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II Expansion Joints

Description

CH AP TER

ݸ¿°¬»® ë Û¨°¿²­·±² Ö±·²¬­ This chapter explains how ÝßÛÍßÎ ×× models various expansion joints

ײ ̸·­ ݸ¿°¬»® Simple Bellows with Pressure Thrust

Complex Model

Tie Rod

Tie Rod Model

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

Û¨°¿²­·±² Ö±·²¬­

Í·³°´» Þ»´´±©­ ©·¬¸ Ю»­­«®» ̸®«­¬ Bellows expansion joints can be modeled with either a zero or a finite length

When finite length bellows are used,

either the bending or the transverse stiffness must be left blank

ÝßÛÍßÎ ×× will calculate the exact stiffness coefficient for the term left blank

For finite length expansion joints,

the user is recommended to leave the Bending Stiffness field blank,

and to enter the lateral stiffness given by the manufacturer into the Transverse Stiffness field on the expansion joint spreadsheet

The lateral stiffness may be computed from the axial stiffness (if not provided) from the equation:

KTR = (3/2) (KAX) (D/L) 2 If the bending stiffness is given,

its value should be approximately (within 1%) equal to:

KBEND =(1/2) (KAX) (D2) ( /180) KAX

For zero length expansion joints:

KBEND = (1/8) (KAX) (D2) ( /180) When a zero length expansion joint is used,

ÝßÛÍßÎ ×× will use either the preceding or the following element to determine the axial direction of the bellows stiffnesses

The preceding element is checked first

Bellows are very fragile under torsional loading

It is recommended that accurate torsional stiffnesses and allowable torsional rotations be obtained from the vendor

Systems using untied bellows should either be of very low pressure or adequately anchored to withstand the possibly large thrust loads developed due to the unrestrained bellows

Bellows and any other miscellaneous weights should be added to flanges on either side of the bellows (or can be added as concentrated forces)

This is particularly true when the bellow is part of a hanger sizing weight calculation

A zero or blank Bellows ID results in a zero pressure thrust

The Bellows ID is the diameter used to find the area for pressure thrust calculations

The total thrust load is applied at the From and To ends of the bellows,

and is used to open the bellows (providing the pressure is positive)

The magnitude of the thrust load is P * A,

where P is the pressure in the pipe above atmospheric,

A = /4 * (Bellows ID) 2 Many manufacturers specify the effective area of the bellows

The Bellows ID for ÝßÛÍßÎ ×× input may be calculated by using the following equation:

Bellows ID =

In the system shown below,

the untied bellows runs between the nodes 8 and 9

The elbow at 11 is anchored to take the thrust load developed in the bellows

The manufacturer's specification for the joint's axial stiffness is 6530 lb

with a transverse stiffness of 3250 lb

The bending stiffness is left blank,

and will be calculated by ÝßÛÍßÎ ×× since the bellows has a finite length

The pump and the baseplate at 5 must be able to withstand the large axial force that may develop due to pressure thrust in the bellows

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ݸ¿°¬»® ë Û¨°¿²­·±² Ö±·²¬­

Þ»´´±©­ ©·¬¸ Ю»­­«®» ̸®«­¬

Aeff= 67

(automatically applied by ÝßÛÍßÎ ××)

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

Û¨°¿²­·±² Ö±·²¬­

Ì·»¼ Þ»´´±©­ ó Í·³°´» ª­ò ݱ³°´»¨ Ó±¼»´ Complex models of expansion joints are much more difficult to build than simple models

Unfortunately there are no hard and fast rules for when to use simple models and when to use complex models

The following guidelines are presented to aid the engineer in making this decision

Complex models are used whenever a failure is being investigated

Complex models are normally used when the pipe diameter and number of convolutions become large

Complex models are used when nuts are only on the outside of the flange,

allowing the tie bars to only carry tension

Complex models give good values for the load distribution in the tie bars

Simple models give no indication of the load distribution

In some cases,

where the tie bars combine to resist relative bending of the joint ends,

one pair of tie-bars can be in compression while the other pair is in tension

This effective redistribution of load in the tie bars will never be observed in a simple model

When this does occur,

and if the tie bars are very long,

buckling of the rods in the complex model should be investigated (evaluate whether the rods can withstand the compressive forces reported in the output report)

The single tied bellows is designed to absorb movement by lateral deflection only

There is no axial deflection or relative bending rotations at the joint ends

Simple models should only be used where the tie bars are either guaranteed to be carrying tension,

or have nuts on either side of the flange and so will carry compression if needed

Be sure to enter the lateral instead of the bending spring rate from the manufacturer’s catalog

