Transportation and Erection of Pressure...

10Transportation and Erection of

Pressure Vessels

Procedure 10-1 Transportation of PressureVessels 632Procedure 10-2 Erection of Pressure Vessels 660Procedure 10-3 Lifting Attachments andTerminology 666Procedure 10-4 Lifting Loads and Forces 675Procedure 10-5 Design of Tail Beams Lugsand Base Ring Details 681Procedure 10-6 Design of Top Head andCone Lifting Lugs 691

Procedure 10-7 Design of Flange Lugs 695

Procedure 10-8 Design of Trunnions 706

Procedure 10-9 Local Loads in Shell Dueto Erection Forces 710

Procedure 10-10 Miscellaneous 713

631

Procedure 10-1 Transportation of Pressure Vessels

The transportation of a pressure vessel by ship bargeroad or rail will subject the vessel to one-time-onlystresses that can bend or permanently deform the vessel ifit is not adequately supported or tied down in the rightlocations The shipping forces must be accounted for toensure that the vessel arrives at its destination withoutdamage

It is very frustrating for all the parties involved to havea load damaged in transit and to have to return it to thefactory for repairs The cost and schedule impacts can bedevastating if a vessel is damaged in transit Certainminimal precautions can avoid the costly mistakes thatoften lead to problems Even when all precautions aremade however there is still the potential for damage dueto unforseen circumstances involved in the shipping andhandling process

Care should be taken to ensure that the size and locationof the shipping saddles tie-downs or lashing are adequateto hold the vessel but not deform the vessel Long thin-walled vessels such as trayed columns are especiallyvulnerable to these shipping forces The important thing toremember is that someone must take the responsibilityThe barge and rail people have their own concerns withregard to loading and lashing These may or may notcoincide with the concerns of the vessel designer

The shipping forces for ships barges trucks and railare contained in this procedure Each method of trans-portation has its own unique load schemes and resultingforces Barge shipping forces will differ from rail due tothe rocking motion of the seas Rail shipments howevergo around corners at high speed In addition rail forcesmust allow for the ldquohumpingrdquo of rail cars when they arejoined with the rest of the train Ocean shipments have toresist storms and waves without breaking free of theirlashings

Whereas horizontal vessels on saddles are designed forsome degree of loading in that position vertical vesselsare not The forces and moments that are used for thedesign of a vertical vessel assume the vessel is in itsoperating position Vertical vessels should generally bedesigned to be put on two saddles in a horizontal positionand transported by various means That is the purpose ofthis procedure Too often the details of transportation anderection are left in the hands of people who though wellversed in their particular field are not pressure vesselspecialists

Often vessels are transported by multiple means Thusthere will be handling operations between each successivemode of transportation Often a vessel must be moved byroad to the harbor and then transferred to a barge or shipOnce it reaches its destination it must be reloaded ontoroad or rail transport to the job site There it will be off-loaded and either stored or immediately erected A finaltransport may be necessary to move the vessel to thelocation where it will be finally erected At each handlingand transport phase there are different sets of forcesexerted on the vessel that must be accounted for

Shipping Saddles

The primary concern of the vessel designer is thelocation and construction of the shipping saddles to takethese forces without overstressing or damaging the vesselIf saddles are to be relocated by the transporter it isimportant that the new locations be reviewed Generallyonly two shipping saddles should be used However thismay not always be possible Remember that the reason forusing two saddles is that more than two saddles createsa statically indeterminate structure You are never assuredthat any given saddle is going to take more than itsapportioned load

Here are some circumstances that would allow for morethan two saddles to be used or for a special location of twosaddles

bull Transporter objects due to load on tiresbull Transporter objects due to load on barge or shipbull Very thin long vesselbull Heavy-walled vessels for spreading load on ship ortransporters

Shipping saddles can be constructed of wood or steel orcombinations The saddles should be attached to the vesselwith straps or bolts so that the vessel can be movedwithout having to reattach the saddle Horizontal vesselsmay be moved on their permanent saddles but should bechecked for the loadings due to shipping forces andclearances for boots and nozzles Shipping saddles shouldhave a minimum contact angle of 120 just like perma-nent saddles Provisions for jacking can be incorporatedinto the design of the saddles to allow loading andhandling operations without a crane(s)

632 Pressure Vessel Design Manual

Shipping saddles should be designed with the vesseland not left up to the transport company In generaltransportation and erection contractors do not havethe capability to design shipping saddles or to checkthe corresponding vessel stresses for the various loadcases

Whenever possible shipping saddles should be locatedadjacent to some major stiffening element Some commonstiffening elements include stiffening rings heads (bothinternal and external) or cones If necessary temporaryinternal spiders can be used and removed after shipment iscomplete

Key factors for shipping saddles to consider

bull Included anglebull Saddle widthbull Type of constructionbull Lashing lugsbull Jacking pocketsbull Method of attachment to the vesselbull Overall shipping height allowabledcheck withshipper

Recommended contact angle and saddle width

Vessel Stresses

The stresses in the vessel shell should be determined bystandard Zickrsquos analysis The location of shipping saddlesshould be determined such that the bending at the midspanand saddles is not excessive Also the stresses due tobending at the horn of the saddle is critical If this stress isexceeded the saddle angle and width of saddle should beincreased Also move the saddle closer to the head ora major stiffening element

Lashing

Vessels are lashed to the deck of ships and bargesIn like manner they must be temporarily fixed to rail-cars trailers and transporters Lashing should berestricted to the area of the saddle locations Vessels

are held in place with longitudinal and transverselashings Lashings should never be attached to smallnozzles or ladder or platform clips In some caseslashing may be attached to lifting lugs and base ringsLashings should not exceed 45 from the horizontalplane

Other Key Factors to Consider

bull Shipping clearancesbull Shipping orientationdpay close attention to lift lugsand nozzles

bull Shipping routebull Lifting orientationbull Type of transportbull Watertight shipment for all water transportationbull Escorts and permitsbull Abnormal loadsdsize and weight restrictionsbull Vessels shipped with a nitrogen purgebull Shippinghandling plan

Organizations That Have a Part in the Transportation andHandling of Pressure Vessels

bull Vessel fabricatorbull Transport companybull Engineering contractorbull Railway authoritiesbull Port authoritiesbull Erectionconstruction companybull Trailertransporter manufacturerbull Ship or barge captainbull Crane companyoperator

Special Considerations for Rail Shipments

1 Any shipment may be subject to advance railroadapproval

2 Any shipment over 10 ft-6 in wide must haverailroad approval

3 A shipping arrangement drawing is required for thefollowinga All multiple carloads (pivot bolster required)b All single carloads over 10 ft-6 in widec All single carloads over 15 ft-0 in ATR (above

top of rail)d All single carloads that overhang the end(s) of

the car and are over 8 ft-0 in ATR

Vessel Diameter Contact Angle

Minimum Saddle

Width

D lt 13 ft-0 in 120 11 in

13 ft-0 in lt D lt 24 ft-0 in 140 17 in

D gt 24 ft-0 in 160 23 in

Transportation and Erection of Pressure Vessels 633

4 Clearances must be checked for the followinga Vessels greater than 9 feet in widthb Vessels greater than 40 feet overall lengthc Vessels greater than 50 tons

5 The railroad will need the following specific dataas a minimum

a Weightb Overall lengthc Method of loadingd Loadpoint locationse Overhang lengthsf Widthg Heighth Routingroute surveysi Center of gravity

6 A swivel (pivot) bolster is required whenever thefollowing conditions exista Two or more cars are requiredb The capacity for a single car is exceededc The overhang of a single car exceeds 15 feet

7 Rated capacities of railcars are based on a uniformlydistributed load over the entire length of the carThe capacity of a car for a concentrated load willonly be a percentage of the rated capacity

8 Rules for loads loading and capacities vary bycarrier Other variables include the types of cars thecarrier runs the availability and the ultimatedestination Verify all information with the specificcarrier before proceeding with the design of ship-ping saddles or locations

9 For vessels that require pivot bolsters the shippingsaddles shall be adequately braced by diagonaltensioncompression rods between the vessel andthe saddle The rods and clips attached to the vesselshell should be designed by the vessel fabricator tosuit the specific requirements of the carrier

10 If requested rail bolsters can be returned to themanufacturer

11 Loading arrangement and tie-downs will have topass inspection by a representative of the railwaysand sometimes by an insurance underwriter prior toshipment

