welding calculations for Base plate

1.OPERATING CONDITIONCALCULATION 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:Nomenclaturebase plate lengthd=3030mmbase plate widthb=260mmweld leg lengthh=30mmhight of saddle fromtrue center lineH=2000mm

length of weldL=3090mmwidth of weldW=320mmArea of weldAw=1.414 x h x (L+W)Aw=144652.2mmSee Example 10.11 R.S Khurmi (14th Ed.)Unit 2nd Moment of areaIu=(d/6)(3b+d)Table 9-2 5829871500mmShigley 9th Edition

2nd Moment of areaBASED ON WELDI=0.7071 x h x IuI=123669064129.5mmtopc: 9-4 Shigley 9th Edition

Wind Operating calculationthis calculation is based on the operating condition wind is acting only as For Transverse Wind LoadsWind PressurePw0.0031bar(g)(THIS VALUE IS FOR OPERATING WIND ONLY)Multiplication factor10000.00Gust factorG0.85Shape Factor(Shell)Cf(sh)1Shape Factor(Saddle)Cf(sa)2Projected shell areaA122.1733mProjected Saddle AreaA20.0869mProjected platorm AreaA35mShape factor (Platform)Cf(p)2Transverse Wind Shear, F (T)=Pw*G*(Cf(sh)*(Proj. shell areaA1) + Cf(sa)*(Proj. saddle areaA2)+Cf(p)*(platform Proj. AreaA3) Eqn#1AF (T)=862.3Kg (f)=8459.62NFor End Wind Condition Saddle AreaAsad1.9465meffective radiusR2.297mShell shape factorCP(s)0.5effective shell areaAs8.2878497166mincluding platformMultiplication factor1.00E+04End Wind Shear on Saddle F(e)= Pw*G*(Cf(shell)*p*Ro^2 + Cf(saddle)*(Proj. saddle area)) Eqn#2A

F(e)=325.2Kg (f)3.19E+03NFrictional Load Operating Weight on One Saddle, W18040KgCoefficiant of friction =0.12Frictional Force F(f)=*W=2164.8Kg(f)=2.12E+04N

This Frictional Load will be added into the End Wind Load ConditionTotal End Force F(E) =F(e)+F(f)Eqn#3A2.49E+03Kg(f)2.44E+04N1. PRIMARY SHEAR STRESS IN WELD DUE TO LONGITUDNAL AND TRANSVERSE LOADSa. SHEAR STRESS IN WELD DUE TO TRANSVERSE LOADS

F(T)=8.46E+03NAw=1.45E+05mmWeld area(T)SHEAR STRESS IN TRANSVERSE CONDITION=F(T)/AwEqn#4A(T)=5.85E-02M Pa.b. SHEAR STRESS IN WELD DUE TO END WIND LOADS

F(E)=2.44E+04NTOTAL LONG. FORCEAw=1.45E+05mmWeld area(L)=F(E)/AwEqn#5A1.69E-01M Pa.

c. COMBINED SHEAR STRESSES DUE TO TRANSVERSE AND LONGITUDNAL LOADS''=COMBINED LOADS''=((T)+(L))Eqn#6A=1.79E-01M Pa.2. SECONDARY SHEAR STRESS IN WELD DUE BENDING DUE TO LONGITUDNAL AND TRANSVERSE LOADSTHIS 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 STRESSa. IN TRANSVERSE CASE

SADDLE HEIGHT H=2000mmCENTEROID x=160mmCENTEROID y=1545mmTransverse Load F(T)=8.46E+03NMOMENT DUE TO TRANSVERSE LOADM(T)=F(T) x HEqn#7A=1.69E+07Nmm'(T)=(M(T) x Y)/IWHERE I IS MOMENT OF INERTIA OF WELDI=123669064129.5mm

'(T)=0.21M Pa.b. IN LONGITUDNAL CASE

END LOAD F(E)=2.44E+04NM(E)=4.89E+07Nmm'(E)=(M(E) x Y)/IEqn#8A0.610322981M Pa.c. COMBINE SECONDARY STRESS'=('(E)+'(T))Eqn#9A=6.46E-01M Pa.3. COMBINED PRIMERY AND SECONDARY SHEAR STRESSESby combining the equation # 9 and equation # 6 will get combined shear acting on the weld=(')+(''))Eqn#10A=6.70E-01M Pa.4. BENDING STRESSES IN WELDSa. IN TRANSVERSE CASE

M(T)=1.69E+07N mmSECTION MODULUS Zt((W x L) + L/3)TABLE 10.7 A TEXT BOOK OF MACHINE DESIGNWHERE t = THROAT THICKNESSBY R.S KHURMIt=0.7071 x h=21.213mmBENDING STRESS b(T)b(T)=M/ZEqn#11AFOR SECTION MODULUSL=3090mmW=320mmZ=8.85E+07mmb(T)0.19M Pa.b. IN LONGITUDNAL CASE

M(E)=4.89E+07NmmZ=8.85E+07mmb(E)=M(E)/ZEqn#12A5.52E-01M Pa.c. TOTAL BENDING STRESSb=(b(E))+(b(T)))Eqn#13A5.84E-01M Pa.4. MAXIMUM NORMAL STRESS THEORY t(max.)0.5 x b +0.5((b)+4)Eqn#14ABENDING STRESS WILL BE TAKEN FROM EQUATION 13 AND SHEAR STRESS SHALL BE TAKEN FROM EQUATION 10 t(max.)=1.02E+00M Pa.5. MAXIMUM SHEAR STRESS THEORY( VON MISES STRESS) (MAX.)=0.5((b)+4)Eqn#15A7.31E-01M Pa.6. VESSEL UP LIFT CHECK VESSEL OPERATING WEIGHT=36246.00Kg355573.26NIF 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)2.44E+04N

TOTAL LOADS THAT WILL CREATE AN UPWARD LIFT.F(UP0)=F(E)2.44E+04NDESIGN CHECKNO 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 Condition31434.00KG.3.08E+05NIN EMPTY CONDITION

UPLIFT FORCE IN EMPTY CASE,FUPEF(E)2.44E+04N

DESIGN CHECKNO UPLIFT OCCURRED IN EMPTY CASE7. ALLOWABLE STRENGTH CHECKFOLLOWING 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 CALCULATIONSHENCEFOR VON MISES STRESSESALLOWABLE STRESS 0.30Sut248.1M Pa.Sa ALLOWABLE IN BASE METAL=0.40SyALLOWABLE IN BASE METAL104M Pa.(260 M Pa yeaild of SA-516 70 N)EQUATION 14 SHALL BE COMPARED ALLOWABLESDESIGN CHECKTRUEDESIGN IS SAFE

CLIENT: TECHNIP - NPCC CONSORTIUMVANDOR: DESCON ENGINEERING FZE. SHARJAH UNITED ARAB EMERITES.DESIGN CALCULATION FOR THE WELD STRENGTH CALCULATION FOR SADDLE BASEPLATE JOINT.EQUIPMENT NO: 566-D-2819A/BEQUIPMENT NAME: NITROGEN RECEVER

2.BLAST LOAD+OPERATINGCALCULATION 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 CASE+BLAST LOAD.

BLAST LOAD CALCULATIONS WILL BE FOLLOWED BY THE ATTACHMENT-1 (ULUTP_HCYL1,ULUTP_AREA3_NITROGEN RECEIVER REV-BAS BLAST LOAD WILL BE CALCULATED AS PER GIVEN SPECIFICATION, AS PER SPECIFICATION TRANSVERSE BLAST LOAD WILL ACT AT Y-AXIS AND Z-AXIS OF THE VESSEL AND LONGITUDNAL BLAST LOAD WILL ACT ALONG THE X-AXIS AS FIG. MENTIONED IN ATTACHMENT1, OF BLAST LOAD SPECIFICATION.BLAST LOAD CALCULATIONS:VESSEL LENGTH, L=15mVESSEL OUTER DIA=4.3mBLAST PRESSURE=0.1335BAR(G)(BLAST + OPERATING WIND PRESSURE)13350.0Pa. OR N/mVESSEL OPERATING WEIGHT OW=36246.0KG355573.3NVESSEL UPLIFT IN OP. CASEVESSEL EMPTY WEIGHT EW=31434.0KG308367.5NVESSEL UPLIFT IN EMPTY CASEA. TRANSVERSE CASE ALONG Z OR Y AXISTOTAL AREA=64.5mABSOLUTE BLAST FORCE=8.61E+05NTHIS TRANSVERSE BLAST LOAD WILL BE RESISTED BY THE WEIGHT OF THE VESSEL EITHER WEIGHT IS LESS THAN OR GREATER THANFOR THE CASE OF UPLIFT, THE BLAST LOAD WILL BE GREATER THAN THE WEIGHT OF THE VESSEL HENCE THE NET BLAST LOAD WILL BE EQUAL TO THE ABSOLUTE TRANSVERSE BLAST LOAD MINUS THE OPERATING WEIGHT OF THE VESSEL.