See the previous discussion for a simple bellows for more information about bellows stiffnesses

The weights of the bellows and associated hardware should be added to the flange weights on either side of the bellows

This is particularly true if the expansion joint is between a hanger to be sized and an anchor

The expansion joint user should be sure to check the displacement limits for the expansion joint once the protected equipment loads are within the allowables

ÝßÛÍßÎ ×× has a processor called EJMA Expansion Joint Rating accessible through the Analysis option of the Main Menu,

which helps the user to compute relative bellows movements for evaluating the bellows distortion

Simple models of single tied bellows are built by entering a large axial stiffness

This axial stiffness simulates the tie bars,

preventing relative axial movement of the bellows

Tie rods may also be modeled with a single rigid element along the centerline of the bellows,

with zero weight and rotational restraints,

prevents the ends of the joint from rotating relative to one another

In reality the tie bars being offset from the centerline prevent this rotation

The complex models are built by running pipe elements whose diameter is equal to the diameter of the tie-bars,

and whose wall thickness is equal to half of the tie-bar diameter,

between rigid elements that extend normal to the pipe axis and from the centerline and to their intersection with the tie-bar centerline (See the following illustration)

Some manufacturers feel that friction at the tie bar ends,

plus other effects serve to limit the overall lateral flexibility of this joint

For lack of a better value,

a 30% increase in lateral stiffness is sometimes used to compensate for these frictional effects

Field situations such as loose nuts on tie-bars,

can be modeled using the complex expansion joint model

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ݸ¿°¬»® ë Û¨°¿²­·±² Ö±·²¬­

Ì·»¼ Þ»´´±©­ Û¨°¿²­·±² Ö±·²¬ ó Í·³°´» Ó±¼»´ Compute the lateral stiffness for the bellows

The flexible length of the bellows is not listed in most expansion joint catalogs

The listed lengths include the rigid end pieces such as flanges or pipe ends

Since the transverse stiffness is based on the flexible length,

the flexible length must be known

A very simple way of pulling this value from the catalog is to examine the incremental increase in overall length of the joint as additional convolutions are added

With all convolutions the same length,

this incremental length can be used to calculate the total flexible length

In this example the total length of a 4 convolution joint is 8 in

and the total length of an 8 convolution joint is 12 in

This means that the extra four convolutions add 4 in

so the length of all twelve convolutions is 12 in

(This also indicates that the rigid end pieces on this joint of 4,

Flexible Convolution Length = 12 in

KTR = =

0)2 885

Ì·»¼ Þ»´´±©­ ó Í·³°´» Ó±¼»´

Zero-weight rigid element (tie rod)

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

Û¨°¿²­·±² Ö±·²¬­

Build the ÝßÛÍßÎ ×× model of the flexible portion of the expansion joint

Note how the rotational restraints between nodes 29 and 30 keep the two flanges parallel

In the field,

the tie bars at four points around the expansion joint will keep the flanges parallel

(The flanges and the tie bars form a parallelogram upon lateral deflection

Ì·»¼ Þ»´´±©­ ó Í·³°´» Ó±¼»´ ݱ²¬·²«»¼ òòò

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ݸ¿°¬»® ë Û¨°¿²­·±² Ö±·²¬­

Ì·»¼ Þ»´´±©­ Û¨°¿²­·±² Ö±·²¬ ó ݱ³°´»¨ Ó±¼»´ In the system shown below the flexible joint is between the nodes 30 and 35

The flanged ends of the joint are modeled as the rigid elements 20 to 30 and 35 to 45

Additional rigid elements,

perpendicular to the pipe axis,

The tie bars are 1-in

The following nodal layout and input is used to build a comprehensive model of the tied bellows

Ì·»¼ Þ»´´±©­ ݱ³°´»¨ Ó±¼»´

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

Û¨°¿²­·±² Ö±·²¬­

Ì·»¼ Þ»´´±©­ ݱ³°´»¨ Ó±¼»´óݱ²¬·²«»¼

Weightless rigid elements extend from the flange centerline to the outside edge of the flanges where the tie rods are attached

Only 2 of eight element inputs shown

Tie Rod will usually be at ambient temperature

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ݸ¿°¬»® ë Û¨°¿²­·±² Ö±·²¬­

˲·ª»®­¿´ Û¨°¿²­·±² Ö±·²¬­ ó Í·³°´» Ó±¼»´­ Please refer to the previous models of bellows expansion joints for specific notes relating to individual bellows designs,

and to some comparisons of simple and complex expansion joint input

The tied universal bellows is designed to absorb movement by lateral deflection only