12 Accelerometers can be installed on the vessel tomonitor shipping forces during transit

13 A rail expediter who accompanies the load shouldbe considered for critical shipments

14 The railroad will allow a fixed time for the cars tobe offloaded cleaned and returned Demurragecharges for late return can be substantial

Outline of Methods of Vessel Shipping and Transportation

1 Roada Trucktractor and trailerb Transportersdsingle or multiple self-propelled

or towedc Specialdbulldozerd Frame adapterse Beams to span trailers or transportersf Rollersg Special

2 Raila Single carb Multiple carsc Special carsd Types of cars

bull Flatcarbull Fishbelly flatcarbull Well carbull Heavy-duty carbull Gondola car

3 Bargea River bargeb Ocean-going bargec Lakes and canals

4 Shipsa Roll-on roll-off typeb Loading and off-loading capabilitiesc In-hull or on-deckd Floating cranes

5 Othera Planeb Helicopterc Bulldozer


Table 10-1Overland shipping limits in the US

State Length ft Width ft Height ft Gross Weight lb

Alabama 150 16 16 180000

Arizona 120 14 16 250000

Arkansas 100 14 14 120000

California 120 14 16 220000

Colorado 130 17 16 228000

Connecticut 120 15 16 250000

Delaware 120 15 15 250000

Florida 150 16 16 250000

Georgia 100 14 14 120000

Idaho 110 16 15rsquo6 200000

Illinois 145 14rsquo6 15 250000

Indiana 110 16 15 250000

Iowa 120 18 16 250000

Kansas 126 16rsquo6 16 250000

Kentucky 110 16 15 250000

Louisiana 125 18 16 250000

Maine 125 16 16 250000

Maryland 100 15rsquo11 15rsquo11 220000

Massachusetts 115 14 14 240000

Michigan 150 16 15 230000

Minnesota 95 14rsquo6 14 250000

Mississippi 100 14 14 120000

Missouri 100 14 14 120000

Montana 110 18 17 240000

Nebraska 120 14 15rsquo6 212000

Nevada 105 17 16 240000

New Hampshire 120 15 16 250000

New Jersey 120 18 16 220000

New Mexico 120 14 16 250000

New York 120 14 14 160000

North Carolina 100 14 14 120000

North Dakota 120 14rsquo6 15rsquo6 150000

Ohio 100 14 14rsquo10 120000

Oklahoma 100 16 16 212000

Oregon 105 14 16 220000

Pennsylvania 120 16 15rsquo6 201000

Rhode Island 90 14 13rsquo6 120000

South Carolina 125 16 14 250000

South Dakota 120 14rsquo6 15rsquo6 150000

Tennessee 120 16 15 250000

Texas 125 20 18rsquo11 252000

Utah 125 15 16rsquo6 250000

Vermont 100 15 14 150000

Virginia 150 14 15 150000

Washington 150 14 16 200000

West Virginia 150 16 15 212000

Wisconsin 150 16 16 250000

Wyoming 110 18 17 252000

Note This information in this Table is for general reference information only and should not be relied upon for any given application These values

change regularly and local state and national regulations should be checked for any given haul application







Table 10-2Barge shipping forces


Pitch

Cases 3a and 3b

Forces in Vessel Due to Pitch

General

F frac14 ma frac14Wg

2pT

2Rqp180

F frac14 00214WRq

T2

f1 frac14 tan1

aR1

Fp frac14 00214WR1q1

T21

Case 3a Fy frac14 Fp sin f1

Fz frac14 Fp cos f1

Case 3b Fy frac14 Fp sin f1

Fz frac14 Fp cos f1

Roll

Case 2a q2 [ 30 max

Case 2b

Forces in Vessel Due to Roll

f2 frac14 tan1ed

R2 frac14 esin f2

FR frac14 00214WR2q2

T22

Case 2a Fy frac14 FR sin f2

Fx frac14 FR cos f2

Case 2b Fy frac14 FR sin f2

Fx frac14 FR cos f2


The job of the designer is to translate the loads resultingfrom the movement of the ship into loads applied to thepressure vessel that is stored either at or below decks Theship itself will rotate about its own center of buoyancy(CB) depending on the direction of the sea and the shiprsquosorientation to that direction of sea The vessel strapped toits deck is in turn affected by its location in relation to theCB of the ship For example if the CG of the vessel islocated near the CB of the ship the forces are minimizedThe farther apart the two are in relation to each other themore pronounced the effect on the vessel

The shiprsquos movement translates into loads on the threeprincipal axes of the vessel Saddles and lashings must bestrong enough to resist these external forces withoutexceeding some allowable stress point in the vessel Thepoint of application of the load is at the CG of the vesselThese loads affect the vessel in the same manner as

seismic forces do In fact the best way to think of theseloads is as vertical and horizontal seismic forces Verticalseismic forces either add or subtract to the weight of thevessel Horizontal seismic forces are either transverse orlongitudinal

The X Y and Z axes translate into and are equivalentto the following loadings in the vessel

X axis horizontal transverseY axis corresponds to vertical loads by either adding orsubtracting from the weight of the vesselZ axis longitudinal axis of the vessel All Z axis loadsare longitudinal loadings

Load Combinations for Sea Forces

1 dead load thorn sway thorn heave thorn wind2 dead load thorn surge thorn heave thorn wind



Axle Loads

The number of axles that must be placed under a load isdetermined by analyzing the weight restrictions andallowable bearing load from local state or national regu-lations The transportation contractor is responsible fordetermining the axle loads based on the equipment usedand the weight and distribution of the loads The author-ities that permit the load will require an analysis of the axleloads to ensure that the roadbed is not overloaded Axleloads include the weight of the vessel transport saddlesbeams hauler (tractor) dollies etc

There are three different methods used to distribute theloads to the axles

1 Flatbed This method uses conventional tractor-trailer assembly with various numbers of axles andwheels under the trailer bed to distribute the weightto the road surface

2 Bolstered This method is used for abnormally longloads in which the sets of axles are attached directlyto the load via the transport saddle Both sets ofaxles will have steering capabilities

3 Bolstered loads using equalizing transportingbeams This method is much the same as thebolstered long vessel load In this case the load istoo heavy for a flatbed yet too short for bolsteredaxles The solution is to utilize beams betweenbolsters to suspend the load

Transporter Stability

There are two types of stability checks that should beperformed on each load The first has to do with thetipping point of the load relative to the roadway as the loadshifts due to the camber of the road The second has to dowith the turning radius of bolstered loads As the load goesaround curves the CG shifts from being in line with thedollies to an eccentric condition In tight curves the

eccentricity of the load can overload the outer set ofwheels to the point of rollover The two cases are

1 Rollover stability due to road camber2 Turning stability due to turning radius

Case 1Due to the camber in roads the load will be subject to

various angles q that will change the location of the centerof gravity of the load On a flat surface the center of gravityis in line with the centerline of the trailer As the roadcamber increases the CG is steadily moved toward theouter set of wheels At some point the wheels are over-loaded on one side and the entire assembly reaches a tippingpoint Beyond this the trailer turns over and the load is lostThis condition has resulted in numerous rollovers

Case 2For bolstered loads the vessel must swivel on the deck

of the trailer in order to accommodate curves in the roadand corners As the curve or corner is negotiated theactual CG gets further away from the projected load pointThis is true whether you have a single pivoting bolster ortwo however the situation is more pronounced with thedouble pivoting case There have been a number of roll-overs as a result of the eccentricity ie shifting the load tothe outer row of wheels until the load becomes unstable

