF THE SADDLEFROM TRUE CL H=

Unit Polar Moment of Inertia of weld Ju = (b+d)³/6

mm³

Polar Moment of Inertia of weld J = 0.7071 x h x Ju mm⁴

r=√x2+y2

τ(To)=Mr/J

N/mm²





BENDING STRESS DUE TO TRANSVERSE LOADSσb(T) = 1.34 M Pa. FROM EQUATION#37

COMBINED SHEAR STRESS ACTING DUE TRANSVERSE LOAD AND TORSIONAL MOMENT

τ= TOTAL SHEAR STRESS DUE TO TRANSVERSE AND TORSIONAL MOMENTτ= = √(τ(TO)²+τ'(T)(Tot.)²) Eqn#52

= 2.354669224 M Pa.


σ t(max.)= 0.5 x σb +0.5((σb)²+4τ²)½ SHEAR STRESS SHALL BE TAKEN FROM EQUATION 23 AND BENDING STRESS SHALL BE TAKEN FROM EQUATION 11

σ t(max.)= 3.118424759 M Pa. Eqn#53


τ (MAX.)= Eqn#54τ (MAX.)= 2.448197618 M Pa.

COMPARING EQN# 24 AND 25 WITH ALLOWABLES

ALLOWABLE STRESS IN WELD S= 0.30Sut36 M Pa.

Sa ALLOWABLE IN BASE METAL=0.40SySa 55.2 M Pa.

DESIGN CHECK

1. FOR WELDING

DESIGN IS SAFE

2. FOR BASE METAL

DESIGN IS SAFEREFERENCE PAGES ARE ATTACHED BELOW

0.5((σb)²+4τ²)½


DESIGN CALCULATION FOR THE TORSION CALCULATION FOR

SADDLE SECTION IN OPERATINGEQUIPMENT NO: 564-D-2160


SADDLE DATA

SADDLE PLATE WIDTH,b2= 180 mmWEB PLATE LENGTH, h1= 1884 mmWEBPLATE THICKNESS, b1= 13 mmSADDLE PLATE THICKNESS h2= 13 mmRIB THICKNESS h3= 13 mmRIB WIDTH b3= 83.5 mm

CALCULATION REFERENCE : MECHANICS OF MATERIALS BY ANDREW PYTELAND JAAN KIUSLAAS 2ND EDITION.

AREAS:AREA OF WEB, A1= 24492AREA OF TOP SADDLE PLATE A2= 2340AREA OF BOT. SADDLE PLATE A3= 2340AREA OF RIB A A4= 1085.5AREA OF RIB B A5= 1085.5AREA OF RIB C A6= 1085.5AREA OF RIB D A7= 1085.5CENTEROID LOCATION FROM X AXIS VALUES FROM FIG 1.CENTEROID OF WEB PLATE Y1= h1/2 942 mmCENTEROID TOP SADDLE PLATE (Y2=W1+h2/2) 1993.105 mm 1986.6047 mmCENTEROID BOT. SADDLE PLATE (Y3=h2/2 6.5 mmCENTEROID OF RIB A = Y4 = (h3/2)+W2 1268.5 mmCENTEROID OF RIB B = Y5= (h3/2)+W3 641.5 mm 1262 mmCENTEROID OF RIB C=Y6== (h3/2)+W2 1268.5 mmCENTEROID OF RIB D= Y7= (h3/2)+W3 641.5 mm

635 mm

CENTEROID LOCATION FROM Y AXIS180.00 mm

CENTEROID OF WEB PLATE X1=(b1/2)+W5 90.00 mmCENTEROID TOP SADDLE PLATE (X2= b2/2) 90.00 mmCENTEROID BOT. SADDLE PLATE (X3=b2/2 90 mm 83.5 mm

CENTEROID OF RIB A = X4 =(b3/2) 41.75 mmCENTEROID OF RIB B = X5=(b3/2) 41.75 mmCENTEROID OF RIB C=X6= W4-(b3/2) 138.25 mmCENTEROID OF RIB D= X7= W4-(b3/2) 138.25 mm

DESIGN CALCULATION FOR THE TORSION DUE TO WIND LOAD IN THE SADDLE SECTION AS ONE SADDLE IS FIXED AND ANOTHER ONE IS FREE (OPERATING CASE)

mm²mm²mm²mm²mm²mm²mm²

DISTANCE FROM X-AXIS TO THE BOTTOM OF AREA A2=W1=(FROM FIG 1)