NET TRANS BLAST FORCE (OPERATING)=5.06E+05NTHIS NET BLAST LOAD (ABSOLUTE BLAST LOAD MINUS OPERATING WEIGHT)NET TRANS BLAST FORCE (EMPTY)=5.53E+05NTHIS NET BLAST LOAD (ABSOLUTE BLAST LOAD MINUS EMPTY WEIGHT)TRANSVERSE BLAST LOAD F(T)B=5.53E+05NGOVERNING CONDITION WILL BE THE WORSE OF EMPTY OR OP.B. LONGITUDNAL BLAST LOAD CASE:AS PER ATTACHMENT-1 LONGITUDNAL OR END BLAST LOAD WILL BE ACT ALONG THE X AXIS OF THE VESSEL SEE ATTACHMENT-1.TOTAL AREA= D/4=14.5220120412mLONGITUDNAL BLAST LOAD F(L)B=1.94E+05NF=P x A

C. TRANSVERSE AND LONGITUDNAL SHEARS FROM OPERATING CASE:

TRANSVERSE SHEAR IN OPERATING, FT(O)=8.46E+03NLONGITUDNAL SHEAR IN OPERATING, FL(O)=2.44E+04N

D. TOTAL TRANSVERSE AND LONGITUDNAL WIND SHEAR (OPERATING+BLAST LOAD)

TOTAL TRANSVERSE SHEAR, F(T)=F(T)O+F(T)B=5.61E+05NTOTAL LONGITUDNAL SHEAR, F(T)=F(T)O+F(T)B=2.18E+05N

This Frictional Load will be added into the End Wind Load Condition

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

F(T)=5.61E+05NAw=144652.2mmWeld area(T)SHEAR STRESS IN TRANSVERSE CONDITION=F(T)/AwEqn#4B(T)=3.8794M Pa.b. SHEAR STRESS IN WELD DUE TO END WIND LOADS

F(E)=2.18E+05NTOTAL LONG. FORCEAw=144652.2mmWeld area(L)=F(E)/AwEqn#5B1.51E+00M Pa.

c. COMBINED SHEAR STRESSES DUE TO TRANSVERSE AND LONGITUDNAL LOADS''=COMBINED LOADS''=((T)+(L))Eqn#6B=4.16E+00M Pa.2. SECONDARY SHEAR STRESS IN WELD DUE BENDING DUE TO LONGITUDNAL AND TRANSVERSE LOADSTHIS 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 STRESSa. IN TRANSVERSE CASE

SADDLE HEIGHT H=2000mmCENTEROID x=160mmCENTEROID y=1545mmTransverse Load F(T)=5.61E+05NMOMENT DUE TO TRANSVERSE LOADM(T)=F(T) x HEqn#7B=1.12E+09Nmm'(T)=(M(T) x Y)/IWHERE I IS MOMENT OF INERTIA OF WELDI=123669064129.5mm


END LOAD F(E)=2.44E+04NM(E)=4.89E+07Nmm'(E)=(M(E) x Y)/IEqn#8B6.10E-01M Pa.c. COMBINE SECONDARY STRESS'=('(E)+'(T))Eqn#9B=1.40E+01M Pa.3. COMBINED PRIMERY AND SECONDARY SHEAR STRESSESby combining the equation # 9 and equation # 6 will get combined shear acting on the weld=(')+(''))Eqn#10B=1.46E+01M Pa.4. BENDING STRESSES IN WELDSa. IN TRANSVERSE CASE

M(T)=1.12E+09N mmSECTION MODULUS Zt((W x L) + L/3)TABLE 10.7 A TEXT BOOK OF MACHINE DESIGNWHERE t = THROAT THICKNESSBY R.S KHURMIt=0.7071 x h=21.213mmBENDING STRESS b(T)b(T)=M/ZEqn#11BFOR SECTION MODULUSL=3090mmW=320mmZ=88490029.5mmb(T)0.19M Pa.b. IN LONGITUDNAL CASE

M(E)=4.89E+07NmmZ=8.85E+07mmb(E)=M(E)/ZEqn#12B5.52E-01M Pa.c. TOTAL BENDING STRESSb=(b(E))+(b(T)))Eqn#13B5.84E-01M Pa.4. MAXIMUM NORMAL STRESS THEORY t(max.)0.5 x b +0.5((b)+4)Eqn#14BBENDING STRESS WILL BE TAKEN FROM EQUATION 13 AND SHEAR STRESS SHALL BE TAKEN FROM EQUATION 10 t(max.)=14.93M Pa.5. MAXIMUM SHEAR STRESS THEORY( VON MISES STRESS) (MAX.)=0.5((b)+4)Eqn#15B14.64M Pa.

7. ALLOWABLE STRENGTH CHECKFOLLOWING 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 CALCULATIONSHENCEFOR VON MISES STRESSESALLOWABLE STRESS 0.30Sut248.1M Pa.Sa ALLOWABLE IN BASE METAL=0.40SyALLOWABLE IN BASE METAL104M Pa.(260 Mpa yeild of SA-516 70 N)EQUATION 14 SHALL BE COMPARED ALLOWABLESDESIGN CHECKTRUEDESIGN IS SAFE

CLIENT: TECHNIP - NPCC CONSORTIUMVANDOR: DESCON ENGINEERING FZE. SHARJAH UNITED ARAB EMERITES.DESIGN CALCULATION FOR THE WELD STRENGTH CALCULATION FOR SADDLE BASEPLATE JOINT.EQUIPMENT NO: 566-D-2819A/BEQUIPMENT NAME: NITROGEN RECEVER

2.Torsion(Operating)TORSION IN FIXED SADDLE DUE TO WIND FORCE

TORSION IN FIXED SADDLE DUE TO WIND FORCE IS CALCULATED IN FOLLOWING MODES:1. OPERATING CONDITION2. STAGE-2 TRANSPORTATION (NPCC TO INSTALLATION SITE)1. OPERATING CONDITIONMethedology: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.

DATA:WIND PRESSURE P =0.0031Bar(g)310N/mEFFECTIVE LENGTH L=11276mmOuter Radius Shell R=3244mmsurface area A=2RL=2.30E+08mmAs wind will be Exerted on the half of the surface area therefore surface area will be divided by 2 called as Effective. Surface Area

A2.30E+02mCOMPLETE 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=1.15E+02mso the force, Fw exerted by wind shall be equal to wind pressure P x effective Surface areaF=P x AeffEqn#16A35624.3934988955NTHE 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 L1Eqn#16AaL1 SHALL BE THE DISTANCE FROM THE CENTER OF FIXED SADDLE TO THE CENTER OF SHADDED AREAL1=5555mm(See Fig 1 attached) 5.555mMt=197893505.886365N-mm197893.505886364N-mWELD STRENGTH CHECK IN TORSIONAL MOMENT

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

DATA:WELD LONG LENGTH d=3090mmWIDTH IN WELD b=320mmWELD LEG LENGTH h=30mmHEIGHT OF THE SADDLEFROM TRUE CL H=2000mm

Unit Polar Moment of Inertia of weld Ju = (b+d)/66.61E+09mmEqn#17A

Polar Moment of Inertia of weld J = 0.7071 x h x Ju1.40E+11mmEqn#18A

CENTEROIDSx=b/2160mmy=D/21545mmr=x2+y21553.2626951034mmEqn#19ANow, the shear stress due to Torsion in welding(To)=Mr/JEqn#20A(To)=2.19N/mm2.19M 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 # 4A AND 7A(T)=5.85E-02M Pa.PRIMARY SHEAR STRESS DUE TO TRANSVERSE LOADS'(T)0.21M Pa.TOTAL SHEAR STRESS DUE TO TRANSVERSE LOADS(T)(Tot.)=((T)+'(T))Eqn#21A(T)(Tot.)=0.2193135493M Pa.Eqn#22ABENDING STRESS DUE TO TRANSVERSE LOADSb(T)=0.19M Pa.FROM EQUATION 11ACOMBINED SHEAR STRESS ACTING DUE TRANSVERSE LOAD AND TORSIONAL MOMENT=TOTAL SHEAR STRESS DUE TO TRANSVERSE AND TORSIONAL MOMENT==((TO)+'(T)(Tot.))Eqn#23A=2.20M Pa.1. MAXIMUM NORMAL STRESS THEORY t(max.)=0.5 x b +0.5((b)+4)SHEAR STRESS SHALL BE TAKEN FROM EQUATION 23 AND BENDING STRESS SHALL BE TAKEN FROM EQUATION 11 t(max.)=2.30M Pa.Eqn#24A2. MAXIMUM SHEAR STRESS THEORY( VON MISES STRESS) (MAX.)=0.5((b)+4)Eqn#25A (MAX.)=2.21M Pa.COMPARING EQN# 24 AND 25 WITH ALLOWABLES ALLOWABLE STRESS IN WELD S=0.30Sut248.1M Pa.Sa ALLOWABLE IN BASE METAL=0.40SySa104M Pa.(260 M Pa yeild of SA-516 70 N)DESIGN CHECK1. FOR WELDINGDESIGN IS SAFE

2. FOR BASE METALDESIGN IS SAFE

&8CLIENT: TECHNIP - NPCC CONSORTIUMVANDOR: DESCON ENGINEERING FZE. SHARJAH UNITED ARAB EMERITES.&8DESIGN CALCULATION FOR THE WELD STRENGTH CALCULATION FOR SADDLE BASEPLATE JOINT.&8EQUPMENT NUMBER:564-D-2160EQUIPMENT NAME:OPEN DRAIN DRUM

2.Torsion(OP+BLAST) TORSION IN FIXED SADDLE DUE TO WIND FORCE

TORSION IN FIXED SADDLE DUE TO WIND FORCE IS CALCULATED IN FOLLOWING MODES:1. OPERATING CONDITION2. STAGE-2 TRANSPORTATION (NPCC TO INSTALLATION SITE)1. OPERATING CONDITION+BLAST LOADMethedology: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.