θ

Figure 101 Rollover stability

DOLLIEe

CG OF LOAD CENTERLINESUSPENDED LOAD

PROJECTEDLOAD POINT

Figure 102 Turning stability




Summary of LoadsForces on Vessels During Transportation

Table 10-3Transportation load coefficients K

Forces Road Rail Barge Ocean

Fx 05 10 095 10

Fy 15 20 13 15

Fz 10 15 15 15

Table 10-4Load per saddle due to transport forces

Due to Load per Saddle Diagram

Fx Q1 frac14 WsL2L1

thorn FxB

2A

Q2 frac14 WsL3L1

thorn FxB

2A

Fy Q1 frac14 ethWs thorn FyTHORNL2L1

Q2 frac14 ethWs thorn FyTHORNL3L1

Fz

Q1 frac14 WsL2L1

thorn FzB

L1

Q2 frac14 WsL3L1

thorn FzB

L1


Load Diagrams for Moments and Forces

Case 1

Note W frac14 weight of vessel plus any impact factors

OAL frac14 L1 thorn L2 thorn L3 w frac14 WOAL

Q1 frac14whethL1 thorn L2THORN2L2

3

i2L1

Q2 frac14 W Q1

M1 frac14 wL22

2

M2 frac14 Q1

Q1

2w L2

M3 frac14 wL23

2

Mx frac14 wethL2 XTHORN22

Mx1 frac14 wethL2 thorn X1THORN22

Q1X1

Mx2 frac14 wethL3 X2THORN22

Case 2

w1 frac14 W1

L2

w2 frac14 W2

L3

Q1 frac14 WL6

L1

Q2 frac14 W Q1

M1 frac14 w1L24

2

M2 frac14 M1 thornM3

2 w1L2

1

8

M3 frac14 w2L23

2


Case 3

Q1 frac14 WL1

2ethL1 thorn L2THORN WL2

2

2L1ethL1 thorn L2THORN

Q2 frac14 WL1

2ethL1 thorn L2THORN thornWL2

L1 thorn L2thorn WL2

2


M1 frac14 Q21ethL1 thorn L2THORN

2W

M2 frac14 Q1L1 WL21

2ethL1 thorn L2THORN

Mx frac14 Q1XWX2


Case 4

w1 frac14 W1

L2

w2 frac14 W2

L3

Q1 frac14 w1L2eth2L1 L2THORN thorn w2L23

2L1


2L1

Moment at any point X from Q1

Mx frac14 Q1X w1X2

2

Moment at any point Y from Q2

My frac14 Q2ethL1 YTHORN w2ethL1 YTHORN22


Transportation-Vertical Vessel on Two Saddles UniformLoad Case With Incorporation of Shipping Factors

NotationF2 frac14 Additional load on Q2 LbsFZ frac14 Longitudinal loading due to shipping forces

LbsFY frac14 Vertical Loading due to shipping forces LbsKZ frac14 Longitudinal impact factorKY frac14 Vertical impact factor

Q1Q2 frac14 Saddle loads without impact factors LbsQ1

0Q20 frac14 Saddle loads with impact factors Lbs

W frac14 Shipping weight of vessel without impactfactors Lbs

WT frac14 Shipping weight with impact factors Lbsw1 frac14 Uniform load without FY LbsFtw2 frac14 Uniform load including FY LbsFt

NOTE Assume that FZ amp FY do not occur at the sametime

Case 1 Adding Load for Fz

bull Longitudinal load FZ

FZ frac14 KZ W

bull Uniform load w1

w1 frac14 W=LT

bull Additional load on saddle F2

F2 frac14 ethFZ BTHORN=L1

bull Saddle loads Q1 Q2 Q10 Q2

0

Q1 frac14 w1

hethL1 thorn L2THORN2L2

3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1

Q20 frac14 Q2 thorn F2

Case 2 Adding Loads for Fy

bull Longitudinal load FY

FY frac14 KY W

bull Vertical load FY

WT frac14 Wthorn FY


w2 frac14 WT=LT

bull Saddle loads Q10 amp Q2

0

Q10 frac14 w2

hethL1 thorn L2THORN2 L2

3

i=2L1

Q20 frac14 WT Q1

0

Select worst case and calculate moments

M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given

B frac14 1575 ftL1 frac14 124 ftL2 frac14 24 ftL3 frac14 21 ftLT frac14 169 ft

LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3

Figure 103 Data for uniform load case


W frac14 741 kipsKY frac14 5KZ frac14 6

Calculation

FZ frac14 KZ W frac14 6 eth741THORN frac14 4446 kips

FY frac14 KY W frac14 5 eth741THORN frac14 3705 kips

WT frac14 Wthorn FY

frac14 741thorn 3705 frac14 1 1115 kips

w1 frac14 W=LT frac14 741=169 frac14 438 kips=ft

w2 frac14 WT=LT frac14 11115=169 frac14 658 kips=ft

F2 frac14 ethFZ BTHORN=L1 frac14 frac124446 eth1575THORN=124frac14 5647 kips

Case 1 Adding Load for FZ


0

Q1 frac14 w1


3

i2 L1

frac14 438heth124thorn 24THORN2212

i2124

frac14 379 kips

Q10 frac14 Q1 F2 frac14 379 565 frac14 3225 kips

Q2 frac14 W Q1 frac14 741 379 frac14 362 kips

Q20 frac14 Q2 thorn F2 frac14 362thorn 565 frac14 4185 kips

Case 2 Adding Load for FY


0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips

Worst case is Case 2Determine moments

M1 frac14 w2 L2

2

2 frac14

658242

2

frac14 1 895 ft kips

M2 frac14 Q10Q1

0=2 w2

L2

frac14 569 eth569=eth2$658THORN 24THORN frac14 10 946 ft kips

M3 frac14 w2 L2

3

2 frac14

658212

2 frac14 1 450 ft kips

Use these moments and loads to determine stresses inshell





Tension Bands on Saddles

NotationAr frac14 area required in2

As frac14 area of bolt in2

Ab frac14 area of band required in2

Aw frac14 allowable load on weld lbinB frac14 saddle height ind frac14 bolt diameter inf frac14 load on weld kipsinFt frac14 allowable stress tension psi

Fx Fy Fz frac14 shipping external forces lbK frac14 maximum band spacing inN frac14 number of bands on one saddlePe frac14 equivalent external pressure psiR frac14 outside vessel radius inT frac14 tension load in band lb

T123 frac14 load cases in bolt and band lbTb frac14 tension load in bolt lbWs frac14 weight of one saddle lbb frac14 angle of tension bands degreessa frac14 stress in bolt psisb frac14 stress in band psi


bull Find tension in band T1 due to shipping forces onsaddle Fx and Fy

T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N

bull Area required for bolt

Ar frac14 T1

Ft

bull Find bolt diameter d

d frac14ffiffiffiffiffiffiffiffi4Ar

p

r

Select nominal bolt diameter

As frac14bull Find maximum stress in bolt due to manualwrenching sa

sa frac14 45 000ffiffiffid

p

bull Maximum tension load in bolts T2

T2 frac14 saAs

bull Load due to saddle weight T3

T3 frac14 Ws

2N

Note Include impact factor in weight of saddle

bull Find maximum load T

T frac14 greater of T1 T2 or T3

bull Load on weld f

f frac14 T4[

bull Determine size of weld from table based on load f

Use w frac14bull Maximum band spacing K

K frac14 4ffiffiffiffiffiRt

p

1285

bull Find area required for tension band Ar

Ar frac14 TFt

Use

bull Check shell stresses due to force T Pe

Pe frac14 4TpRK

lt ASME factor ldquoBrdquo

Table 10-5Allowable load weld

Weld Size w E60XX E70XX

31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure

Tension Band Notation

N frac14 Number of bands on one saddleQ frac14 Total load on one saddle LbsR frac14 Outside vessel radius inT frac14 Tension load in band Lbsb frac14 Angle of tension bands degrees

K1 frac14 Transverse shipping coefficientK2 frac14 Vertical shipping coefficientX frac14 Horizontal distance to centroid of saddle

reaction inY frac14 Vertical distance to centroid of saddle in

Calculation

bull Vertical distance to centroid of saddle Y

Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R

bull Horizontal distance to centroid of saddle X

X frac14 R sina

bull Tension load in band T

T frac14 frac12Q frac12ethK1 Y=XTHORN thorn K2=frac122N cos b

Notes

1 Vertical reaction can be a result of longitudinal loadUse largest value

2 Use K2 frac14 0 for transverse case3 Use K1 frac14 0 for longitudinal case4 Use worst case of T1 or T2 and design the balance of

components per previous method

Example

K1 frac14 25K2 frac14 5Q frac14 500 kipsR frac14 925 inq frac14 75 frac14 1308 radb frac14 75N frac14 2

Y frac14 R sin qqY frac14 925 sin 75 1308 frac14 683 in

a frac14 cos-1 (Y R)a frac14 cos-1 (683 925) frac14 424o

X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in

Transverse (K2 frac14 0)