There is no axial deflection or relative bending rotations at the joint ends

Lateral instead of the bending spring rates from the manufacturer’s catalog should be entered

See the “simple bellows” discussion for more information about bellows stiffnesses

Manufacturers publish a wide variety of data for universal expansion joints

In most cases the published spring rates are for the universal joint as a whole assembly

When the lateral stiffness is given for the whole assembly the simple or complex models of single bellows can be used

In this case the manufacturer must also provide a cumulative assembly displacement limit so that the piping designer can verify that neither of the bellows are over-extended

Many universal expansion joint assemblies have stops along the tie-bars that are connected to the center spool-piece

These stops are designed to prevent over-extension of the bellows and can be modeled in the complex universal joint model

For the simple universal joint models,

the user must check the results to verify that the stops are not engaged

Stops should typically be considered a safety feature,

and should not be included as a working part of the design,

unless particular attention is paid to the design surrounding the stop components

The expansion joint user should be sure to check the displacement limits for each of the expansion joints once the protected equipment loads are within the allowables

ÝßÛÍßÎ ×× has a program called Expansion Joint Rating

- EJMA,

which helps the user compute relative bellows movements for evaluating the convolution’s strength

This program only works on single bellows,

and so the user would need to model and then check each bellows in the universal assembly

Some manufacturers feel that friction at the tie bar ends,

plus other effects serve to limit the overall lateral flexibility of this joint

For lack of a better value,

a 10% increase in overall lateral stiffness is sometimes used to compensate for these frictional effects

The complex models are built by running pipe elements,

whose diameter is equal to the diameter of the tie-bars,

and whose wall thickness is equal to half of the tie-bar diameter,

between rigid elements that extend normal to the pipe axis and from the centerline and to their intersection with the tie-bar centerline

The weights of the bellows and associated hardware should be added to the flange weights on either side of the bellows

This is particularly true if the expansion joint is between a hanger to be sized and an anchor

In-situ field effects like loose nuts on tie-bars,

can be modeled using the complex expansion joint model

Descriptions of various universal models display in the following figures

The two models shown also have example inputs given

Simple models should only be used when the user knows that both ends of the tie-bars will be fixed to the flanges,

when there are nuts on both sides of the flange

(The top drawing shows nuts on only one side of the flange at the left end

This configuration should be modeled with a complex joint model unless the user is sure that all tie-bars will remain in tension

) The top model is used when the analyst is provided with global assembly data for the universal,

the assembly lateral stiffness

The second model is used when the analyst is given angular spring rates for each of the two bellows used in the model

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ëóïð

Û¨°¿²­·±² Ö±·²¬­

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ݸ¿°¬»® ë Û¨°¿²­·±² Ö±·²¬­

When provided individual bellows angular stiffness:

˲·ª»®­¿´ Û¨°¿²­·±² Ö±·²¬­ ó Í·³°´» Ó±¼»´­

Note: This model does not show the addition of any extra hardware or bellows weights which could affect load distribution and spring hanger design in the area

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ëóïï

ëóïî

Û¨°¿²­·±² Ö±·²¬­

When provided individual bellows angular stiffness:

˲·ª»®­¿´ Û¨°¿²­·±² Ö±·²¬­ ó Í·³°´» Ó±¼»´­ ײ¼·ª·¼«¿´ Þ»´´±©­

Note: The rigid tie bar(s) should be modeled at the ambient temperature

Note: Pressure thrust is contained by double-nutted tie rods

Effective ID and axial restraints can be eliminated

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ݸ¿°¬»® ë Û¨°¿²­·±² Ö±·²¬­

ëóïí

˲·ª»®­¿´ Û¨°¿²­·±² Ö±·²¬­ ó Í·³°´» Ó±¼»´­ ײ¼·ª·¼«¿´ Þ»´´±©­

Note: This model does not show the addition of any extra hardware or bellows weights,

which could affect weight load distribution and spring hanger design in the area

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ëóïì

Û¨°¿²­·±² Ö±·²¬­

˲·ª»®­¿´ Ö±·²¬ ó ݱ³°®»¸»²­·ª» Ì·» α¼ The comprehensive universal joint model involves defining,

all tie rods and connections between tie rods and end plates

The following groups illustrate the method used in constructing the universal expansion joint model shown above

—Rigid Elements (Flanges) — 15-17 / 31-33 —Rigid Elements normal to the pipe axis and between the pipe and tie bar centerlines