DISTANCE FROM X-AXIS TO THE BOTTOM OF RIB A AND RIB C W2= (FROM FIG 1)

DISTANCE FROM X-AXIS TO THE BOTTOM OF RIB B AND RIB D W3= (FROM FIG 1)

DISTANCE FROM Y-AXIS TO THE EDGE OF RIB C AND RIB D W4= (FROM FIG 1)

DISTANCE FROM Y-AXIS TO THE NEAREST OUTER EDGE OF WEB W5= (FROM FIG 1)





ABOVE DATA IS BASED BY CONSIDERING FIG 1 ATTACHED WHICH IS THE SECTION OF THE SADDLE RIBSAND WEB PLATE, POALR MOMENT OF INERTIA WILL BE CALCULATED AT THE CENTEROIDAL AXIS JxxBY CALCULATING Ixx and Iyy (MOMENT OF INTERTIA WRT TO XX AND YY AXIS).

Y= CENTEROID LOCATION THROUGH X AXISΣ(Ai x yi)/Ai i=1,2,3…. Eqn#55X= CENTEROID LOCATION THROUGH Y AXISΣ(Ai x xi)/Ai i=1,2,3,4…. Eqn#56Y= 951.76 mm CENTEROID FROM X-AXIS X= 90.00 mm CENTEROID FROM Y-AXIS

MOMENT OF INERTIA ALONG XX AXIS USING PARALLEL AXIS THEOREM (neutral axis)

Eqn#57 Eqn#58(Ixx)1= M.O.I OF WEB AREA A1 7.24E+09(Ixx)2= M.O.I OF TOP SADDLE PLATE AREA A2 3.30E+04(Ixx)3= M.O.I OF BOT. SADDLE PLATE AREA A3 3.30E+04(Ixx)4= MO.I OF RIB A 1.53E+04(Ixx)5= MO.I OF RIB B 1.53E+04(Ixx)6= MO.I OF RIB C 1.53E+04(Ixx)7= MO.I OF RIB D 1.53E+04AS ALL RIBS AND TOP AND BOTTOM SADDLE PLATES HAVE SAME DIMENSIONS WILL HAVE THE SAME MOMENT OF INERTIA

A1(Y-Y1)²= 2.33E+06 WEB AREA A1A2(Y-Y2)²= 2.54E+09 TOP SADDLE PLATE AREA A2A3(Y-Y3)²= 2.09E+09 BOT. SADDLE PLATE AREA A3A4(Y-Y4)²= 1.09E+08 RIB AA5(Y-Y5)²= 1.04E+08 RIB BA6(Y-Y6)²= 1.09E+08 RIB CA7(Y-Y7)²= 1.04E+08 RIB DAS THE TERM Ai(Y-Yi)² FOR RIBS A,C AND B,D ARE SAME THEREFORE THE FORMULLA AND THE LOCAL MOMENT OF INERTIA OF ALL RIBS ARE SAME THEREFORE THE FORMULLA OF Ixx WILL BE

Ixx = [(Ixx)1+A1(Y-Y1)²]+[(Ixx)2+A2(Y-Y2)²]+[(Ixx)3+A3(Y-Y3)²]+2 x [(Ixx)4+ A4(Y-Y4)²]+2 x [(Ixx)5+ A5(Y-Y5)²]

Ixx = 1.23020E+10 Eqn#57MOMENT OF INERTIA ALONG XX AXIS (neutral axis) NOT USING PARRALEL AXIS THEOREM

Ixx = 7.24E+09 Ixx = Σ[(Ixx)i]Eqn#59

AS Ixx = Σ[(Ixx)i+Ai(Y-Yi)²] WHERE i=1,2,3,4 WHERE Ixx=bh³/12

mm⁴mm⁴mm⁴mm⁴mm⁴mm⁴mm⁴


mm⁴

mm⁴





MOMENT OF INERTIA ALONG YY AXIS USING PARALLEL AXIS THEOREM (neutral axis)