DATA:WIND PRESSURE P =0.1335Bar(g)13350N/mEFFECTIVE LENGTH L=11276mmOuter Radius Shell R=3244mmsurface area A=2RL=229834796.767068mmAs wind will be Exerted on the half of the surface area therefore surface area will be divided by 2 called as Effective. Surface Area

A229.8347967671mCOMPLETE 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=114.9173983835mso the force, Fw exerted by wind shall be equal to wind pressure P x effective Surface areaF=P x AeffEqn#16B1534147.26842018NTHE 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 L1Eqn#16BaL1 SHALL BE THE DISTANCE FROM THE CENTER OF FIXED SADDLE TO THE CENTER OF SHADDED AREAL1=5555mm(See Fig 1 attached) 5.555mMt=8522188076.07409N-mm8.52E+06N-mWELD STRENGTH CHECK IN TORSIONAL MOMENT



Unit Polar Moment of Inertia of weld Ju = (b+d)/66.61E+09mmEqn#17B

Polar Moment of Inertia of weld J = 0.7071 x h x Ju1.40E+11mmEqn#18B

CENTEROIDSx=b/2160mmy=D/21545mmr=x2+y21553.2626951034mmEqn#19BNow, the shear stress due to Torsion in welding(To)=Mr/JEqn#20B(To)=9.44E+01N/mm94.42M 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 # 4B AND 7B(T)=3.8794230236M Pa.PRIMARY SHEAR STRESS DUE TO TRANSVERSE LOADS'(T)14.02M Pa.TOTAL SHEAR STRESS DUE TO TRANSVERSE LOADS(T)(Tot.)=((T)+'(T))Eqn#21B(T)(Tot.)=14.55M Pa.Eqn#22BBENDING STRESS DUE TO TRANSVERSE LOADSb(T)=0.19M Pa.FROM EQUATION 11BCOMBINED SHEAR STRESS ACTING DUE TRANSVERSE LOAD AND TORSIONAL MOMENT=TOTAL SHEAR STRESS DUE TO TRANSVERSE AND TORSIONAL MOMENT==((TO)+'(T)(Tot.))Eqn#23B=95.54M Pa.1. MAXIMUM NORMAL STRESS THEORY t(max.)=0.5 x b +0.5((b)+4)SHEAR STRESS SHALL BE TAKEN FROM EQUATION 23 AND BENDING STRESS SHALL BE TAKEN FROM EQUATION 11 t(max.)=95.63M Pa.Eqn#24B2. MAXIMUM SHEAR STRESS THEORY( VON MISES STRESS) (MAX.)=0.5((b)+4)Eqn#25B (MAX.)=95.54M Pa.COMPARING EQN# 24B AND 25B WITH ALLOWABLES ALLOWABLE STRESS IN WELD S=0.30Sut248.1M Pa.(260 M Pa yeild of SA-516 70 N)Sa ALLOWABLE IN BASE METAL=0.40SySa104M Pa.DESIGN CHECK1. FOR WELDINGDESIGN IS SAFE



3.Transportation CaseCALCULATION 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 IN PV-ELITEcompress.Input Data:Nomenclaturebase plate lengthd=3030mmbase plate widthb=260mmweld leg lengthh=30mmhight of saddle fromtrue center lineH=2000mm

length of weldL=3090mmwidth of weldW=320mmArea of weldAw=1.414 x h x (L+W)Aw=144652.2mmSee Example 10.11 R.S Khurmi (14th Ed.)EQN#26Unit 2nd Moment of areaIu=(d/6)(3b+d)Table 9-2 EQN#275829871500mmShigley 9th Edition

2nd Moment of areaBASED ON WELDI=0.7071 x h x IuI=123669064129.5mmtopc: 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)292KN292000Nthese values has been taken from the calculations done for Stage-2 Transportation in PV Elite basedon G Values provided by client

F (T)=29775.532Kg (f)=292000.00741972NFor End Wind Condition Max. Longitudnal loads shall be taken from the calculations as attached Max. Long. Load F(e)=91KN91000Nthese values has been taken from the calculations done for Stage-2 Transportation in PV Elite basedon G Values provided by clientEnd Wind Shear on Saddle F(e)= Pw*G*(Cf(shell)*p*Ro^2 + Cf(saddle)*(Proj. saddle area)) Eqn#2

F(e)=9279.361Kg (f)91000.00231231NFrictional Load Operating Weight on One Saddle, W18040KgCoefficiant of friction =0.12Frictional Force F(f)=*W=2164.8Kg(f)=21236.688N

This Frictional Load will be added into the End Wind Load ConditionTotal End Force F(E) =F(e)+F(f)EQN#2811444.161Kg(f)112267.21941N1. PRIMARY SHEAR STRESS IN WELD DUE TO LONGITUDNAL AND TRANSVERSE LOADSa. SHEAR STRESS IN WELD DUE TO TRANSVERSE LOADS

F(T)=292000.00741972NAw=144652.2mmWeld area(T)SHEAR STRESS IN TRANSVERSE CONDITION=F(T)/AwEQN#29(T)=2.0186M Pa.b. SHEAR STRESS IN WELD DUE TO END WIND LOADS

F(E)=112267.21941NTOTAL LONG. FORCEAw=144652.2mmWeld area(L)=F(E)/AwEQN#300.78M Pa.

c. COMBINED SHEAR STRESSES DUE TO TRANSVERSE AND LONGITUDNAL LOADS(COMBINING EQUATION 29 AND 30)''=COMBINED LOADS''=((T)+(L))EQN#31=2.16M Pa.2. SECONDARY SHEAR STRESS IN WELD DUE BENDING DUE TO LONGITUDNAL AND TRANSVERSE LOADSTHIS 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 STRESSa. IN TRANSVERSE CASE

SADDLE HEIGHT H=2000mmCENTEROID x=160mmCENTEROID y=1545mmTransverse Load F(T)=292000.00741972NMOMENT DUE TO TRANSVERSE LOADM(T)=F(T) x HEQN#32=5.84E+08Nmm'(T)=(M(T) x Y)/IWHERE I IS MOMENT OF INERTIA OF WELDI=123669064129.5mm


END LOAD F(E)=112267.21941NM(E)=224534438.82Nmm'(E)=(M(E) x Y)/IEQN#332.8051130686M Pa.c. COMBINE SECONDARY STRESS'=('(E)+'(T))EQN#34=7.82M Pa.3. COMBINED PRIMERY AND SECONDARY SHEAR STRESSESby combining the equation # 31 and equation # 34 will get combined shear acting on the weld=(')+(''))EQN#35=8.11M Pa.4. BENDING STRESSES IN WELDSa. IN TRANSVERSE CASE

M(T)=5.84E+08N mmSECTION MODULUS Zt((W x L) + L/3)TABLE 10.7 A TEXT BOOK OF MACHINE DESIGNWHERE t = THROAT THICKNESSBY R.S KHURMIEQN#36t=0.7071 x h=21.213mmBENDING STRESS b(T)b(T)=M/ZEQN#37FOR SECTION MODULUSL=3090mmW=320mmZ=88490029.5mmb(T)6.60M Pa.b. IN LONGITUDNAL CASE

M(E)=224534438.82NmmZ=88490029.5mmb(E)=M(E)/ZEQN#382.54M Pa.C. TOTAL BENDING STRESSb=(b(E))+(b(T)))EQN#397.1M Pa.4. MAXIMUM NORMAL STRESS THEORY t(max.)0.5 x b +0.5((b)+4)EQN#40BENDING STRESS WILL BE TAKEN FROM EQUATION 39 AND SHEAR STRESS SHALL BE TAKEN FROM EQUATION 35 t(max.)=12.3825973101M Pa.5. MAXIMUM SHEAR STRESS THEORY( VON MISES STRESS) (MAX.)=0.5((b)+4)EQN#418.8473006101M Pa.6. VESSEL UP LIFT CHECK VESSEL OPERATING WEIGHT=36233Kg355445.73NIF 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)112267.21941N

TOTAL LOADS THAT WILL CREATE AN UPWARD LIFT.F(UP0)=F(E)112267.21941NDESIGN CHECKNO LIFT WILL OCCURHence No uplift shall be considered only shear stress due to longitudnal load shall be taken into account.Vessel weight in empty Condition36233KG.355445.73NIN EMPTY CONDITION

UPLIFT FORCE IN EMPTY CASE,FUPEF(E)112267.21941N

DESIGN CHECKNO UPLIFT OCCURRED IN EMPTY CASE7. ALLOWABLE STRENGTH CHECKFOLLOWING 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 CALCULATIONSHENCEFOR VON MISES STRESSESALLOWABLE STRESS 0.30SutEQN#42248.1M Pa.Sa ALLOWABLE IN BASE METAL=0.40SyEQN#43ALLOWABLE IN BASE METAL104M Pa.EQUATION 14 SHALL BE COMPARED ALLOWABLESDESIGN CHECKTRUEDESIGN IS SAFE

&9CLIENT: TECHNIP - NPCC CONSORTIUMVANDOR: DESCON ENGINEERING FZE. SHARJAH UNITED ARAB EMERITES.&9DESIGN CALCULATION FOR THE WELD STRENGTH CALCULATION FOR SADDLE BASEPLATE JOINT.&9EQUIPMENT NO: 564-D-2160EQUIPMENT NAME: OPEN DRAIN DRUM