At the end where there are nuts on either side of the flange,

fixing the tie-bar to the flange: 33-1033 / 33-2033 / 33-3033 —Rigid Elements normal to the pipe axis,

and between the pipe and tie-bar centerlines

At the end where there are nuts only on the backside of the flange: 15-1015 / 15-2015 / 15-3015 ——Intermediate lateral tee supports (Rigid) — 23-1023 / 23-2023 / 23-3023 25-1025 / 25-2025 / 25-3025 ——Tie-bars — 1033-1034-1035-1036 2033-2034-2035-2036 3033-3034-3035-3036 — Restraints with connecting nodes at the tension-only flange end

—— RESTR NODE = 1036 CNODE = 1015 TYPE =

Z RESTR NODE = 2036 CNODE = 2015 TYPE =

Z RESTR NODE = 3036 CNODE = 3015 TYPE =

Z — Restraints with connecting nodes at the intermediate support points

RESTR NODE = 1035 CNODE = 1023 TYPE = Y ,

Z RESTR NODE = 2035 CNODE = 2023 TYPE = Y ,

Z RESTR NODE = 3035 CNODE = 3023 TYPE = Y ,

Z RESTR NODE = 1034 CNODE = 1025 TYPE = Y ,

Z RESTR NODE = 2034 CNODE = 2025 TYPE = Y ,

Z RESTR NODE = 3034 CNODE = 3025 TYPE = Y ,

©©©ò½¿¼º¿³·´§ò½±³ ÛÓ¿·´æ½¿¼­»®ªîïั¬³¿·´ò½±³ ̸» ¼±½«³»²¬ ·­ º±® ­¬«¼§ ±²´§ô·º ¬±®¬ ¬± §±«® ®·¹¸¬­ô°´»¿­» ·²º±®³ «­ô©» ©·´´ ¼»´»¬»

ݸ¿°¬»® ë Û¨°¿²­·±² Ö±·²¬­

ëóïë

˲·ª»®­¿´ Ö±·²¬ É·¬¸ Ô¿¬»®¿´ ݱ²¬®±´ ͬ±°­ ó ݱ³°®»¸»²­·ª» Ì·» α¼ Ó±¼»´ Double-acting restraints with connecting nodes and gaps are used to model stop gaps along the tie bars

Stops along the tiebars are installed to restrict lateral motion at each end of the universal joint

The following groups illustrate the method used in constructing the universal joint with lateral stops shown above

Only the right side tie rod elements are shown below

— Standard pipe elements — 34-36 / 36-38 — Rigid flange elements — 30-32 / 40-42 — Bellows elements — 32-34 / 38-40 — Rigid elements from the pipe to the tie-bar centerline — (Normal to the pipe axis) 30-1030 / 36-1036 / 42-1042 — Tie-bar elements — 1003-1002 / 1002-1001 — Restraints with connecting nodes — RESTR NODE=1001 CNODE = 1042 TYPE = +Y ,

Z RESTR NODE=1002 CNODE = 1036 TYPE = Y w/gap=1

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Ø·²¹»¼ Ö±·²¬ The hinged joint is defined using a zero length expansion joint with axial,

and torsional stiffnesses rigid

The bending stiffness is set equal to the bending stiffness of the hinge

Hinge directions are defined using restraints and connecting nodes

The restraint line of action is always normal to the hinge axis

Hinged joints are designed to take pressure thrust

The analyst should make sure that the joint manufacturer is aware of the design loads in the hinges

Some expansion joint manufacturers believe that the hinge friction can provide considerable additional resistance to bending

Certainly as the axial load the hinge is to carry becomes large,

this “hinge friction” effect will increase

Approximations to this increase in bending stiffness can be made by increasing the stiffness of the bellows in proportion to the axial load on the hinge

The expansion joint manufacturer can hopefully provide assistance here

Several typical geometries for hinged expansion joints are shown in the figures below:

In the example that follows,

the hinged joint is zero length and is defined between nodes 45 and 46

“X” is the hinge axis,

all relative rotations are permitted between 45 and 46 about the X axis

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The following figures display the coding of the hinged joint for the model shown on the bottom of the previous page

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Í´±¬¬»¼ Ø·²¹» Ö±·²¬ ó Í·³°´» Ó±¼»´ The hinged joint is defined using a zero length expansion joint and rigid elements with zero weight to define the interaction of the hinge geometry

Hinge directions are defined using restraints with connecting nodes

The restraint line of action is always normal to the hinge axis

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Elements from 10 to 15 and from 16 to 20 are weightless 9 in