T1 frac14 [Q [(K1 Y X) thorn K2]] [2N cos b]T1 frac14 [500 [(025$683624) thorn 0]] [2$2 cos 75] frac14

3435 kips

Longitudinal (K1 frac14 0)

T2 frac14 [Q [(K1 Y X) thorn K2]] [2N cos b]T2 frac14 [500 [0 thorn 05]] [2$2 cos 75] frac14 6304 kips

K2 W

K1 W

B

AX

TY

R

β α

θ

Figure 10-4 Dimensions of shipping saddle for alternatecase



Procedure 10-2 Erection of Pressure Vessels

The designer of pressure vessels and similar equipmentwill ultimately become involved in the movement trans-portation and erection of that equipment The degree ofthat involvement will vary due to the separation of dutiesand responsibilities of the parties concerned It is prudenthowever for the designer to plan for the eventuality ofthese events and to integrate these activities into theoriginal design If this planning is done properly thereis seldom a problem when the equipment gets to itsfinal destination Conversely there have been numerousproblems encountered when proper planning has notbeen done

There is also an economic benefit in including thelifting attachments in the base vessel bid and designThese lifting attachments are relatively inexpensive incomparison to the overall cost of the vessel and minusculecompared to the cost of the erection of the equipment Theerection alone for a major vessel can run into millions ofdollars If these attachments are added after the purchaseorder is awarded they can become expensive extras

There are also the consequences to life property andschedules if this activity is not carried out to a successfulconclusion Compared to the fabricated cost of the liftingattachments the consequences to life property andschedule are too important to leave the design of thesecomponents and their effect on the vessel to those not fullyversed in the design and analysis of pressure vessels

In addition it is important that the designer of the liftingattachments be in contact with the construction organi-zation that will be executing the lift This ensures that alllifting attachments meet the requirements imposed by thelifting equipment There are so many different methodsand techniques for the erection of vessels and the relatedcosts of each that a coordinated effort between thedesigner and erector is mandatory To avoid surprisesneither the designer nor the erector can afford to work ina vacuum To this end it is not advisable for the vesselfabricator to be responsible for the design if the fabricatoris not the chief coordinator of the transport and erection ofthe vessel

Vessels and related equipment can be erected ina variety of ways Vessels are erected by means of singlecranes multiple cranes gin poles jacking towers andother means The designer of the lifting attachmentsshould not attempt to dictate the erection method by thetypes of attachments that are designed for the vessel The

selection of one type of attachment versus another couldvery well do just that

Not every vessel needs to be designed for erection orhave lifting attachments Obviously the larger the vesselthe more complex the vessel the more expensive thevessel the more care and concern that should be taken intoaccount when designing the attachments and coordinatingthe lift The following listing will provide some guidelinesfor the provision of special lifting attachments and a liftinganalysis to be done In general provide lifting attachmentsfor the following cases

bull Vessels over 50000 lb (25 tons)bull Vessels with LD ratios greater than 5bull Vertical vessels greater than 8 ft in diameter or 50 ftin length

bull Vessels located in a structure or supported bya structure

bull High-alloy or heat-treated vessels (since it would notbe advisable for the field to be doing welding onthese vessels after they arrive on site and wire ropeslings could contaminate the vessel material)

bull Flare stacksbull Vessels with special transportation requirements

At the initial pick point when the vessel is still hori-zontal the load is shared between the lifting lugs and thetail beam or lug based on their respective distances to thevessel center of gravity As the lift proceeds a greaterpercentage of the load is shifted to the top lugs or trun-nions until the vessel is vertical and all of the load is thenon the top lugs At this point the tail beam or shackle canbe removed

During each degree of rotation the load on the lugstrunnions tailing device base ring and vessel shell arecontinually varying The loads on the welds attachingthese devices will also change The designer shouldevaluate these loadings at the various lift angles todetermine the worst coincident case

The worst case is dependent on the type of vessel andthe type of attachments For example there are threetypes of trunnions described in this procedure There isthe bare trunnion (Type 3) where the wire rope slidesaround the trunnion itself While the vessel is in thehorizontal position (initial pick point) the load producesa circumferential moment on the shell Once the vesselis in the upright position the same load produces


a longitudinal moment in the shell At all the interme-diate angles of lift there is a combination of circumfer-ential and longitudinal moments The designer shouldcheck the two worst cases at 0 and 90 and severalcombinations in between

The same trunnion could have a lifting lug welded tothe end of the trunnion (Type 1) This lug also producescircumferential and longitudinal moments in the shellHowever in addition this type of lug will producea torsional moment on the shell that is maximum at 0 andzero at 90 of angular rotation The rotating lug (Type 2)eliminates any torsional moment

There is one single lift angle that will produce themaximum stress in the vessel shell but no lift angle that isthe worst for all vessels The worst case is dependent onthe type of lift attachments distances weights and posi-tion relative to the center of gravity

The minimum lift location is the lowest pick point thatdoes not overstress the overhanging portion of the vesselThe maximum lift location is the highest pick point thatdoes not overstress the vessel between the tail and pickpoints These points become significant when locating thelift points to balance the stress at the top lug the overhangand the midspan stress

The use of side lugs can sometimes provide anadvantage by reducing the buckling stress at midspan andthe required lift height Side lugs allow for shorter boomlengths on a two-crane lift or gin poles A shorter boomlength in turn allows a higher lift capacity for the cranesThe lower the lug location on the shell the shorter the liftand the higher the allowable crane capacity This cantranslate into dollars as crane capacity is affected Thechallenge from the vessel side is the longitudinal bendingdue to the overhang and increased local shell stresses Allof these factors must be balanced to determine the lowestoverall cost of an erected vessel

Requirement for Erection and Setting of Vertical Vessels

The following is a brief synopsis of general recom-mendations regarding the setting leveling and shimmingof vertical vessels The following should be considered asguidelines only There are no codes or standards that areapplied In general company specifications containcontract requirements for the contractors scope of supplyor duties The following lists help to clarify generalconstruction practices with regard to the setting of verticalvessels and towers

Contractor Duties

1 Prepare tops of foundations (bush hammer ifrequired)

2 Perform surveying as required to establish center-lines sole plate or shim elevations at bottom of baseof equipment

3 Shimming4 Erect equipment5 Levelplumbing6 Final alignment7 Grouting8 Bolting

Tolerances

Out of vertical tolerance for vertical vessels unlessspecified otherwise shall be 01 of the vessel heightor about frac14 inch for every 20 feet to a maximum offrac34 inches

Soleplates (also called bearing pads levelingplates or embedments)

Soleplates are stainless steel plates 05 inches to075 inches thick set in grout on top of the foundation atthe exact height of the underside of the base plate Asa rule two soleplates should be installed per anchor boltone on each side of the bolt Depending on the towerdiameter and the distance between the anchor boltsanother soleplate may be installed between adjacentanchor bolts The dimensions of the soleplates will varyaccording to the width of the vessel base ring and vesselweight Soleplates are supported in place by a mixture ofPortland cement and sand in proportions 13 The vesselshould not be erected until the soleplates have been inplace for 28 days to allow for concrete curing Shims andsoleplates will remain in place after the grouting operation

Shims

Shims are used to provide precise leveling of the vesselShim packs may be grouted into the foundation in lieu ofsole plates but this practice is unusual Typically shimsare used on top of the sole plates for the leveling opera-tion Special shims may be required for unique applica-tions such as a large vessel supported on a braced framestructure with minimal contactbearing at each support


point The following are some guidelines for the use ofshims

1 Shims If left in place shall be stainless steel2 Shims must have rounded corners3 Shims will be fixed in place4 Shims shall be deburred5 Shims shall be full bearing6 Shims may be horseshoe type7 Shims thinner than 0001 inches are not allowed8 Shims with holes are not allowed9 Shims should be the full width of the base plate

LevelingStraightnessPlumbness

After the vessel has been placed on its foundation itmust be checked to be certain it is vertical and plumbLeveling is normally checked by use of two theodilites90 degrees apart The theodilites shall be spaced anadequate distance from the vessel to allow visual field ofthe entire height of the vessel Adjustments can be made tothe vessel alignment by means of wedges either poweredor not and then shimmed The wedges should not be leftin place after shimming