AS Iyy = Σ[(Iyy)i+Ai(X-Xi)²] WHERE i=1,2,3,4 WHERE Iyy=b³h/12

Eqn#60 Eqn#61(Iyy)1= M.O.I OF WEB AREA A1 3.45E+05(Iyy)2= M.O.I OF TOP SADDLE PLATE AREA A2 6.32E+06(Iyy)3= M.O.I OF BOT. SADDLE PLATE AREA A3 6.32E+06(Iyy)4= MO.I OF RIB A 6.31E+05(Iyy)5= MO.I OF RIB B 6.31E+05(Iyy)6= MO.I OF RIB C 6.31E+05(Iyy)7= MO.I OF RIB D 6.31E+05AS ALL RIBS AND TOP AND BOTTOM SADDLE PLATES HAVE SAME DIMENSIONS WILL HAVE THE SAME MOMENT OF INERTIA

A1(X-X1)²= 0.00E+00A2(X-X2)²= 0.00E+00A3(X-X3)²= 0.00E+00A4(X-X4)²= 2.53E+06A5(X-X5)²= 2.53E+06A6(X-X6)²= 2.53E+06A7(X-X7)²= 2.53E+06

AS Iyy = [(Iyy)1+A1(X-X1)²]+ [(Iyy)2+A2(X-X2)²]+ [(Iyy)3+A3(X-X3)²]+ [(Iyy)4+A4(X-X4)²]+ [(Iyy)5+A5(X-X5)²]

Iyy = 2.56E+07 Eqn#60

MOMENT OF INERTIA ALONG YY AXIS (neutral axis) NOT USING PARRALEL AXIS THEOREM

Iyy = 1.55E+07 Eqn#61 Iyy = Σ[(Iyy)i]

POLAR MOMENT OF INERTIA Jo

Jo=Ixx+Iyy Eqn#62

POLAR MOMENT OF INERTIA USING INERTIAS FROM PARALLEL AXIS THEOREM

FROM EQUATION 60 AND 57



mm⁴

mm⁴





Jo= 1.23E+10 Eqn#62a

POLAR MOMENT OF INERTIA USING INERTIAS FROM SUMMATION OF LOCAL MOMENT OF INERTIAS

Jo=Ixx+Iyy Eqn#62(b)FROM EQUATION 59 AND 61Jo= 7.26E+09

WE WILL CONSIDER EQUATION NO 62(B) AS OF STRINGENT CONDITIONS IN THE MAXIMUM SHEAR STRESSFORMULLA AND WILL BE COMPARED WITH THE ALLOWABLE STRESS OF SADDLE MATERIAL

MOMENT SHALL BE TAKEN FROM THE EQUATION 16 a

Mt= TORSIONAL MOMENT 3.57E+07 N-mm

AS τ(MAX.)= MAX. SHEAR STRESS IN TORSIONτ(MAX.)= (MT x r)/Jo N/mm2

WHERE r IS THE CENTEROID ON WHICH THE MOMENT IS ACTING IN THIS CASE Y WILL BE THE CENTEROI

THEREFORE τ(MAX.)= 4.681552 N/mm2

ALLOWABLE CHECK

SADDLE ALLOWABLE STRESS Sa = 138 M Pa

POLAR MOMENT OF INERTIA OF SADDLE SECTION IS SUFFICIENT TO SUSTAIN TORSION

mm⁴

mm⁴









SADDLE DATA

SADDLE PLATE WIDTH,b2= 180 mmWEB PLATE LENGTH, h1= 1884 mmWEBPLATE THICKNESS, b1= 13 mmSADDLE PLATE THICKNESS h2= 13 mmRIB THICKNESS h3= 13 mmRIB WIDTH b3= 83.5 mm

CALCULATION REFERENCE : MECHANICS OF MATERIALS BY ANDREW PYTELAND JAAN KIUSLAAS 2ND EDITION.

AREAS:AREA OF WEB, A1= 24492AREA OF TOP SADDLE PLATE A2= 2340AREA OF BOT. SADDLE PLATE A3= 2340AREA OF RIB A A4= 1085.5AREA OF RIB B A5= 1085.5AREA OF RIB C A6= 1085.5AREA OF RIB D A7= 1085.5CENTEROID LOCATION FROM X AXIS VALUES FROM FIG 1.CENTEROID OF WEB PLATE Y1= h1/2 942 mmCENTEROID TOP SADDLE PLATE (Y2=W1+h2/2) 1993.105 mm 1986.6047 mmCENTEROID BOT. SADDLE PLATE (Y3=h2/2 6.5 mmCENTEROID OF RIB A = Y4 = (h3/2)+W2 1268.5 mmCENTEROID OF RIB B = Y5= (h3/2)+W3 641.5 mm 1262 mmCENTEROID OF RIB C=Y6== (h3/2)+W2 1268.5 mmCENTEROID OF RIB D= Y7= (h3/2)+W3 641.5 mm