4.Torsion(TRANS)TORSION IN FIXED SADDLE DUE TO WIND FORCE

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).DATA:WIND PRESSURE P =0.0031Bar(g)310N/mEFFECTIVE LENGTH L=11276mmTOTAL LENGTH OF VESSEL FROM FIXED SADDLE CL.Outer Radius Shell R=3244mmsurface area A=2RL=229834796.767068mmAs wind will be Exerted on the half of the surface area therefore surface area will be divided by 2 called as Effective. Surface Area

A229.83mCOMPLETE SURFACE AREAMOMENT ARM "L" IS THE DISTANCE OF THE CENTER OF FIXED SADDLE TO THE OUTER OF OPPOSITE HEADEFFECTIVE SURFACE AREA SHALL BE Aeff=114.92mso the force, Fw exerted by wind shall be equal to wind pressure P x effective Surface areaF(WIND)=P x Aeff=35624.39NEqn#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.)292000.00741972NEqn#44(a)HENCE THE TOTAL FORCE WILL BE THE SUMM OF THESE TWO FORCES F(TOR.)=(F(WIND))+(F(TRANS)))Eqn#45F(TOR.)=2.94E+05NTHE MOMENT M ON FIXED SADDLE CREATED BY TOTAL FORCE SHALL BE EQUAL TO WIND FORCE INTO MOMENT ARMTORSIONAL 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=3486.56mmCOG IN TRANSPORTATION CONDITION(FROM DATUM)5336.56mm3.48656mDISTANCE OF FIXED SADDLE FROM DATUM1850mmMt=1.03E+09N-mmEFFECTIVE LENGTH L13486.56mm1.03E+06N-mWELD STRENGTH CHECK IN TORSIONAL MOMENT



Unit Polar Moment of Inertia of weld Ju = (b+d)/66.61E+09mmEqn#46

Polar Moment of Inertia of weld J = 0.7071 x h x Ju1.40E+11mmEqn#47

CENTEROIDSx=b/2160mmy=D/21545mmr=x2+y21553.3mmEqn#48Now, the shear stress due to Torsion in welding(To)=Mr/JEqn#49(To)=11.4N/mm11.4M 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(T)=2.02M Pa.PRIMARY SHEAR STRESS DUE TO TRANSVERSE LOADS'(T)7.30M Pa.TOTAL SHEAR STRESS DUE TO TRANSVERSE LOADS(T)(Tot.)=((T)+'(T))Eqn#50(T)(Tot.)=7.5700321258M Pa.Eqn#51BENDING STRESS DUE TO TRANSVERSE LOADSb(T)=6.60M Pa.FROM EQUATION#37 COMBINED SHEAR STRESS ACTING DUE TRANSVERSE LOAD AND TORSIONAL MOMENT=TOTAL SHEAR STRESS DUE TO TRANSVERSE AND TORSIONAL MOMENT==((TO)+'(T)(Tot.))Eqn#52=13.65M Pa.1. MAXIMUM NORMAL STRESS THEORY t(max.)=0.5 x b +0.5((b)+4)SHEAR STRESS SHALL BE TAKEN FROM EQUATION 23 AND BENDING STRESS SHALL BE TAKEN FROM EQUATION 11 t(max.)=17.35M Pa.Eqn#532. MAXIMUM SHEAR STRESS THEORY( VON MISES STRESS) (MAX.)=0.5((b)+4)Eqn#54 (MAX.)=14.0473294179M Pa.COMPARING EQN# 24 AND 25 WITH ALLOWABLES ALLOWABLE STRESS IN WELD S=0.30Sut248.1M Pa.Sa ALLOWABLE IN BASE METAL=0.40SySa104M Pa.DESIGN CHECK1. FOR WELDINGDESIGN IS SAFE


REFERENCE PAGES ARE ATTACHED BELOW


5.TORSION IN SADDLE (OP)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)

(SEE FIG. 3 FOR REFERENCE)SADDLE DATASADDLE PLATE WIDTH,b2=230mmWEB PLATE LENGTH, h1=2847mmWEBPLATE THICKNESS, b1=30mmSADDLE PLATE THICKNESS h2=30mmRIB THICKNESS h3=30mmRIB WIDTH b3=100mmCALCULATION REFERENCE :MECHANICS OF MATERIALS BY ANDREW PYTELAND JAAN KIUSLAAS 2ND EDITION.AREAS:VALUES FROM FIG 3.AREA OF WEB, A1=85410mmDISTANCE FROM X-AXIS TO THE BOTTOM OF AREA A2=W1=(FROM FIG 1)2877mmAREA OF TOP SADDLE PLATE A2=6900mmAREA OF BOT. SADDLE PLATE A3=6900mmAREA OF RIBS A A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM OF RIB A AND RIB C W2= (FROM FIG 1)2616mmAREA OF RIBS B A4=3000mmAREA OF RIBS C A4=3000mmAREA OF RIBS D A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM OF RIB B AND RIB D W3= (FROM FIG 1)2354mmAREA OF RIBS E A4=3000mmAREA OF RIBS F A4=3000mmAREA OF RIBS G A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB E AND RIB G W4= (FROM FIG 1)2093mmAREA OF RIBS H A4=3000mmAREA OF RIBS I A4=3000mmAREA OF RIBS J A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB FAND H W5= (FROM FIG 1)1831AREA OF RIBS K A4=3000mmAREA OF RIBS L A4=3000mmAREA OF RIBS M A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB IAND J W6= (FROM FIG 1)1570mmAREA OF RIBS N A4=3000mmAREA OF RIBS O A4=3000mmAREA OF RIBS P A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB KAND L W7= (FROM FIG 1)1308mmAREA OF RIBS Q A4=3000mmAREA OF RIBS R A4=3000mmAREA OF RIBS S A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB MAND N W8= (FROM FIG 1)1046mmAREA OF RIBS T A4=3000mmTOTAL AREA OF RIBS=A4'=60000mmCENTEROID LOCATION FROM X AXISDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB OAND P W9= (FROM FIG 1)785mmCENTEROID OF WEB PLATE Y1= h1/21423.5mmCENTEROID TOP SADDLE PLATE (Y2=W1+h2/2)2892mmCENTEROID BOT. SADDLE PLATE (Y3=h2/215mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB Q AND R W10= (FROM FIG 1)523mmCENTEROID OF RIBS A = Y4 = (h3/2)+W22631mmCENTEROID OF RIBS C = Y5= (h3/2)+W22631mmCENTEROID OF RIBS B=Y6= (h3/2)+W32369mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB S AND T W11= (FROM FIG 1)262mmCENTEROID OF RIBS D= Y7= (h3/2)+W32369mmCENTEROID OF RIBS E= Y8= (h3/2)+W42108mmCENTEROID OF RIBS G= Y9= (h3/2)+W42108mmDISTANCE FROM Y-AXIS TO THE FAR EDGE OF RIBS C,D,G,H,J,L,N,P,R AND T W12= (FROM FIG 1)230mmCENTEROID OF RIBS F= Y10= (h3/2)+W51846mmCENTEROID OF RIBS H= Y11= (h3/2)+W51846mmCENTEROID OF RIBS I= Y12= (h3/2)+W61585mmDISTANCE FROM Y-AXIS TO THE OUTER EDGE OF WEBW13= (FROM FIG 1)100mmCENTEROID OF RIBS J= Y13= (h3/2)+W61585mmCENTEROID OF RIBS K= Y14= (h3/2)+W71323mmCENTEROID OF RIBS L= Y15= (h3/2)+W71323mmCENTEROID OF RIBS M= Y16= (h3/2)+W81061mmCENTEROID OF RIBS N= Y17= (h3/2)+W81061mmCENTEROID OF RIBS O= Y18= (h3/2)+W9800mmCENTEROID OF RIBS P= Y19= (h3/2)+W9800mmCENTEROID OF RIBS Q= Y20= (h3/2)+W10538mmCENTEROID OF RIBS R= Y21= (h3/2)+W10538mmCENTEROID OF RIBS S= Y22= (h3/2)+W11277mmCENTEROID OF RIBS T= Y23= (h3/2)+W11277mm

CENTEROID LOCATION FROM Y AXIS

CENTEROID OF WEB PLATE X1=(b1/2)+W13115.00mmCENTEROID TOP SADDLE PLATE (X2= b2/2)115.00mmCENTEROID BOT. SADDLE PLATE (X3=b2/2115mmCENTEROID OF RIB A = X4 =(b3/2)50mmCENTEROID OF RIB B = X5=(b3/2)50mmCENTEROID OF RIB C=X6= W12-(b3/2)180mmCENTEROID OF RIB D= X7= W12-(b3/2)180mmCENTEROID OF RIB E=X8=(b3/2)50mmCENTEROID OF RIB F=X9=(b3/2)50mmCENTEROID OF RIB G=X10=W12-(b3/2)180mmCENTEROID OF RIB H=X11=W12-(b3/2)180mmCENTEROID OF RIB I=X12=(b3/2)50mmCENTEROID OF RIB J=X13=W12-(b3/2)180mmCENTEROID OF RIB K=X14=(b3/2)50mmCENTEROID OF RIB L=X15=W12-(b3/2)180mmCENTEROID OF RIB M=X16=(b3/2)50mmCENTEROID OF RIB N=X17=W12-(b3/2)180mmCENTEROID OF RIB O=X18=(b3/2)50mmCENTEROID OF RIB P=X19=W12-(b3/2)180mmCENTEROID OF RIB Q=X20=(b3/2)50mmCENTEROID OF RIB R=X21=W12-(b3/2)180mmCENTEROID OF RIB S=X22=(b3/2)50mmCENTEROID OF RIB T=X23=W12-(b3/2)180mmABOVE 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 AXIS=(Ai x yi)/Aii=1,2,3.Eqn#55X=CENTEROID LOCATION THROUGH Y AXIS=(Ai x xi)/Aii=1,2,3,4.Eqn#56Y=1437.52mmCENTEROID FROM X-AXIS X=83.52mmCENTEROID FROM Y-AXIS MOMENT OF INERTIA ALONG XX AXIS USING PARALLEL AXIS THEOREM (neutral axis)AS Ixx = [(Ixx)i+Ai(Y-Yi)] WHERE i=1,2,3,4WHERE Ixx=bh/12