The vessel may be heated by the sun to a highertemperature on one side than the other This can createa slight ldquobananardquo effect which should be taken intoaccount when checking levelness The equation forcalculating the deflection from this effect is as follows

2 frac14 p D2 t H2 aDT

8 I

where2 frac14 Deflection inD frac14 Diameter ftT frac14 Thickness inH frac14 Height fta frac14 Coefficient of thermal expansion ininF

DT frac14 Temperature difference from one side of thecolumn to the other F

I frac14 Moment of inertia of vessel cross section ft4

Bolting

After the vessel is aligned and shimmed the nuts on theanchor bolts must be tightened The vessel should not be

left standing without the crane attached unless all anchorbolts have been tightened

The anchor bolts should not be tightened to theirmaximum load until the drypacking under the base plateis complete At this stage the base plate is suspendedbetween the soleplates until the drypack is installedSince the soleplates straddle the anchor bolts there isa chance of deforming the baseplate prior to the instal-lation of the drypack if the anchor bolts are overtightened

After drypacking the anchor bolts should be tight-ened to the correct torque to produce the maximumallowable bolt stress The anchor bolts should not betightened beyond the point of maximum allowable boltstress

Note that the initial anchor bolt tension does notincrease the maximum bolt tension caused by wind orearthquake This initial tension will only clamp the basering to the concrete Both are in equal compression untilthe external load is applied The external load reduces thecompression in the concrete before additional load isapplied to the bolts After the external load overcomes allthe compression in the concrete the stress in the bolt willincrease to the value it would have been had there been noinitial tension

Grouting

Grout under base plates shall provide full uniform loadtransfer between the bottom of the base plate and the topof the foundation Load transfer to the foundation must bethrough the grout not through the shims or soleplate

Prior to setting of the vessel the top of the foundationshould be bush hammered and cleaned This ensures thatthe grout will adhere to the surface of the foundation Bushhammering may be done strictly under the base plate oracross the entire top of the foundation

Once the vessel is leveled shimmed and bolted it isready to be grouted Grouting shall consist of filling thevoid area between the top of foundation and the undersideof base plate with cementatious grout The grout shall beinstalled in accordance with the manufacturers recom-mendations and any applicable contract specifications

Depending on the type of grout to be used grout damsmay be used


18

132 mm times 15 m SLINGSWL = 120 TON (SP)DOUBLE (2 PCS)SWL = 500 T times 40 m LENGTHSPREADER BEAM

SWL = 400T SHACKLE (4 NOS)

SWL = 400T SHACKLE (1 NO)

100 mm times 20 m SLINGSWL = 90 TON (SP)SWL = 737 TON (BH)DOUBLE (2 PCS)


CounterWeight = 40ton


BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m

Figure 10-5 Typical example of erection study

Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design

Given the overall weight and geometry of the vesseland the location of the center of gravity based on theerected weight apply the following steps to eithercomplete the design or analyze the design

Step 1 Select the type of lifting attachments as aninitial starting point

Lift end (also referred to as the ldquopick endrdquo)a Head lug Usually the simplest and most

economical and produces the least stressb Cone lug Similar to a head lug but located at

a conical transition section of the vesselc Side lug Complex and expensived Top flange lug The choice for high-pressure

vessels where the top center flange and head arevery rigid This method is uneconomical foraverage applications

e Side flange lug Rarely used because it requiresa very heavy nozzle and shell reinforcement

f Trunnions Simple and economical Used ona wide variety of vessels

g OtherTail end

a Tail beamb Tail lugc Choker (cinch) see later commentary

Tailing a column during erection with a wire ropechoker on the skirt above the base ring is a fairly commonprocedure Most experienced erectors are qualified toperform this procedure safely There are several advan-tages to using a tailing choker

bull Saves material design detailing and fabricationbull Simplifies concerns with lug and shippingorientations

bull May reduce overall height during transportation

There are situations and conditions that could make theuse of a tailing choker impractical costly and possiblyunsafe Provide tailing lugs or a tailing beam if

bull The column is more than about 10 feet in diameterThe larger the diameter the more difficult it is for thewire rope to cinch down and form a good choke onthe column

bull The tail load is so great that it requires the use ofslings greater than about 1frac12 inches in diameter Thelarger the diameter of the rope the less flexible it isand the more likely that it could slip up unexpectedlyduring erection

Step 2 Determine the forces T and P for all angles oferection

Step 3 Designcheck the lifting attachments for thetailing force T and pick force P

Step 4 Designcheck the base ring assembly forstresses due to tailing force T

Step 5 Determine the base ring stiffening configura-tion if required and design struts

Step 6 Check shell stresses due to bending during liftThis would include midspan as well as anyoverhang

Step 7 Analyze local loads in vessel shell and skirt dueto loads from attachments

Allowable Stresses

Per AISC

Tension

Ft frac14 06Fy on gross area

frac14 05Fu on effective net area

frac14 045Fy for pin-connected members

Compression(for short members only)

Fc frac14 Use buckling value

frac14 for vessel shell 133 ASME Factor B

Shear

Fs frac14 Net area of pin hole 045Fy

frac14 other than pin-connected members 04Fy

frac14 fillet welds in shear

E60XX 9600 lb=in or 13 600 psi

E70XX 11 200 lb=in or 15 800 psi

Bending

Fb frac14 06Fy to 075Fy depending on the shape of themember

Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1


Tension compression and bendingsa

Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1

Note Custom-designed lifting devices that supportlifted loads are generally governed by ASME B3020ldquoBelow the hook lifting devicesrdquoUnder this specificationdesign stresses are limited to Fy3 The use of AISCallowables with a load factor of 18 or greater willgenerally meet this requirement

Notation

A frac14 area in2

Aa frac14 area available in2

Ab frac14 area bolt in2

An frac14 net cross-sectional area of lug in2

Ap frac14 area pin hole in2

Ar frac14 area required in2

As frac14 area strut in2 or shear area of boltsC frac14 lug dimension see sketchDo frac14 diameter vessel OD inD1 frac14 diameter lift hole inD2 frac14 diameter pin inD3 frac14 diameter pad eye inDsk frac14 diameter skirt inDm frac14 mean vessel diameter inE frac14 modulus of elasticity psifr frac14 tail end radial force lbfL frac14 tail end longitudinal force lbfs frac14 shear load lb or lbinFa frac14 allowable stress combined loading psiFb frac14 allowable stress bending psiFc frac14 allowable stress compression psiFp frac14 allowable stress bearing pressure psiFs frac14 allowable stress shear psiFt frac14 allowable stress tension psiFy frac14 minimum specified yield stress psiI frac14 moment of inertia in4

Jw frac14 polar moment of inertia of weld in4

K frac14 end connection coefficientKL frac14 overall load factor combining impact and safety

factors 15ndash20Ki frac14 impact factor 025ndash05Kr frac14 internal moment coefficient in circular ring due

to radial loadKs frac14 safety factorKT frac14 internal tensioncompression coefficient in

circular ring due to radial loadLs frac14 length of skirtbase stiffenerstrut in

M frac14 moment in-lbMb frac14 bending moment in-lbMC frac14 circumferential moment in-lbML frac14 longitudinal moment in-lbMT frac14 torsional moment in-lbNb frac14 number of bolts used in tail beam or flange

lugN frac14 width of flange of tail beam with a web stiffener

(N frac14 10 without web stiffener)nL frac14 number of head or side lugsP frac14 pick end load lbPe frac14 equivalent load lbPL frac14 longitudinal load per lug lbPr frac14 radial load lbPT frac14 transverse load per lug lbRb frac14 radius of base ring to neutral axis inr frac14 radius of gyration of strut in

Rc frac14 radius of bolt circle of flange inSu frac14 minimum specified tensile stress of bolts psitb frac14 thickness of base plate intg frac14 thickness of gusset intL frac14 thickness of lug intP frac14 thickness of pad eye ints frac14 thickness of shell inT frac14 tail end load lbTb frac14 bolt pretension load lbsTt frac14 tangential force lbw1 frac14 fillet weld size shell to re-padw2 frac14 fillet weld size re-pad to shellw3 frac14 fillet weld size pad eye to lugw4 frac14 fillet weld size base plate to skirtw5 frac14 uniform load on vessel lbinWE frac14 design erection weight lbWL frac14 erection weight lbZ frac14 section modulus in3

a frac14 angular position for moment coefficients in basering clockwise from 0

b frac14 angle between parallel beams degreess frac14 stress combined psisb frac14 stress bending psisp frac14 stress bearing psisc frac14 stress compression psiscr frac14 critical buckling stress psisT frac14 stress tension psis frac14 shear stress psisT frac14 torsional shear stress psiq frac14 lift angle degrees

qB frac14 minimum bearing contact angle degreesqH frac14 sling angle to lift line horizontal degreesqv frac14 sling angle to lift line vertical degrees


