Caesar II Application Guide

CAESAR II Applications Guide Version 2018 (10.0) July 2017

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CAESAR II Applications II Applications Guide

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Contents Overview ....................................................................................................................................................... 8 Bends ............................................................................................................................................................ 9

Bend Definition Defin ition ................... .................. .................. ................... .................. .................. ................... ....... 9 Single and Double Flanged Bends or Stiffened Bends.................. Bends........................... .................. ................. ................. .................. .................. ........... .. 10 180º Return Fitting-to-Fitting 90º Bends ............................................................................................... 11 Mitered Bends ................. .................. .................. ................... .................. .................. ................... ........ 12 Closely-Spaced Mitered Bend ........................................................................................................ 13 Widely-Spaced Mitered Bend ......................................................................................................... 14 Elbows - Different Wall Thickness ........................................................................................................ 17 Bend Flexibility Flexibil ity Factor ................ ................... .................. .................. ................... .................. ............... 18 Restr ain ts .................. ........................... .................. .................. .................. .................. .................. ................. ................. .................. .................. .................. .................. .................. .................. ............. .... 19

Anchors .................. .......................... ................. .................. .................. .................. .................. .................. .................. .................. .................. .................. .................. .................. ................. ........... ... 20 Anchors with Displacements .................. .......................... ................. .................. .................. .................. .................. .................. .................. .................. ................. ........21 Flexible Flexib le Anchors Anchor s .................. ................... .................. .................. ................... .................. ............... 22 Flexible Anchors with Predefined Displacements ................. .......................... .................. .................. .................. .................. .................. ............ ... 23 Flexible Nozzle - WRC Bulletin 297 ...................................................................................................... 24 Flexible Nozzle with Predefined Displacements .................. ........................... ................. ................. .................. .................. .................. .............. .....26 Flexible Nozzle with Complete Vessel Model................. Model.......................... .................. ................. ................. .................. .................. .................. ........... .. 27 Double-Acting Restraints ...................................................................................................................... 30 Single-Direct Single -Directional ional Restraints ................... .................. .................. ................... .................. .................. ... 31 Guides ................................................................................................................................................... 32 Limit Stops ............................................................................................................................................ 34 Windows ................................................................................................................................................ 36 Rotational Directional Restraints with Gaps ................. .......................... .................. .................. .................. .................. .................. .................. ................. .......... 37 Single-Directional Restraint with Predefined Displacement ................. .......................... .................. .................. .................. .................. ............ ... 38 Single-Directional R estraint and Guide Guid e with Gap and Predefined Disp lacement .................. ........................... ............... ......39 Restraint Settlement ............................................................................................................................. 40 Skewed Double-Acting Restraint with Gap ........................................................................................... 41 Skewed Single-Directional Restraint..................................................................................................... 42 Restraint between Two Pipes Using CNodes ................. .......................... ................. ................. .................. .................. .................. .................. ................. ........43 Restraint between Vessel and Pipe Models ................. .......................... .................. .................. .................. .................. .................. .................. .................. .........44 Restraints on a Bend at 45º .................................................................................................................. 45 Restraints on a Bend at 30º and 60º ..................................................................................................... 46 Vertical Dummy Leg on Bends ............................................................................................................. 46 Vertical Leg Attachment Angle .............................................................................................................. 50 Horizontal Dummy Leg on Bends ......................................................................................................... 50 Large Rotation Rods - Basic Model ...................................................................................................... 51 Large Rotation Rods - Chain Supports ................................................................................................. 53 Bi-Linear Restraints .............................................................................................................................. 54 Static Snubbers .................. .................. .................. ................... .................. .................. ................... ..... 56 Plastic Hinges ....................................................................................................................................... 56 Sway Brace Assemblies ....................................................................................................................... 57


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Contents

Hangers ...................................................................................................................................................... 60

Hanger Design Basics .......................................................................................................................... 61 Simple Hanger Design .......................................................................................................................... 63 Single Can Design ................................................................................................................................ 64 Constant Effort Support Design ............................................................................................................ 65 Constant Effort Supports - No Design .................................................................................................. 66 Existing Springs - No Design ................................................................................................................ 67 Multiple Multipl e Can Design ................ ................... .................. .................. ................... .................. .................. 68 Old Spring Redesign ............................................................................................................................. 69 Pipe and Hanger Supported From Vessel ............................................................................................ 70 Hanger Design with Support Thermal Movement ............ ..................... .................. .................. .................. .................. .................. .................. ............... ......70 Hanger Between Two Pipes ................................................................................................................. 71 Hanger Design with Anchors in the Vicinity ....................... ................................ .................. ................. ................. .................. .................. .................. .............. .....72 Hanger Design with User-Specified Operating Load ................... ............................ .................. .................. .................. .................. .................. ............ ... 74 Simple Bottomed Out Spring ................................................................................................................ 75 Lift Off Spring Can ................................................................................................................................ 76 Modeling Spring Cans with Friction ...................................................................................................... 77 Expan si on Joi nt s .................. ........................... .................. .................. .................. .................. ................. ................. .................. .................. .................. .................. .................. .................. ........... 78