Eqn#57Eqn#58(Ixx)1=M.O.I OF WEB AREA A15.77E+10mm(Ixx)2=M.O.I OF TOP SADDLE PLATE AREA A25.18E+05mm(Ixx)3=M.O.I OF BOT. SADDLE PLATE AREA A35.18E+05mm(Ixx)4=MO.I OF RIB A2.25E+05mm(Ixx)5=MO.I OF RIB B2.25E+05mm(Ixx)6=MO.I OF RIB C2.25E+05mm(Ixx)7=MO.I OF RIB D2.25E+05mm(Ixx)8=MO.I OF RIB E2.25E+05mm(Ixx)9=MO.I OF RIB F2.25E+05mm(Ixx)10=MO.I OF RIB G2.25E+05mm(Ixx)11=MO.I OF RIB H2.25E+05mm(Ixx)12=MO.I OF RIB I2.25E+05mm(Ixx)13=MO.I OF RIB J2.25E+05mm(Ixx)14=MO.I OF RIB K2.25E+05mm(Ixx)15=MO.I OF RIB L2.25E+05mm(Ixx)16=MO.I OF RIB M2.25E+05mm(Ixx)17=MO.I OF RIB N2.25E+05mm(Ixx)18=MO.I OF RIB O2.25E+05mm(Ixx)19=MO.I OF RIB P2.25E+05mm(Ixx)20=MO.I OF RIB Q2.25E+05mm(Ixx)21=MO.I OF RIB R2.25E+05mm(Ixx)22=MO.I OF RIB S2.25E+05mm(Ixx)23=MO.I OF RIB T2.25E+05mmAS ALL RIBS AND TOP AND BOTTOM SADDLE PLATES HAVE SAME DIMENSIONS WILL HAVE THE SAME MOMENT OF INERTIA

A1(Y-Y1)=1.68E+07mmWEB AREA A1A2(Y-Y2)=1.46E+10mmTOP SADDLE PLATE AREA A2A3(Y-Y3)=1.40E+10mm BOT. SADDLE PLATE AREA A3A4(Y-Y4)=4.27E+09mmRIB AA4(Y-Y5)=4.27E+09mmRIB CA4(Y-Y6)=2.60E+09mmRIB BA4(Y-Y7)=2.60E+09mmRIB DA4(Y-Y8)=1.35E+09mmRIB EA4(Y-Y9)=1.35E+09mmRIB GA4(Y-Y10)=5.01E+08mmRIB FA4(Y-Y11)=5.01E+08mmRIB HA4(Y-Y12)=6.53E+07mmRIB IA4(Y-Y13)=6.53E+07mmRIB JA4(Y-Y14)=3.93E+07mmRIB KA4(Y-Y15)=3.93E+07mmRIB LA4(Y-Y16)=4.25E+08mmRIB MA4(Y-Y17)=4.25E+08mmRIB NA4(Y-Y18)=1.22E+09mmRIB OA4(Y-Y19)=1.22E+09mmRIB PA4(Y-Y20)=2.43E+09mmRIB QA4(Y-Y21)=2.43E+09mmRIB RA4(Y-Y22)=4.04E+09mmRIB SA4(Y-Y23)=4.04E+09mmRIB TAS THE TERM Ai(Y-Yi) 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)]+[(Ixx)2+A2(Y-Y2)]+[(Ixx)3+A3(Y-Y3)]+2 x [(Ixx)4+ A4(Y-Y4)]+2 x [(Ixx)5+ A5(Y-Y5)]

Ixx =1.20157E+11mmEqn#57MOMENT OF INERTIA ALONG XX AXIS (neutral axis) NOT USING PARRALEL AXIS THEOREMIxx =5.77E+10mm Ixx = [(Ixx)i]Eqn#59

MOMENT OF INERTIA ALONG YY AXIS USING PARALLEL AXIS THEOREM (neutral axis)AS Iyy = [(Iyy)i+Ai(X-Xi)] WHERE i=1,2,3,4WHERE Iyy=bh/12

Eqn#60Eqn#61(Iyy)1=M.O.I OF WEB AREA A16.41E+06mm(Iyy)2=M.O.I OF TOP SADDLE PLATE AREA A23.04E+07mm(Iyy)3=M.O.I OF BOT. SADDLE PLATE AREA A33.04E+07mm(Iyy)4=MO.I OF RIB A2.50E+06mm(Iyy)5=MO.I OF RIB B2.50E+06mm(Iyy)6=MO.I OF RIB C2.50E+06mm(Iyy)7=MO.I OF RIB D2.50E+06mm(Iyy)8=MO.I OF RIB E2.50E+06mm(Iyy)9=MO.I OF RIB F2.50E+06mm(Iyy)10=MO.I OF RIB G2.50E+06mm(Iyy)11=MO.I OF RIB H2.50E+06mm(Iyy)12=MO.I OF RIB I2.50E+06mm(Iyy)13=MO.I OF RIB J2.50E+06mm(Iyy)14=MO.I OF RIB K2.50E+06mm(Iyy)15=MO.I OF RIB L2.50E+06mm(Iyy)16=MO.I OF RIB M2.50E+06mm(Iyy)17=MO.I OF RIB N2.50E+06mm(Iyy)18=MO.I OF RIB O2.50E+06mm(Iyy)19=MO.I OF RIB P2.50E+06mm(Iyy)20=MO.I OF RIB Q2.50E+06mm(Iyy)21=MO.I OF RIB R2.50E+06mm(Iyy)22=MO.I OF RIB S2.50E+06mm(Iyy)23=MO.I OF RIB T2.50E+06mmAS ALL RIBS AND TOP AND BOTTOM SADDLE PLATES HAVE SAME DIMENSIONS WILL HAVE THE SAME MOMENT OF INERTIAA1(X-X1)=8.46E+07mmA2(X-X2)=6.84E+06mmA3(X-X3)=6.84E+06mmA4(X-X4)=3.37E+06mmA4(X-X5)=3.37E+06mmA4(X-X6)=2.79E+07mmA4(X-X7)=2.79E+07mmA4(X-X8)=3.37E+06mmA4(X-X9)=3.37E+06mmA4(X-X10)=2.79E+07mmA4(X-X11)=2.79E+07mmA4(X-X12)=3.37E+06mmA4(X-X13)=2.79E+07mmA4(X-X14)=3.37E+06mmA4(X-X15)=2.79E+07mmA4(X-X16)=3.37E+06mmA4(X-X17)=2.79E+07mmA4(X-X18)=3.37E+06mmA4(X-X19)=2.79E+07mmA4(X-X20)=3.37E+06mmA4(X-X21)=2.79E+07mmA4(X-X22)=3.37E+06mmA4(X-X23)=2.79E+07mm

AS Iyy = [(Iyy)1+A1(X-X1)]+ [(Iyy)2+A2(X-X2)]+ [(Iyy)3+A3(X-X3)]+ [(Iyy)4+A4(X-X4)]+ [(Iyy)5+A5(X-X5)] Iyy =5.28E+08mmEqn#60MOMENT OF INERTIA ALONG YY AXIS (neutral axis) NOT USING PARRALEL AXIS THEOREM Iyy =1.17E+08mmEqn#61 Iyy = [(Iyy)i]POLAR MOMENT OF INERTIA JoJo=Ixx+IyyEqn#62POLAR MOMENT OF INERTIA USING INERTIAS FROM PARALLEL AXIS THEOREMFROM EQUATION 60 AND 57Jo=1.21E+11mmEqn#62aPOLAR MOMENT OF INERTIA USING INERTIAS FROM SUMMATION OF LOCAL MOMENT OF INERTIASJo=Ixx+IyyEqn#62(b)FROM EQUATION 59 AND 61Jo=5.78E+10mmWE 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 MATERIALMOMENT SHALL BE TAKEN FROM THE EQUATION 16 (OPERATING+BLAST LOAD) CONDITION CONSIDEREDMt=TORSIONAL MOMENT8.52E+09N-mmAS(MAX.)=MAX. SHEAR STRESS IN TORSION(MAX.)=(MT x r)/JoN/mm2WHERE r IS THE CENTEROID ON WHICH THE MOMENT IS ACTING IN THIS CASE Y WILL BE THE CENTEROID THEREFORE(MAX.)=105.95N/mm2Eqn#62(c)ALLOWABLE CHECK

SADDLE ALLOWABLE STRESSSa=138M Pa.POLAR MOMENT OF INERTIA OF SADDLE SECTION IS SUFFICIENT TO SUSTAIN TORSION

&8CLIENT: TECHNIP - NPCC CONSORTIUMVANDOR: DESCON ENGINEERING FZE. SHARJAH UNITED ARAB EMERITES.&8DESIGN CALCULATION FOR THE TORSION CALCULATION FOR SADDLE SECTION IN OPERATING&8EQUIPMENT NO:564-D-2160EQUIPMENT NAME: OPEN DRAIN DRUM