Procedure 10-3 Lifting Attachments and Terminology







Tailing Trunnion

Utilizes reinforced openings in skirt with through pipePipe is removed after erection and the openings used asskirt manways

Shell Flange Lug

Lifting Device Utilizing Top Body Flanges



Miscellaneous Lugs WL lt 60 kips

Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear

tL frac14 PTethA D1THORNFs

Due to tension

tL frac14 PLethA D1THORNFt

Notes

1 Table 10-6 is based on an allowable stress of 137ksi

2 Design each lug for a 21 safety factor3 Design each lug for a minimum 10 side force

Hertzian Stress Bearing

sp frac14 0418

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiE

PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy

Shear Load in WeldType 1 greater of following

sw frac14 6PTB

2A2

sw frac14 PL2A

Type 2 Use design for top head lug

Table 10-6Lug dimensions

WL kips A D1 B C tL w1 WL kips A D1 B C tL w1

4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088

Figure 10-6 Dimensions and forces


Procedure 10-1 Transportation of Pressure Vessels

The transportation of a pressure vessel by ship bargeroad or rail will subject the vessel to one-time-onlystresses that can bend or permanently deform the vessel ifit is not adequately supported or tied down in the rightlocations The shipping forces must be accounted for toensure that the vessel arrives at its destination withoutdamage

It is very frustrating for all the parties involved to havea load damaged in transit and to have to return it to thefactory for repairs The cost and schedule impacts can bedevastating if a vessel is damaged in transit Certainminimal precautions can avoid the costly mistakes thatoften lead to problems Even when all precautions aremade however there is still the potential for damage dueto unforseen circumstances involved in the shipping andhandling process

Care should be taken to ensure that the size and locationof the shipping saddles tie-downs or lashing are adequateto hold the vessel but not deform the vessel Long thin-walled vessels such as trayed columns are especiallyvulnerable to these shipping forces The important thing toremember is that someone must take the responsibilityThe barge and rail people have their own concerns withregard to loading and lashing These may or may notcoincide with the concerns of the vessel designer

The shipping forces for ships barges trucks and railare contained in this procedure Each method of trans-portation has its own unique load schemes and resultingforces Barge shipping forces will differ from rail due tothe rocking motion of the seas Rail shipments howevergo around corners at high speed In addition rail forcesmust allow for the ldquohumpingrdquo of rail cars when they arejoined with the rest of the train Ocean shipments have toresist storms and waves without breaking free of theirlashings

Whereas horizontal vessels on saddles are designed forsome degree of loading in that position vertical vesselsare not The forces and moments that are used for thedesign of a vertical vessel assume the vessel is in itsoperating position Vertical vessels should generally bedesigned to be put on two saddles in a horizontal positionand transported by various means That is the purpose ofthis procedure Too often the details of transportation anderection are left in the hands of people who though wellversed in their particular field are not pressure vesselspecialists

Often vessels are transported by multiple means Thusthere will be handling operations between each successivemode of transportation Often a vessel must be moved byroad to the harbor and then transferred to a barge or shipOnce it reaches its destination it must be reloaded ontoroad or rail transport to the job site There it will be off-loaded and either stored or immediately erected A finaltransport may be necessary to move the vessel to thelocation where it will be finally erected At each handlingand transport phase there are different sets of forcesexerted on the vessel that must be accounted for

Shipping Saddles

The primary concern of the vessel designer is thelocation and construction of the shipping saddles to takethese forces without overstressing or damaging the vesselIf saddles are to be relocated by the transporter it isimportant that the new locations be reviewed Generallyonly two shipping saddles should be used However thismay not always be possible Remember that the reason forusing two saddles is that more than two saddles createsa statically indeterminate structure You are never assuredthat any given saddle is going to take more than itsapportioned load

Here are some circumstances that would allow for morethan two saddles to be used or for a special location of twosaddles

bull Transporter objects due to load on tiresbull Transporter objects due to load on barge or shipbull Very thin long vesselbull Heavy-walled vessels for spreading load on ship ortransporters

Shipping saddles can be constructed of wood or steel orcombinations The saddles should be attached to the vesselwith straps or bolts so that the vessel can be movedwithout having to reattach the saddle Horizontal vesselsmay be moved on their permanent saddles but should bechecked for the loadings due to shipping forces andclearances for boots and nozzles Shipping saddles shouldhave a minimum contact angle of 120 just like perma-nent saddles Provisions for jacking can be incorporatedinto the design of the saddles to allow loading andhandling operations without a crane(s)