Universal Expansion Joints - Simple Model ............ ..................... .................. .................. .................. .................. .................. .................. .................. ............... ...... 78 Tied Bellows Expansion Joint - Simple Model ................... ............................ .................. ................. ................. .................. .................. .................. .............. .....82 Universal Joint with Lateral Controls Stops - Comprehensive Tie Rod Ro d Model ................ ......................... .................. ............ ... 85 Hinged Joint .......................................................................................................................................... 86 Slotted Hinge Joint - Simple Model ....................................................................................................... 88 Slotted Hinge Joint - Comprehensive Model .................. ........................... ................. ................. .................. .................. .................. .................. ................. ........89 Tied Bellows - Simple vs. Complex Model .......... ................... .................. ................. ................. .................. .................. .................. .................. .................. ........... .. 90 Slip Joint................. .................. .................. ................... .................. .................. ................... ................. 91 Gimbal Joints ........................................................................................................................................ 92 Tied Bellows Expansion Joint - Complex Model ............. ...................... .................. ................. ................. .................. .................. .................. ................. ........93 Dual Gimbal .......................................................................................................................................... 95 Simple Bellows with Pressure Thrust.................................................................................................... 97 Pressure-Balanced Tees and Elbows ................................................................................................. 100 Universal Joint - Comprehensive Tie Rod ................. .......................... .................. .................. .................. .................. .................. .................. ................. .......... .. 101 Mis cell aneou s Mod els ................. .......................... .................. .................. .................. .................. .................. .................. .................. .................. .................. .................. .................. .........103

Reducers ............................................................................................................................................. 103 Ball Joints .................. .................. .................. ................... .................. .................. ................... ............ 104 Jacketed Pipe...................................................................................................................................... 106 Cold Spring ......................................................................................................................................... 108 Connecting Connect ing Equipment Equipm ent .................. ................... .................. .................. ................... .................. .......... 108 Vertical Vertica l Vessels Vessel s ................ ................... .................. .................. ................... .................. ................ 108 Horizontal Horizon tal Vessels ................... .................. .................. ................... .................. .................. .......... 114 Tut or ial A .................. .................. .................. ................... .................. .................. ................... .................. 118

Review the piping system drawing...................................................................................................... 120 Configure the software ........................................................................................................................ 121 Define the first pipe element ............................................................................................................... 122 Define the vertical pipe run ................................................................................................................. 127


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Contents Define the horizontal pipe run ............................................................................................................. 131 Define the 6-inch bypass .................................................................................................................... 136 Review the piping input ....................................................................................................................... 142 Check for errors .................................................................................................................................. 144 Check the static load cases ................................................................................................................ 144 Run the static analysis ........................................................................................................................ 146 Review the analysis results ................................................................................................................. 146 Static Analysis Output Listing ....................................................................................................... 150 Conclusions ......................................................................................................................................... 157

Tut or ial B .................................................................................................................................................. 158

Evaluating Pump Discharge Loads ..................................................................................................... 158 Creating Accurate Models ................................................................................................................... 166 WRC 297 Calculations Completed at the End of Error Checking....................................................... 169 Checking Nozzle Loads ...................................................................................................................... 175 System Redesign ................................................................................................................................ 177 Conclusion .......................................................................................................................................... 185 Examples .................................................................................................................................................. 187