COMB. AT SADDLE SEC.(OPERAT (2COMBINATION OF THE SHEAR STRESSES, BENDING STRESS, (OPERATING CASE)THIS SHEET WILL DEAL WITH THE COMBINATION OF SHEAR STRESSES DUE TO TORSIONAND BENDING.(OPERATING CASE)

SHEAR STRESS DUE TO TORSION (T)=105.95M Pa(FROM EQ 62B)VALUE 1TOTAL PRIMARY SHEAR STRESS IN TRANS/LONGCASE=4.16E+00M Pa(FROM EQ 6B)

TOTAL SECONDARY SHEAR STRESS IN TRANS./LONG. CASE=1.40E+01M PaFROM Eqn#9BVALUE 2COMBINED SHEAR STRESS IN TRANS./LONG. CASE T(combined)=14.64M Pa

AS TO COMBINE THE EFFECT OF BENDING STRESSES AND THE SHEAR STRESSES ALL THE SHEAR STRESSES OBTAINED FROM TRANSVERSE, LONGITUDNAL AND TORSIONAL CASES WILL BE COMBINED WITH THE BENDING MOMENT COMING ON THE SADDLE DUE TOWIND, (OPERARTING)

BENDING MOMENT AT THE FIXED SADDLEM(b)=5.88E+07N mmFROM COMPRESSCALCULATIONS.

BENDING STRESS(b)=M(B) x Y/ITHE VALUE OF BENDING MOMENT ON SADDLE IS OBTAINED FROM COMPRESS CALCSWHERE Y= CENTEROID AND IS CALCULATED IN PREVIOUS CALCULATION OF SADDLE SECTIONI IS THE TOTAL MOMENT OF INERTIA CALCULATED IN PREVIOUS

(b)=41.92M PaNOW COMBINING THE VALUE 1 AND VALUE 2 TO GET THE MAXIMUM SHEAR STRESSES(MAX)=106.9584993644M Pa.NOW USING MAX. SHEAR STRESS THEORY AND MAXIMUM NORMAL STRESS THEORY AND MAX. NORMAL STRESS THRORY TO COMBINE RESULTANT SHEAR AND BENDING LOADS1. MAX. NORMAL STRESS THEORY t(max.)=0.5 x b +0.5((b)+4)129.9515689468M Pa.2. MAXIMUM SHEAR STRESS THEORY (MAX.)=0.5((b)+4) (MAX.)=108.9926465975M Pa.VESSEL DESIGN CHECKALLOWABLE STRESS, S=138M Pa.DESIGN IS SAFE

&8CLIENT: TECHNIP - NPCC CONSORTIUMVANDOR: DESCON ENGINEERING FZE. SHARJAH UNITED ARAB EMERITES.&8DESIGN CALCULATION FOR THE COMBINATION OF STRESSES FOR SADDLE SECTION IN OPERATING&8EQUIPMENT NO:566-V-2819 A/BEQUIPMENT NAME:NIRTOGEN RECEIVER

COMB. AT SADDLE SEC.(transport)COMBINATION OF THE SHEAR STRESSES, BENDING STRESS, (OPERATING CASE)THIS SHEET WILL DEAL WITH THE COMBINATION OF SHEAR STRESSES DUE TO TORSIONAND BENDING.(TRANSPORTATION)

SHEAR STRESS DUE TO TORSION (T)=11.36M Pa(FROM EQ 62B)VALUE 1TOTAL PRIMARY SHEAR STRESS IN TRANS/LONGCASE=2.16E+00M Pa(FROM EQ 6B)

TOTAL SECONDARY SHEAR STRESS IN TRANS./LONG. CASE=7.82E+00M PaFROM Eqn#9BVALUE 2COMBINED SHEAR STRESS IN TRANS./LONG. CASE T(combined)=8.11M Pa

AS TO COMBINE THE EFFECT OF BENDING STRESSES AND THE SHEAR STRESSES ALL THE SHEAR STRESSES OBTAINED FROM TRANSVERSE, LONGITUDNAL AND TORSIONAL CASES WILL BE COMBINED WITH THE BENDING MOMENT COMING ON THE SADDLE DUE TOWIND, (OPERARTING)

BENDING MOMENT AT THE FIXED SADDLEM(b)=4.63E+07N mmFROM COMPRESSCALCULATIONS.

BENDING STRESS(b)=M(B) x Y/ITHE VALUE OF BENDING MOMENT ON SADDLE IS OBTAINED FROM COMPRESS CALCSWHERE Y= CENTEROID AND IS CALCULATED IN PREVIOUS CALCULATION OF SADDLE SECTIONI IS THE TOTAL MOMENT OF INERTIA CALCULATED IN PREVIOUS

(b)=32.96M PaNOW COMBINING THE VALUE 1 AND VALUE 2 TO GET THE MAXIMUM SHEAR STRESSES(MAX)=13.9610084638M Pa.NOW USING MAX. SHEAR STRESS THEORY AND MAXIMUM NORMAL STRESS THEORY AND MAX. NORMAL STRESS THRORY TO COMBINE RESULTANT SHEAR AND BENDING LOADS1. MAX. NORMAL STRESS THEORY t(max.)=0.5 x b +0.5((b)+4)38.0767460182M Pa.2. MAXIMUM SHEAR STRESS THEORY (MAX.)=0.5((b)+4) (MAX.)=21.5978059663M Pa.VESSEL DESIGN CHECKALLOWABLE STRESS, S=138M Pa.DESIGN IS SAFE

&8CLIENT: TECHNIP - NPCC CONSORTIUMVANDOR: DESCON ENGINEERING FZE. SHARJAH UNITED ARAB EMERITES.&8DESIGN CALCULATION FOR THE COMBINATION OF STRESSES FOR SADDLE SECTION IN OPERATING&8EQUIPMENT NO:566-V-2819 A/BEQUIPMENT NAME:NIRTOGEN RECEIVER

5.TORSION IN SADDLE (TRANS)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)

(SEE FIG. 3 FOR REFERENCE)SADDLE DATASADDLE PLATE WIDTH,b2=230mmWEB PLATE LENGTH, h1=2847mmWEBPLATE THICKNESS, b1=30mmSADDLE PLATE THICKNESS h2=30mmRIB THICKNESS h3=30mmRIB WIDTH b3=100mmCALCULATION REFERENCE :MECHANICS OF MATERIALS BY ANDREW PYTELAND JAAN KIUSLAAS 2ND EDITION.AREAS:VALUES FROM FIG 3.AREA OF WEB, A1=85410mmDISTANCE FROM X-AXIS TO THE BOTTOM OF AREA A2=W1=(FROM FIG 1)2877mmAREA OF TOP SADDLE PLATE A2=6900mmAREA OF BOT. SADDLE PLATE A3=6900mmAREA OF RIBS A A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM OF RIB A AND RIB C W2= (FROM FIG 1)2616mmAREA OF RIBS B A4=3000mmAREA OF RIBS C A4=3000mmAREA OF RIBS D A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM OF RIB B AND RIB D W3= (FROM FIG 1)2354mmAREA OF RIBS E A4=3000mmAREA OF RIBS F A4=3000mmAREA OF RIBS G A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB E AND RIB G W4= (FROM FIG 1)2093mmAREA OF RIBS H A4=3000mmAREA OF RIBS I A4=3000mmAREA OF RIBS J A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB FAND H W5= (FROM FIG 1)1831AREA OF RIBS K A4=3000mmAREA OF RIBS L A4=3000mmAREA OF RIBS M A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB IAND J W6= (FROM FIG 1)1570mmAREA OF RIBS N A4=3000mmAREA OF RIBS O A4=3000mmAREA OF RIBS P A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB KAND L W7= (FROM FIG 1)1308mmAREA OF RIBS Q A4=3000mmAREA OF RIBS R A4=3000mmAREA OF RIBS S A4=3000mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB MAND N W8= (FROM FIG 1)1046mmAREA OF RIBS T A4=3000mmTOTAL AREA OF RIBS=A4'=60000mmCENTEROID LOCATION FROM X AXISDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB OAND P W9= (FROM FIG 1)785mmCENTEROID OF WEB PLATE Y1= h1/21423.5mmCENTEROID TOP SADDLE PLATE (Y2=W1+h2/2)2892mmCENTEROID BOT. SADDLE PLATE (Y3=h2/215mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB Q AND R W10= (FROM FIG 1)523mmCENTEROID OF RIBS A = Y4 = (h3/2)+W22631mmCENTEROID OF RIBS C = Y5= (h3/2)+W22631mmCENTEROID OF RIBS B=Y6= (h3/2)+W32369mmDISTANCE FROM X-AXIS TO THE BOTTOM EDGE OF RIB S AND T W11= (FROM FIG 1)262mmCENTEROID OF RIBS D= Y7= (h3/2)+W32369mmCENTEROID OF RIBS E= Y8= (h3/2)+W42108mmCENTEROID OF RIBS G= Y9= (h3/2)+W42108mmDISTANCE FROM Y-AXIS TO THE FAR EDGE OF RIBS C,D,G,H,J,L,N,P,R AND T W12= (FROM FIG 1)230mmCENTEROID OF RIBS F= Y10= (h3/2)+W51846mmCENTEROID OF RIBS H= Y11= (h3/2)+W51846mmCENTEROID OF RIBS I= Y12= (h3/2)+W61585mmDISTANCE FROM Y-AXIS TO THE OUTER EDGE OF WEBW13= (FROM FIG 1)100mmCENTEROID OF RIBS J= Y13= (h3/2)+W61585mmCENTEROID OF RIBS K= Y14= (h3/2)+W71323mmCENTEROID OF RIBS L= Y15= (h3/2)+W71323mmCENTEROID OF RIBS M= Y16= (h3/2)+W81061mmCENTEROID OF RIBS N= Y17= (h3/2)+W81061mmCENTEROID OF RIBS O= Y18= (h3/2)+W9800mmCENTEROID OF RIBS P= Y19= (h3/2)+W9800mmCENTEROID OF RIBS Q= Y20= (h3/2)+W10538mmCENTEROID OF RIBS R= Y21= (h3/2)+W10538mmCENTEROID OF RIBS S= Y22= (h3/2)+W11277mmCENTEROID OF RIBS T= Y23= (h3/2)+W11277mm

