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
'(T)=14.02M Pa.b. IN LONGITUDNAL CASE
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
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#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
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
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
'(T)=7.30M Pa.b. IN LONGITUDNAL CASE
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
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#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
2. FOR BASE METALDESIGN IS SAFE
REFERENCE PAGES ARE ATTACHED BELOW
&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
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