635 mm

CENTEROID LOCATION FROM Y AXIS180.00 mm

CENTEROID OF WEB PLATE X1=(b1/2)+W5 90.00 mmCENTEROID TOP SADDLE PLATE (X2= b2/2) 90.00 mmCENTEROID BOT. SADDLE PLATE (X3=b2/2 90 mm 83.5 mm

CENTEROID OF RIB A = X4 =(b3/2) 41.75 mmCENTEROID OF RIB B = X5=(b3/2) 41.75 mmCENTEROID OF RIB C=X6= W4-(b3/2) 138.25 mmCENTEROID OF RIB D= X7= W4-(b3/2) 138.25 mm

DESIGN CALCULATION FOR THE TORSION DUE TO WIND LOAD IN THE SADDLE SECTION AS ONE SADDLE IS FIXED AND ANOTHER ONE IS FREE (NPCC TO SITE TRANSPORTATION CASE)

mm²mm²mm²mm²mm²mm²mm²

DISTANCE FROM X-AXIS TO THE BOTTOM OF AREA A2=W1=(FROM FIG 1)

DISTANCE FROM X-AXIS TO THE BOTTOM OF RIB A AND RIB C W2= (FROM FIG 1)

DISTANCE FROM X-AXIS TO THE BOTTOM OF RIB B AND RIB D W3= (FROM FIG 1)

DISTANCE FROM Y-AXIS TO THE EDGE OF RIB C AND RIB D W4= (FROM FIG 1)

DISTANCE FROM Y-AXIS TO THE NEAREST OUTER EDGE OF WEB W5= (FROM FIG 1)





ABOVE DATA IS BASED BY CONSIDERING FIG 1 ATTACHED WHICH IS THE SECTION OF THE SADDLE RIBSAND WEB PLATE, POALR MOMENT OF INERTIA WILL BE CALCULATED AT THE CENTEROIDAL AXIS JxxBY CALCULATING Ixx and Iyy (MOMENT OF INTERTIA WRT TO XX AND YY AXIS).

Y= CENTEROID LOCATION THROUGH X AXISΣ(Ai x yi)/Ai i=1,2,3…. Eqn#55X= CENTEROID LOCATION THROUGH Y AXISΣ(Ai x xi)/Ai i=1,2,3,4…. Eqn#56Y= 951.76 mm CENTEROID FROM X-AXIS X= 90.00 mm CENTEROID FROM Y-AXIS

MOMENT OF INERTIA ALONG XX AXIS USING PARALLEL AXIS THEOREM (neutral axis)

Eqn#57 Eqn#58(Ixx)1= M.O.I OF WEB AREA A1 7.24E+09(Ixx)2= M.O.I OF TOP SADDLE PLATE AREA A2 3.30E+04(Ixx)3= M.O.I OF BOT. SADDLE PLATE AREA A3 3.30E+04(Ixx)4= MO.I OF RIB A 1.53E+04(Ixx)5= MO.I OF RIB B 1.53E+04(Ixx)6= MO.I OF RIB C 1.53E+04(Ixx)7= MO.I OF RIB D 1.53E+04AS ALL RIBS AND TOP AND BOTTOM SADDLE PLATES HAVE SAME DIMENSIONS WILL HAVE THE SAME MOMENT OF INERTIA

A1(Y-Y1)²= 2.33E+06 WEB AREA A1A2(Y-Y2)²= 2.54E+09 TOP SADDLE PLATE AREA A2A3(Y-Y3)²= 2.09E+09 BOT. SADDLE PLATE AREA A3A4(Y-Y4)²= 1.09E+08 RIB AA5(Y-Y5)²= 1.04E+08 RIB BA6(Y-Y6)²= 1.09E+08 RIB CA7(Y-Y7)²= 1.04E+08 RIB DAS THE TERM Ai(Y-Yi)² FOR RIBS A,C AND B,D ARE SAME THEREFORE THE FORMULLA AND THE LOCAL MOMENT OF INERTIA OF ALL RIBS ARE SAME THEREFORE THE FORMULLA OF Ixx WILL BE