CENTEROID LOCATION FROM Y AXIS

CENTEROID OF WEB PLATE X1=(b1/2)+W13115.00mmCENTEROID TOP SADDLE PLATE (X2= b2/2)115.00mmCENTEROID BOT. SADDLE PLATE (X3=b2/2115mmCENTEROID OF RIB A = X4 =(b3/2)50mmCENTEROID OF RIB B = X5=(b3/2)50mmCENTEROID OF RIB C=X6= W12-(b3/2)180mmCENTEROID OF RIB D= X7= W12-(b3/2)180mmCENTEROID OF RIB E=X8=(b3/2)50mmCENTEROID OF RIB F=X9=(b3/2)50mmCENTEROID OF RIB G=X10=W12-(b3/2)180mmCENTEROID OF RIB H=X11=W12-(b3/2)180mmCENTEROID OF RIB I=X12=(b3/2)50mmCENTEROID OF RIB J=X13=W12-(b3/2)180mmCENTEROID OF RIB K=X14=(b3/2)50mmCENTEROID OF RIB L=X15=W12-(b3/2)180mmCENTEROID OF RIB M=X16=(b3/2)50mmCENTEROID OF RIB N=X17=W12-(b3/2)180mmCENTEROID OF RIB O=X18=(b3/2)50mmCENTEROID OF RIB P=X19=W12-(b3/2)180mmCENTEROID OF RIB Q=X20=(b3/2)50mmCENTEROID OF RIB R=X21=W12-(b3/2)180mmCENTEROID OF RIB S=X22=(b3/2)50mmCENTEROID OF RIB T=X23=W12-(b3/2)180mmABOVE 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 AXIS=(Ai x yi)/Aii=1,2,3.Eqn#55X=CENTEROID LOCATION THROUGH Y AXIS=(Ai x xi)/Aii=1,2,3,4.Eqn#56Y=1437.52mmCENTEROID FROM X-AXIS X=83.52mmCENTEROID FROM Y-AXIS MOMENT OF INERTIA ALONG XX AXIS USING PARALLEL AXIS THEOREM (neutral axis)AS Ixx = [(Ixx)i+Ai(Y-Yi)] WHERE i=1,2,3,4WHERE Ixx=bh/12

Eqn#57Eqn#58(Ixx)1=M.O.I OF WEB AREA A15.77E+10mm(Ixx)2=M.O.I OF TOP SADDLE PLATE AREA A25.18E+05mm(Ixx)3=M.O.I OF BOT. SADDLE PLATE AREA A35.18E+05mm(Ixx)4=MO.I OF RIB A2.25E+05mm(Ixx)5=MO.I OF RIB B2.25E+05mm(Ixx)6=MO.I OF RIB C2.25E+05mm(Ixx)7=MO.I OF RIB D2.25E+05mm(Ixx)8=MO.I OF RIB E2.25E+05mm(Ixx)9=MO.I OF RIB F2.25E+05mm(Ixx)10=MO.I OF RIB G2.25E+05mm(Ixx)11=MO.I OF RIB H2.25E+05mm(Ixx)12=MO.I OF RIB I2.25E+05mm(Ixx)13=MO.I OF RIB J2.25E+05mm(Ixx)14=MO.I OF RIB K2.25E+05mm(Ixx)15=MO.I OF RIB L2.25E+05mm(Ixx)16=MO.I OF RIB M2.25E+05mm(Ixx)17=MO.I OF RIB N2.25E+05mm(Ixx)18=MO.I OF RIB O2.25E+05mm(Ixx)19=MO.I OF RIB P2.25E+05mm(Ixx)20=MO.I OF RIB Q2.25E+05mm(Ixx)21=MO.I OF RIB R2.25E+05mm(Ixx)22=MO.I OF RIB S2.25E+05mm(Ixx)23=MO.I OF RIB T2.25E+05mmAS ALL RIBS AND TOP AND BOTTOM SADDLE PLATES HAVE SAME DIMENSIONS WILL HAVE THE SAME MOMENT OF INERTIA

A1(Y-Y1)=1.68E+07mmWEB AREA A1A2(Y-Y2)=1.46E+10mmTOP SADDLE PLATE AREA A2A3(Y-Y3)=1.40E+10mm BOT. SADDLE PLATE AREA A3A4(Y-Y4)=4.27E+09mmRIB AA4(Y-Y5)=4.27E+09mmRIB CA4(Y-Y6)=2.60E+09mmRIB BA4(Y-Y7)=2.60E+09mmRIB DA4(Y-Y8)=1.35E+09mmRIB EA4(Y-Y9)=1.35E+09mmRIB GA4(Y-Y10)=5.01E+08mmRIB FA4(Y-Y11)=5.01E+08mmRIB HA4(Y-Y12)=6.53E+07mmRIB IA4(Y-Y13)=6.53E+07mmRIB JA4(Y-Y14)=3.93E+07mmRIB KA4(Y-Y15)=3.93E+07mmRIB LA4(Y-Y16)=4.25E+08mmRIB MA4(Y-Y17)=4.25E+08mmRIB NA4(Y-Y18)=1.22E+09mmRIB OA4(Y-Y19)=1.22E+09mmRIB PA4(Y-Y20)=2.43E+09mmRIB QA4(Y-Y21)=2.43E+09mmRIB RA4(Y-Y22)=4.04E+09mmRIB SA4(Y-Y23)=4.04E+09mmRIB TAS THE TERM Ai(Y-Yi) 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)]+[(Ixx)2+A2(Y-Y2)]+[(Ixx)3+A3(Y-Y3)]+2 x [(Ixx)4+ A4(Y-Y4)]+2 x [(Ixx)5+ A5(Y-Y5)]

Ixx =1.20157E+11mmEqn#57MOMENT OF INERTIA ALONG XX AXIS (neutral axis) NOT USING PARRALEL AXIS THEOREMIxx =5.77E+10mm Ixx = [(Ixx)i]Eqn#59

MOMENT OF INERTIA ALONG YY AXIS USING PARALLEL AXIS THEOREM (neutral axis)AS Iyy = [(Iyy)i+Ai(X-Xi)] WHERE i=1,2,3,4WHERE Iyy=bh/12

Eqn#60Eqn#61(Iyy)1=M.O.I OF WEB AREA A16.41E+06mm(Iyy)2=M.O.I OF TOP SADDLE PLATE AREA A23.04E+07mm(Iyy)3=M.O.I OF BOT. SADDLE PLATE AREA A33.04E+07mm(Iyy)4=MO.I OF RIB A2.50E+06mm(Iyy)5=MO.I OF RIB B2.50E+06mm(Iyy)6=MO.I OF RIB C2.50E+06mm(Iyy)7=MO.I OF RIB D2.50E+06mm(Iyy)8=MO.I OF RIB E2.50E+06mm(Iyy)9=MO.I OF RIB F2.50E+06mm(Iyy)10=MO.I OF RIB G2.50E+06mm(Iyy)11=MO.I OF RIB H2.50E+06mm(Iyy)12=MO.I OF RIB I2.50E+06mm(Iyy)13=MO.I OF RIB J2.50E+06mm(Iyy)14=MO.I OF RIB K2.50E+06mm(Iyy)15=MO.I OF RIB L2.50E+06mm(Iyy)16=MO.I OF RIB M2.50E+06mm(Iyy)17=MO.I OF RIB N2.50E+06mm(Iyy)18=MO.I OF RIB O2.50E+06mm(Iyy)19=MO.I OF RIB P2.50E+06mm(Iyy)20=MO.I OF RIB Q2.50E+06mm(Iyy)21=MO.I OF RIB R2.50E+06mm(Iyy)22=MO.I OF RIB S2.50E+06mm(Iyy)23=MO.I OF RIB T2.50E+06mmAS ALL RIBS AND TOP AND BOTTOM SADDLE PLATES HAVE SAME DIMENSIONS WILL HAVE THE SAME MOMENT OF INERTIAA1(X-X1)=8.46E+07mmA2(X-X2)=6.84E+06mmA3(X-X3)=6.84E+06mmA4(X-X4)=3.37E+06mmA4(X-X5)=3.37E+06mmA4(X-X6)=2.79E+07mmA4(X-X7)=2.79E+07mmA4(X-X8)=3.37E+06mmA4(X-X9)=3.37E+06mmA4(X-X10)=2.79E+07mmA4(X-X11)=2.79E+07mmA4(X-X12)=3.37E+06mmA4(X-X13)=2.79E+07mmA4(X-X14)=3.37E+06mmA4(X-X15)=2.79E+07mmA4(X-X16)=3.37E+06mmA4(X-X17)=2.79E+07mmA4(X-X18)=3.37E+06mmA4(X-X19)=2.79E+07mmA4(X-X20)=3.37E+06mmA4(X-X21)=2.79E+07mmA4(X-X22)=3.37E+06mmA4(X-X23)=2.79E+07mmmmAS Iyy = [(Iyy)1+A1(X-X1)]+ [(Iyy)2+A2(X-X2)]+ [(Iyy)3+A3(X-X3)]+ [(Iyy)4+A4(X-X4)]+ [(Iyy)5+A5(X-X5)] Iyy =5.28E+08mmEqn#60MOMENT OF INERTIA ALONG YY AXIS (neutral axis) NOT USING PARRALEL AXIS THEOREM Iyy =1.17E+08mmEqn#61 Iyy = [(Iyy)i]POLAR MOMENT OF INERTIA JoJo=Ixx+IyyEqn#62POLAR MOMENT OF INERTIA USING INERTIAS FROM PARALLEL AXIS THEOREMFROM EQUATION 60 AND 57Jo=1.21E+11mmEqn#62aPOLAR MOMENT OF INERTIA USING INERTIAS FROM SUMMATION OF LOCAL MOMENT OF INERTIASJo=Ixx+IyyEqn#62(b)FROM EQUATION 59 AND 61Jo=5.78E+10mmWE 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 MATERIALMOMENT SHALL BE TAKEN FROM THE EQUATION (MOMENT IN THE TRANSPORTATION)Mt=TORSIONAL MOMENT1.03E+09N-mmAS(MAX.)=MAX. SHEAR STRESS IN TORSION(MAX.)=(MT x r)/JoN/mm2r=y/2=718.7596099491mmWHERE r IS THE CENTEROID ON WHICH THE MOMENT IS ACTING IN THIS CASE Y WILL BE THE CENTEROID THEREFORE(MAX.)=12.75N/mm2ALLOWABLE CHECK