Note: In this model,

the relative rotation at the hinge about the “Y” axis is assumed to be zero

The slots on either side will provide some limit to this Y rotation

In most applications of this type,

the relative Y rotation is zero because the problem is kept planar using guides

A good first pass can be made using the model shown,

then if the analysis shows that the RY restraint between nodes 15 and 16 is supporting load,

a further refinement to the model can be made

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Í´±¬¬»¼ Ø·²¹» Ö±·²¬ ó ݱ³°®»¸»²­·ª» Ó±¼»´ This model is somewhat different from the previous model because of the need to provide for the non-hinge axis rotation due to the slots on either side of the joint

The schematic below illustrates the extra input required to incorporate this effect

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Zero weight rigid elements defining the hinge assembly are listed below: 10 10 55 55 15 35 50 30

Normal to pipe axis to centerline of hinge assy

" " " Parallel to pipe axis to centerline of hinge axis

The finite length bellows must be defined accurately between nodes 10 and 55

This typically means entering the correct flexible length and using the manufacturer’s axial and lateral spring rates

Remember that manufacturer’s angular spring rates should not be used in finite length expansion joint models

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Í´·° Ö±·²¬ Large slip joints are usually difficult to install and difficult to accurately model

Smaller diameter slip joints are telescoping,

axial displacement devices that permit considerable axial displacement of the slip joint ends and moderately rigid resistance to pipe bending

Smaller slip joints are usually categorized by having two annular packing glands separated axially along the joint by a dead air space,

The following figure shows the cross-section of a typical large slip joint

The stiffnesses between nodes 15 and 25 are a function of the packing stiffness for transverse and rotational relative deformation and of packing stiffness and tightening for axial relative deformation

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Ò±¬» ïæ Typical delta dimensions are: 5

(The same values would also be used for 25

or the travel expected plus 4",

or a 12" estimate if nothing else is known

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K1 is the spring stiffness for forces below the yield force,

Ò±¬» íæ K2 is the spring stiffness (for joint compression) for forces greater than FY

The best estimate for this resistance is cumulative friction effects of guides and supports,

K2 =((100)N/(a)

(Approximation)

Where (N) is the nominal pipe diameter in inches,

and (a) is the thermal expansion at the operating temperature in inches per 100 ft

Ò±¬» ìæ Fy is the joint friction thrust from the vendor catalog

Typical values are given as 400 lbs times the nominal pipe size

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Ù·³¾¿´ Ö±·²¬­ Gimballed joints are designed to resist pressure thrust

The analyst should make sure that the joint manufacturer is aware of the design loads on the gimbals

The angular-only gimbal can be input as a zero length expansion joint with rigid axial,

The bending stiffness is set equal to the rotational stiffness specified in the manufacturer's catalog

Angular and Offset gimbals should probably be thoroughly modeled as shown in the following figures

Angular and Offset gimballed joints are usually installed in large diameter lines where lumped property assumptions for the bellows may not be within reasonable engineering accuracy

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Rigid elements between nodes 105 and 110 and nodes 111 and 115 each contain half the weight of the hinge mechanism

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Ü«¿´ Ù·³¾¿´ Dual gimbal joints are two,

gimballed joints in series in the pipeline

Putting two (or three) angularonly gimballed joints together provides for an ability to absorb lateral and possibly axial deformation

A Pipe Flexibility program will never be able to model the axial-only component of the possible deformation because it requires large rotation of the expansion joint components—something not considered in such programs

The single “angular deformation only” gimbal should always be used in series with at least one other gimballed joint

It is only in series that the “angular deformation only” gimbal provides for any lateral movement

Gimballed joints are designed to take pressure thrust

Analysts should verify that the joint manufacturer is aware of the design loads on the gimbal

Each individual angular-only gimbal joint should be modeled as a zero length expansion joint with rigid axial,

The bending stiffness should be equal to the manufacturer's published rotational stiffness term

The minimum required distance “L” between adjacent single gimballed joints (shown as 8-7 in the following example),

is principally a function of the angular and rotational deformation to be absorbed,

and the number of convolutions per joint

The following figure shows a dual gimbal comprised of two angular-only gimbals

The bending stiffness for each gimballed joint is 490

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Note: Both expansion joints use the same stiffness values,

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Ю»­­«®»óÞ¿´¿²½»¼ Ì»»­ ¿²¼ Û´¾±©­ Pressure balanced tees and elbows are used primarily to absorb axial displacements at a change in direction,

without any associated pressure thrust

Pressure balanced tees can also be used in universal type configurations to absorb axial and lateral movement

The example below shows briefly the coding of a pressure-balanced tee in a turbine exhaust line

The bottom side of the tee is blanked off

The tee is a standard unreinforced fabricated tee

The tie bars will only act in tension

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