Vessel Stresses


Lashing















Minimum Saddle

Width

D lt 13 ft-0 in 120 11 in


D gt 24 ft-0 in 160 23 in

























Alabama 150 16 16 180000

Arizona 120 14 16 250000

Arkansas 100 14 14 120000

California 120 14 16 220000

Colorado 130 17 16 228000

Connecticut 120 15 16 250000

Delaware 120 15 15 250000

Florida 150 16 16 250000

Georgia 100 14 14 120000

Idaho 110 16 15rsquo6 200000


Indiana 110 16 15 250000

Iowa 120 18 16 250000

Kansas 126 16rsquo6 16 250000

Kentucky 110 16 15 250000

Louisiana 125 18 16 250000

Maine 125 16 16 250000



Michigan 150 16 15 230000


Mississippi 100 14 14 120000

Missouri 100 14 14 120000

Montana 110 18 17 240000


Nevada 105 17 16 240000


New Jersey 120 18 16 220000

New Mexico 120 14 16 250000

New York 120 14 14 160000



Ohio 100 14 14rsquo10 120000

Oklahoma 100 16 16 212000

Oregon 105 14 16 220000





Tennessee 120 16 15 250000

Texas 125 20 18rsquo11 252000

Utah 125 15 16rsquo6 250000

Vermont 100 15 14 150000

Virginia 150 14 15 150000

Washington 150 14 16 200000


Wisconsin 150 16 16 250000

Wyoming 110 18 17 252000











Pitch

Cases 3a and 3b


General


2pT

2Rqp180

F frac14 00214WRq

T2

f1 frac14 tan1

aR1


T21


Fz frac14 Fp cos f1


Fz frac14 Fp cos f1

Roll

Case 2a q2 [ 30 max

Case 2b


f2 frac14 tan1ed

R2 frac14 esin f2


T22


Fx frac14 FR cos f2


Fx frac14 FR cos f2











Axle Loads














θ


DOLLIEe


PROJECTEDLOAD POINT








Fx 05 10 095 10

Fy 15 20 13 15

Fz 10 15 15 15



Fx Q1 frac14 WsL2L1

thorn FxB

2A

Q2 frac14 WsL3L1

thorn FxB

2A



Fz

Q1 frac14 WsL2L1

thorn FzB

L1

Q2 frac14 WsL3L1

thorn FzB

L1



Case 1




3

i2L1

Q2 frac14 W Q1

M1 frac14 wL22

2

M2 frac14 Q1

Q1

2w L2

M3 frac14 wL23

2



Q1X1


Case 2

w1 frac14 W1

L2

w2 frac14 W2

L3

Q1 frac14 WL6

L1

Q2 frac14 W Q1

M1 frac14 w1L24

2


2 w1L2

1

8

M3 frac14 w2L23

2


Case 3

Q1 frac14 WL1


2


Q2 frac14 WL1



2



2W

M2 frac14 Q1L1 WL21


Mx frac14 Q1XWX2


Case 4

w1 frac14 W1

L2

w2 frac14 W2

L3


2L1


2L1


Mx frac14 Q1X w1X2

2














FZ frac14 KZ W


w1 frac14 W=LT




0

Q1 frac14 w1


3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1




FY frac14 KY W


WT frac14 Wthorn FY


w2 frac14 WT=LT


0

Q10 frac14 w2


3

i=2L1

Q20 frac14 WT Q1

0


M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given


LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088








Vessel Stresses


Lashing















Minimum Saddle

Width

D lt 13 ft-0 in 120 11 in


D gt 24 ft-0 in 160 23 in

























Alabama 150 16 16 180000

Arizona 120 14 16 250000

Arkansas 100 14 14 120000

California 120 14 16 220000

Colorado 130 17 16 228000

Connecticut 120 15 16 250000

Delaware 120 15 15 250000

Florida 150 16 16 250000

Georgia 100 14 14 120000

Idaho 110 16 15rsquo6 200000


Indiana 110 16 15 250000

Iowa 120 18 16 250000

Kansas 126 16rsquo6 16 250000

Kentucky 110 16 15 250000

Louisiana 125 18 16 250000

Maine 125 16 16 250000



Michigan 150 16 15 230000


Mississippi 100 14 14 120000

Missouri 100 14 14 120000

Montana 110 18 17 240000


Nevada 105 17 16 240000


New Jersey 120 18 16 220000

New Mexico 120 14 16 250000

New York 120 14 14 160000



Ohio 100 14 14rsquo10 120000

Oklahoma 100 16 16 212000

Oregon 105 14 16 220000





Tennessee 120 16 15 250000

Texas 125 20 18rsquo11 252000

Utah 125 15 16rsquo6 250000

Vermont 100 15 14 150000

Virginia 150 14 15 150000

Washington 150 14 16 200000


Wisconsin 150 16 16 250000

Wyoming 110 18 17 252000











Pitch

Cases 3a and 3b


General


2pT

2Rqp180

F frac14 00214WRq

T2

f1 frac14 tan1

aR1


T21


Fz frac14 Fp cos f1


Fz frac14 Fp cos f1

Roll

Case 2a q2 [ 30 max

Case 2b


f2 frac14 tan1ed

R2 frac14 esin f2


T22


Fx frac14 FR cos f2


Fx frac14 FR cos f2











Axle Loads














θ


DOLLIEe


PROJECTEDLOAD POINT








Fx 05 10 095 10

Fy 15 20 13 15

Fz 10 15 15 15



Fx Q1 frac14 WsL2L1

thorn FxB

2A

Q2 frac14 WsL3L1

thorn FxB

2A



Fz

Q1 frac14 WsL2L1

thorn FzB

L1

Q2 frac14 WsL3L1

thorn FzB

L1



Case 1




3

i2L1

Q2 frac14 W Q1

M1 frac14 wL22

2

M2 frac14 Q1

Q1

2w L2

M3 frac14 wL23

2



Q1X1


Case 2

w1 frac14 W1

L2

w2 frac14 W2

L3

Q1 frac14 WL6

L1

Q2 frac14 W Q1

M1 frac14 w1L24

2


2 w1L2

1

8

M3 frac14 w2L23

2


Case 3

Q1 frac14 WL1


2


Q2 frac14 WL1



2



2W

M2 frac14 Q1L1 WL21


Mx frac14 Q1XWX2


Case 4

w1 frac14 W1

L2

w2 frac14 W2

L3


2L1


2L1


Mx frac14 Q1X w1X2

2














FZ frac14 KZ W


w1 frac14 W=LT




0

Q1 frac14 w1


3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1




FY frac14 KY W


WT frac14 Wthorn FY


w2 frac14 WT=LT


0

Q10 frac14 w2


3

i=2L1

Q20 frac14 WT Q1

0


M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given


LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088


























Alabama 150 16 16 180000

Arizona 120 14 16 250000

Arkansas 100 14 14 120000

California 120 14 16 220000

Colorado 130 17 16 228000

Connecticut 120 15 16 250000

Delaware 120 15 15 250000

Florida 150 16 16 250000

Georgia 100 14 14 120000

Idaho 110 16 15rsquo6 200000


Indiana 110 16 15 250000

Iowa 120 18 16 250000

Kansas 126 16rsquo6 16 250000

Kentucky 110 16 15 250000

Louisiana 125 18 16 250000

Maine 125 16 16 250000



Michigan 150 16 15 230000


Mississippi 100 14 14 120000

Missouri 100 14 14 120000

Montana 110 18 17 240000


Nevada 105 17 16 240000


New Jersey 120 18 16 220000

New Mexico 120 14 16 250000

New York 120 14 14 160000



Ohio 100 14 14rsquo10 120000

Oklahoma 100 16 16 212000

Oregon 105 14 16 220000





Tennessee 120 16 15 250000

Texas 125 20 18rsquo11 252000

Utah 125 15 16rsquo6 250000

Vermont 100 15 14 150000

Virginia 150 14 15 150000

Washington 150 14 16 200000


Wisconsin 150 16 16 250000

Wyoming 110 18 17 252000











Pitch

Cases 3a and 3b


General


2pT

2Rqp180

F frac14 00214WRq

T2

f1 frac14 tan1

aR1


T21


Fz frac14 Fp cos f1


Fz frac14 Fp cos f1

Roll

Case 2a q2 [ 30 max

Case 2b


f2 frac14 tan1ed

R2 frac14 esin f2


T22


Fx frac14 FR cos f2


Fx frac14 FR cos f2











Axle Loads














θ


DOLLIEe


PROJECTEDLOAD POINT








Fx 05 10 095 10

Fy 15 20 13 15

Fz 10 15 15 15



Fx Q1 frac14 WsL2L1

thorn FxB

2A

Q2 frac14 WsL3L1

thorn FxB

2A



Fz

Q1 frac14 WsL2L1

thorn FzB

L1

Q2 frac14 WsL3L1

thorn FzB

L1



Case 1




3

i2L1

Q2 frac14 W Q1

M1 frac14 wL22

2

M2 frac14 Q1

Q1

2w L2

M3 frac14 wL23

2



Q1X1


Case 2

w1 frac14 W1

L2

w2 frac14 W2

L3

Q1 frac14 WL6

L1

Q2 frac14 W Q1

M1 frac14 w1L24

2


2 w1L2

1

8

M3 frac14 w2L23

2


Case 3

Q1 frac14 WL1


2


Q2 frac14 WL1



2



2W

M2 frac14 Q1L1 WL21


Mx frac14 Q1XWX2


Case 4

w1 frac14 W1

L2

w2 frac14 W2

L3


2L1


2L1


Mx frac14 Q1X w1X2

2














FZ frac14 KZ W


w1 frac14 W=LT




0

Q1 frac14 w1


3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1




FY frac14 KY W


WT frac14 Wthorn FY


w2 frac14 WT=LT


0

Q10 frac14 w2


3

i=2L1

Q20 frac14 WT Q1

0


M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given


LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088










Pitch

Cases 3a and 3b


General


2pT

2Rqp180

F frac14 00214WRq

T2

f1 frac14 tan1

aR1


T21


Fz frac14 Fp cos f1


Fz frac14 Fp cos f1

Roll

Case 2a q2 [ 30 max

Case 2b


f2 frac14 tan1ed

R2 frac14 esin f2


T22


Fx frac14 FR cos f2


Fx frac14 FR cos f2











Axle Loads














θ


DOLLIEe