SADDLE ALLOWABLE STRESSSa=138M Pa.POLAR MOMENT OF INERTIA OF SADDLE SECTION IS SUFFICIENT TO SUSTAIN TORSION

&8CLIENT: TECHNIP - NPCC CONSORTIUMVANDOR: DESCON ENGINEERING FZE. SHARJAH UNITED ARAB EMERITES.&8DESIGN CALCULATION FOR THE TORSION CALCULATION FOR SADDLE SECTION IN OPERATING&8EQUIPMENT NO:564-D-2160EQUIPMENT NAME: OPEN DRAIN DRUM

7.DEFLECTION IN SADDLE (OP)DEFLECTION IN VESSEL BETWEEN THE LENGTH OF FIXED SADDLE TO SLIDING SADDLE (OPERATING CASE)METHEDOLOGY:AS THESE VESSELS HAVE ON SADDLE FIXED AND ANOTHER FREE THEREFORE WIND IN TRANSVERSE CASE SHALL TEND TO DEFLECT VESSEL SO IF VESSEL DEFLECTS IT MAY STRIKE THE STOPPER PLATE TO VALIDATE IS THIS STRIKE IS POSSIBLE OR NOT WE HAVE TO CALCULATE THE TOTAL DEFLECTION OFVESSEL BETWEEN THE FIXED AND FREE SADDLE CENTER LINES (ONLY TRANSEVERSE CASE OF OPERATING CONDITION) SHALL AFFECT THE DEFLECTION.(REFERENCE DIAGRAM: FIG 2 ATTACHED)TO EVALUATE THE DEFLECTION WE WILL CONSIDER THE MOMENT ARM LENGTH AS THE DISTANCE BETWEEN THE FIXED SADDLE AND THE CENTER POINT OF THE DISTANCE BETWEEN THE FIXED AND FREE SADDLE.DATA:1.SHELL INTERNAL DIAMETER DI=3200mm2. SHELL OUTER DIAMETER Do=3244mm3.SHELL THICKNESS t=22mm4.YOUNG'S MODULUS E=2.02E+05N/mm5. SADDLE SPACING L8600mm6. DISTANCE OF FIXED SADDLETO CENTEROID OF SHADDED AREA L1=8600mm7.Moment of intertia I =pi(Do-Di)/642.89E+11mm

TRANSVERSE FORCE F(T)5.61E+05NTOTAL TRANSVERSE FORCE IN OP+BLAST CONDITIONDEFLECTION,=FL/3EIDEFLECTION,=2.04E+00mmANGLE OF TWIST IN FIXED SADDLE DUE TO TORSIONAL MOMENTPOLAR MOMENT OF INERTIA J5.78E+10mmFROM EQUATION NUMBER 62 (B)POISSON'S RATIO v =0.333SHEAR MODULUS G=E/2(1+v)1.35E+05N/mmTorsional Moment T=1.03E+09N-mmHeight of Saddle L2=2000mmtwist =TL/GJradians2.64E-04radians0.0150995392DegreesDISTANCE BETWEEN FIXED AND FREE SADDLES L=8600mmAS TAN= P/bp= PERPENDICULAR DISTANCE THROUGH WHICH ANGULAR TWIST OCCURSb=DISTANCE BETWEEN FIXED AND FREE SADDLESp=bTAN2.27E+00mm

THEREFORE THE TOTAL DEFLECTION IN THE SPAN OF FIXED AND FREE SADDLES ISTOTAL DEFLECTION,(T)=0.00430mmDISTANCE BETWEEN BASE PLATE TIP AND STOPPER PLATE,X29mmRESULT:DEFLECTION IS VERY SMALL THAT STOPPER PLATE WILL NOT BE UNDER THE AFFECT OF VESSEL DEFLECTION

&8VESSEL TAG:564-D-2160VESSEL NAME:OPEN DRAIN DRUM&8DEFLECTION IN VESSELS DUE TO WIND LOAD ACTING TRANSVERSE&8CLIENT:TECHNIP NPCC CONSORTIUMPROJECT NUMBER:14615

7.DEFLECTION IN SADDLE (TRANS)DEFLECTION IN VESSEL BETWEEN THE LENGTH OF FIXED SADDLE TO SLIDING SADDLE (OPERATING CASE)METHEDOLOGY:AS THESE VESSELS HAVE ON SADDLE FIXED AND ANOTHER FREE THEREFORE WIND IN TRANSVERSE CASE SHALL TEND TO DEFLECT VESSEL SO IF VESSEL DEFLECTS IT MAY STRIKE THE STOPPER PLATE TO VALIDATE IS THIS STRIKE IS POSSIBLE OR NOT WE HAVE TO CALCULATE THE TOTAL DEFLECTION OFVESSEL BETWEEN THE FIXED AND FREE SADDLE CENTER LINES (ONLY TRANSEVERSE CASE OF OPERATING CONDITION) SHALL AFFECT THE DEFLECTION.(REFERENCE DIAGRAM: FIG 2 ATTACHED)TO EVALUATE THE DEFLECTION WE WILL CONSIDER THE MOMENT ARM LENGTH AS THE DISTANCE BETWEEN THE FIXED SADDLE AND THE CENTER POINT OF THE DISTANCE BETWEEN THE FIXED AND FREE SADDLE.DATA:1.SHELL INTERNAL DIAMETER DI=3200mm2. SHELL OUTER DIAMETER Do=3244mm3.SHELL THICKNESS t=22mm4.YOUNG'S MODULUS E=2.02E+05N/mm5. SADDLE SPACING L8600mm6. DISTANCE OF FIXED SADDLETO CENTEROID OF SHADDED AREA L1=8600mm7.Moment of intertia I =pi(Do-Di)/642.89E+11mm

TRANSVERSE FORCE F(T)292000.01NTOTAL TRANSVERSE FORCE TRANSPORTATION STAGE-2DEFLECTION,=FL/3EIDEFLECTION,=1.06E+00mmANGLE OF TWIST IN FIXED SADDLE DUE TO TORSIONAL MOMENTPOLAR MOMENT OF INERTIA J5.78E+10mmFROM EQUATION NUMBER 62 (B)POISSON'S RATIO v =0.333SHEAR MODULUS G=E/2(1+v)1.35E+05N/mmTorsional Moment T=1.03E+09N-mmHeight of Saddle L2=2000mmtwist =TL/GJradians2.64E-04radians0.0150995392DegreesDISTANCE BETWEEN FIXED AND FREE SADDLES L=8600mmAS TAN= P/bp= PERPENDICULAR DISTANCE THROUGH WHICH ANGULAR TWIST OCCURSb=DISTANCE BETWEEN FIXED AND FREE SADDLESp=bTAN2.27E+00mm

THEREFORE THE TOTAL DEFLECTION IN THE SPAN OF FIXED AND FREE SADDLES ISTOTAL DEFLECTION,(T)=0.00333mmDISTANCE BETWEEN BASE PLATE TIP AND STOPPER PLATE,X29mmRESULT:DEFLECTION IS VERY SMALL THAT STOPPER PLATE WILL NOT BE UNDER THE AFFECT OF VESSEL DEFLECTION

&8VESSEL TAG:564-D-2160VESSEL NAME:OPEN DRAIN DRUM&8DEFLECTION IN VESSELS DUE TO WIND LOAD ACTING TRANSVERSE&8CLIENT:TECHNIP NPCC CONSORTIUMPROJECT NUMBER:14615

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