Ixx = [(Ixx)1+A1(Y-Y1)²]+[(Ixx)2+A2(Y-Y2)²]+[(Ixx)3+A3(Y-Y3)²]+2 x [(Ixx)4+ A4(Y-Y4)²]+2 x [(Ixx)5+ A5(Y-Y5)²]

Ixx = 1.23020E+10 Eqn#57MOMENT OF INERTIA ALONG XX AXIS (neutral axis) NOT USING PARRALEL AXIS THEOREM

Ixx = 7.24E+09 Ixx = Σ[(Ixx)i]Eqn#59

AS Ixx = Σ[(Ixx)i+Ai(Y-Yi)²] WHERE i=1,2,3,4 WHERE Ixx=bh³/12



mm⁴

mm⁴





MOMENT OF INERTIA ALONG YY AXIS USING PARALLEL AXIS THEOREM (neutral axis)

AS Iyy = Σ[(Iyy)i+Ai(X-Xi)²] WHERE i=1,2,3,4 WHERE Iyy=b³h/12

Eqn#60 Eqn#61(Iyy)1= M.O.I OF WEB AREA A1 3.45E+05(Iyy)2= M.O.I OF TOP SADDLE PLATE AREA A2 6.32E+06(Iyy)3= M.O.I OF BOT. SADDLE PLATE AREA A3 6.32E+06(Iyy)4= MO.I OF RIB A 6.31E+05(Iyy)5= MO.I OF RIB B 6.31E+05(Iyy)6= MO.I OF RIB C 6.31E+05(Iyy)7= MO.I OF RIB D 6.31E+05AS ALL RIBS AND TOP AND BOTTOM SADDLE PLATES HAVE SAME DIMENSIONS WILL HAVE THE SAME MOMENT OF INERTIA

A1(X-X1)²= 0.00E+00A2(X-X2)²= 0.00E+00A3(X-X3)²= 0.00E+00A4(X-X4)²= 2.53E+06A5(X-X5)²= 2.53E+06A6(X-X6)²= 2.53E+06A7(X-X7)²= 2.53E+06

AS Iyy = [(Iyy)1+A1(X-X1)²]+ [(Iyy)2+A2(X-X2)²]+ [(Iyy)3+A3(X-X3)²]+ [(Iyy)4+A4(X-X4)²]+ [(Iyy)5+A5(X-X5)²]

Iyy = 2.56E+07 Eqn#60

MOMENT OF INERTIA ALONG YY AXIS (neutral axis) NOT USING PARRALEL AXIS THEOREM

Iyy = 1.55E+07 Eqn#61 Iyy = Σ[(Iyy)i]

POLAR MOMENT OF INERTIA Jo

Jo=Ixx+Iyy Eqn#62

POLAR MOMENT OF INERTIA USING INERTIAS FROM PARALLEL AXIS THEOREM

FROM EQUATION 60 AND 57



mm⁴

mm⁴





Jo= 1.23E+10 Eqn#62a

POLAR MOMENT OF INERTIA USING INERTIAS FROM SUMMATION OF LOCAL MOMENT OF INERTIAS

Jo=Ixx+Iyy Eqn#62(b)FROM EQUATION 59 AND 61Jo= 7.26E+09

WE WILL CONSIDER EQUATION NO 62(B) AS OF STRINGENT CONDITIONS IN THE MAXIMUM SHEAR STRESSFORMULLA AND WILL BE COMPARED WITH THE ALLOWABLE STRESS OF SADDLE MATERIAL

MOMENT SHALL BE TAKEN FROM THE EQUATION 44 a

Mt= TORSIONAL MOMENT 6.01E+04 N-mm

AS τ(MAX.)= MAX. SHEAR STRESS IN TORSIONτ(MAX.)= (MT x r)/Jo N/mm2

WHERE r IS THE CENTEROID ON WHICH THE MOMENT IS ACTING IN THIS CASE Y WILL BE THE CENTEROI

THEREFORE τ(MAX.)= 7.88E-03 N/mm2

ALLOWABLE CHECK

SADDLE ALLOWABLE STRESS Sa = 138 M Pa

POLAR MOMENT OF INERTIA OF SADDLE SECTION IS SUFFICIENT TO SUSTAIN TORSION

mm⁴

mm⁴





Documents

564 D 2160 Welding Calculations