PROJECTEDLOAD POINT








Fx 05 10 095 10

Fy 15 20 13 15

Fz 10 15 15 15



Fx Q1 frac14 WsL2L1

thorn FxB

2A

Q2 frac14 WsL3L1

thorn FxB

2A



Fz

Q1 frac14 WsL2L1

thorn FzB

L1

Q2 frac14 WsL3L1

thorn FzB

L1



Case 1




3

i2L1

Q2 frac14 W Q1

M1 frac14 wL22

2

M2 frac14 Q1

Q1

2w L2

M3 frac14 wL23

2



Q1X1


Case 2

w1 frac14 W1

L2

w2 frac14 W2

L3

Q1 frac14 WL6

L1

Q2 frac14 W Q1

M1 frac14 w1L24

2


2 w1L2

1

8

M3 frac14 w2L23

2


Case 3

Q1 frac14 WL1


2


Q2 frac14 WL1



2



2W

M2 frac14 Q1L1 WL21


Mx frac14 Q1XWX2


Case 4

w1 frac14 W1

L2

w2 frac14 W2

L3


2L1


2L1


Mx frac14 Q1X w1X2

2














FZ frac14 KZ W


w1 frac14 W=LT




0

Q1 frac14 w1


3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1




FY frac14 KY W


WT frac14 Wthorn FY


w2 frac14 WT=LT


0

Q10 frac14 w2


3

i=2L1

Q20 frac14 WT Q1

0


M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given


LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088












Axle Loads














θ


DOLLIEe


PROJECTEDLOAD POINT








Fx 05 10 095 10

Fy 15 20 13 15

Fz 10 15 15 15



Fx Q1 frac14 WsL2L1

thorn FxB

2A

Q2 frac14 WsL3L1

thorn FxB

2A



Fz

Q1 frac14 WsL2L1

thorn FzB

L1

Q2 frac14 WsL3L1

thorn FzB

L1



Case 1




3

i2L1

Q2 frac14 W Q1

M1 frac14 wL22

2

M2 frac14 Q1

Q1

2w L2

M3 frac14 wL23

2



Q1X1


Case 2

w1 frac14 W1

L2

w2 frac14 W2

L3

Q1 frac14 WL6

L1

Q2 frac14 W Q1

M1 frac14 w1L24

2


2 w1L2

1

8

M3 frac14 w2L23

2


Case 3

Q1 frac14 WL1


2


Q2 frac14 WL1



2



2W

M2 frac14 Q1L1 WL21


Mx frac14 Q1XWX2


Case 4

w1 frac14 W1

L2

w2 frac14 W2

L3


2L1


2L1


Mx frac14 Q1X w1X2

2














FZ frac14 KZ W


w1 frac14 W=LT




0

Q1 frac14 w1


3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1




FY frac14 KY W


WT frac14 Wthorn FY


w2 frac14 WT=LT


0

Q10 frac14 w2


3

i=2L1

Q20 frac14 WT Q1

0


M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given


LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088








Fx 05 10 095 10

Fy 15 20 13 15

Fz 10 15 15 15



Fx Q1 frac14 WsL2L1

thorn FxB

2A

Q2 frac14 WsL3L1

thorn FxB

2A



Fz

Q1 frac14 WsL2L1

thorn FzB

L1

Q2 frac14 WsL3L1

thorn FzB

L1



Case 1




3

i2L1

Q2 frac14 W Q1

M1 frac14 wL22

2

M2 frac14 Q1

Q1

2w L2

M3 frac14 wL23

2



Q1X1


Case 2

w1 frac14 W1

L2

w2 frac14 W2

L3

Q1 frac14 WL6

L1

Q2 frac14 W Q1

M1 frac14 w1L24

2


2 w1L2

1

8

M3 frac14 w2L23

2


Case 3

Q1 frac14 WL1


2


Q2 frac14 WL1



2



2W

M2 frac14 Q1L1 WL21


Mx frac14 Q1XWX2


Case 4

w1 frac14 W1

L2

w2 frac14 W2

L3


2L1


2L1


Mx frac14 Q1X w1X2

2














FZ frac14 KZ W


w1 frac14 W=LT




0

Q1 frac14 w1


3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1




FY frac14 KY W


WT frac14 Wthorn FY


w2 frac14 WT=LT


0

Q10 frac14 w2


3

i=2L1

Q20 frac14 WT Q1

0


M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given


LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088




Case 1




3

i2L1

Q2 frac14 W Q1

M1 frac14 wL22

2

M2 frac14 Q1

Q1

2w L2

M3 frac14 wL23

2



Q1X1


Case 2

w1 frac14 W1

L2

w2 frac14 W2

L3

Q1 frac14 WL6

L1

Q2 frac14 W Q1

M1 frac14 w1L24

2


2 w1L2

1

8

M3 frac14 w2L23

2


Case 3

Q1 frac14 WL1


2


Q2 frac14 WL1



2



2W

M2 frac14 Q1L1 WL21


Mx frac14 Q1XWX2


Case 4

w1 frac14 W1

L2

w2 frac14 W2

L3


2L1


2L1


Mx frac14 Q1X w1X2

2














FZ frac14 KZ W


w1 frac14 W=LT




0

Q1 frac14 w1


3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1




FY frac14 KY W


WT frac14 Wthorn FY


w2 frac14 WT=LT


0

Q10 frac14 w2


3

i=2L1

Q20 frac14 WT Q1

0


M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given


LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088



Case 3

Q1 frac14 WL1


2


Q2 frac14 WL1



2



2W

M2 frac14 Q1L1 WL21


Mx frac14 Q1XWX2


Case 4

w1 frac14 W1

L2

w2 frac14 W2

L3


2L1


2L1


Mx frac14 Q1X w1X2

2














FZ frac14 KZ W


w1 frac14 W=LT




0

Q1 frac14 w1


3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1




FY frac14 KY W


WT frac14 Wthorn FY


w2 frac14 WT=LT


0

Q10 frac14 w2


3

i=2L1

Q20 frac14 WT Q1

0


M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given


LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088













FZ frac14 KZ W


w1 frac14 W=LT




0

Q1 frac14 w1


3

i2 L1

Q10 frac14 Q1 F2

Q2 frac14 W Q1




FY frac14 KY W


WT frac14 Wthorn FY


w2 frac14 WT=LT


0

Q10 frac14 w2


3

i=2L1

Q20 frac14 WT Q1

0


M1 frac14 wn L22=2

M2 frac14 Q10Q1

0=2 wn

L2

M3 frac14 wn L23=2

Sample Problem

Given


LT

M2

M1 M3

WT

w1 or w2

W

L2

Q1 Q2

Fz

B

L1 L3




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088




Calculation



WT frac14 Wthorn FY







0

Q1 frac14 w1


3

i2 L1


i2124

frac14 379 kips






0

Q10 frac14 w2


3

i2 L1


i2124

frac14 569 kips

Q10 frac14 WT Q1

0 frac14 11115 569

frac14 5425 kips


M1 frac14 w2 L2

2

2 frac14

658242

2


M2 frac14 Q10Q1

0=2 w2

L2


M3 frac14 w2 L2

3

2 frac14

658212
















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088















T1 frac14 cos b

FxB4RN

thorn Fy Ws

4N


Ar frac14 T1

Ft



p

r




p


T2 frac14 saAs


T3 frac14 Ws

2N




bull Load on weld f

f frac14 T4[




p

1285


Ar frac14 TFt

Use


Pe frac14 4TpRK




31616 in 239 278

frac14 in 318 37151616 in 398 464

⅜ in 477 55771616 in 556 650

Kipsin of weld


Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088



Alternate Procedure





Calculation


Y frac14 R sin q=q

bull Find angle a

a frac14 cos1 Y=R


X frac14 R sina



Notes




Example




X frac14 R sin a

X frac14 925 (sin 424) frac14 624 in



3435 kips



K2 W

K1 W

B

AX

TY

R

β α

θ



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088



























Contractor Duties




Tolerances




Shims









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088









8 I




Bolting






Grouting






18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088



18








BOO

M 9

6m

90009000

620002000017000 100

0

100

0 13000

MAST 24m

oslash3840

oslash4160

1692

1829

1026

987

4

BOO

M 36m


Transporta

tionandErectio

nofPressure

Vessels

663

Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088



Steps in Design









g OtherTail end














Allowable Stresses

Per AISC

Tension







Shear





E70XX 11 200 lb=in or 15 800 psi

Bending


Bearing

Fp frac14 09Fy

Combined

Shear and tension

sa

Fathorn sFs

1



Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088




Fathorn sb

Fb 1 or

sT

FTthorn sb

Fb 1


Notation

A frac14 area in2



























Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088










Tailing Trunnion


Shell Flange Lug





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088





Calculations

Due to bending

tL frac14 6PTB

A2Fb

Due to shear


Due to tension


Notes




sp frac14 0418


PtL

ethR1 R2THORN

13

R1R2

vuuutlt 2Fy


sw frac14 6PTB

2A2

sw frac14 PL2A




4 3 088 15 2 05 025 20 7 175 3 3 1 038

6 35 1 163 2 063 025 25 7 238 4 4 1 044

8 4 113 175 2 063 025 35 8 238 4 4 1125 05

10 45 125 2 2 075 025 40 8 238 4 4 1125 063

12 5 138 213 3 088 025 45 8 288 4 4 1125 063

14 55 15 238 3 1 038 50 10 288 4 4 125 075

16 65 163 25 3 1 038 55 10 288 4 4 125 075

18 7 175 275 3 1 038 60 10 288 4 4 125 088



Transportation and Erection of Pressure...

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