Harmonic Analysis (TABLE) ............................................................................................................... 187 Harmonic Analysis of this System ................................................................................................ 190 Relief Valve Loads (RELIEF) .............................................................................................................. 192 Relief Valve Example Problem Setup .......................................................................................... 193 Relief Valve Loading Output ......................................................................................................... 197 Dynamic Analysis of Water Hammer Loads (HAMMER) ....................................................................201 Notes for Analyzing Water Hammer Loads .................................................................................. 207 Water Hammer Loading Output ................................................................................................... 210 Solution Summary ........................................................................................................................ 212 Dynamic Analysis of Independent Support Earthquake Excitation (CRYNOS) ..................................213 Results .......................................................................................................................................... 224 Structural Analysis (FRAME) .............................................................................................................. 225 Dynamic Analysis (NUREG9) ............................................................................................................. 232 NRC Example NUREG 9 .............................................................................................................. 233 NRC Example Problem 2A ........................................................................................................... 233 NRC Example Problem 2B ........................................................................................................... 236 NRC Example Problem 2C ........................................................................................................... 238 Omega Loop Modelling (OMEGA) ...................................................................................................... 240 Jacketed Piping (JACKET) ................................................................................................................. 245 Modeling Plan ............................................................................................................................... 246 Node Layout ................................................................................................................................. 247 Core Piping Input .......................................................................................................................... 249 Jacket Input (1st Half)................................................................................................................... 249 Jacket Input (2nd Half) ................................................................................................................. 253 WRC 107............................................................................................................................................. 255 Convert Forces/Moments from CAESAR II Global Coordinates to WRC 107 Local Axes ..............................................................................................................................................256 NEMA SM23 ....................................................................................................................................... 265 Nozzle Results for PT69M ............................................................................................................ 266 Nozzle Load Summation Report .................................................................................................. 267

CAESAR II Applications Guide

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Contents

Index ......................................................................................................................................................... 268


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SECTION 1

Overview The CAESAR II Application Guide shows the application of CAESAR II. Refer to this manual for examples of specific piping components, as well as examples of complete systems. The following sections illustrate the techniques and methods used to model individual piping components, restraints, and attached equipment. 

Bends (on page 9)



Restraints (on page 19)



Expansion Joints (on page 78)



Miscellaneous Models (on page 103)

Refer to these sections when you are modeling seldom-used components or unusual geometries. The component modeling examples are especially useful for modeling techniques and gaining a general u nderstanding of the software. Examples (on page 187) contains examples that illustrate the application of CAESAR II to various piping problems. These examples show modelin g, problem solving, and software operation. This section provides engineering guidelines and indicates where assumptions must be made in attempting to solve real-world problems. Tutorial A (on page 118) and Tutorial B (on page 158) contain tutorials that walk through the modeling and analysis of a complete system. Work through these sections if you have n ot previously attempted a particular analysis.


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SECTION 2

Bends This section provides guidelines and recommended best practices for the modeling various bend geometries in CAESAR II. For detailed information about using the software to define bend geometry, see "Component Information" in the CAESAR II User's Guide.

Bend Definition A bend is defined by the element entering the bend and the element leaving the bend. The bend curvature is always physically at the To end of the element entering the bend. The input for the element leaving the bend must follow the elem ent entering the bend. The bend angle is defined by these two elements. The default bend radius is 1-1/2 times the pipe nominal diameter (long radius), but it can be changed to any other value. When you specify a bend, two additional intermediate nodes are automatically generated--one at the 0º location and one at the bend midpoint (M). For stress and displacement output, the To node of the element entering the bend is located geometrically at the far-point on the bend. The far-point is at the weld line of the bend, adjacent to the straight element leaving the bend. The 0º point on the bend is at the weld line of the bend, adjacent to the straight element entering the ben d. The From point on the element is located at the 0º point of the bend (and no 0º node point is generated) if the total length of the element as specified by DX, DY, and DZ is equal to:

R t an ( b / 2) Where b is the bend angle, and R is the bend radius of curvature to the bend centerline. Nodes defined by the Angl e and Node properties are placed at the given angle on the bend curvature. The angle starts with zero degrees at the near-point on the bend and g oes to b degrees at the far-point of the bend. Angles are always entered in degrees. Entering the letter M as the angle designates the bend midpoint. Nodes on the bend curvature cannot be placed closer together than the angle distance specified by Minimum Angle to Adjacent Bend in the Geometry category of the Tools > Configure/Setup command. This includes the spacing between the nodes on the bend curvature and the near- and far-points of the bend. The minimum and maximum total bend angle is specified by the Minimum Allowable Bend An gl e and Maximum Allow able Bend Angle properties, also in the Geometry category of the Tools > Configu re Setup command.


9

Bends

Single and Double Flanged Bends or Stiffened Bends Single- and double-flanged bend specifications affect onl y the stress intensification and flexibility of the bend. There is no automatic rigid element (or change in weight) generated for the end of the bend. Single- and double-flanged bends are indicated by typing a 1 or 2 to define the Type in the bend auxiliary in put. Rigid elements defined before or after the bend do not alter the stiffness of the bend or its stress intensification factors. When specifying single-flanged bends, it does not matter on which end of the bend you place the flange. If you want to include the weight of the rigid flange at the bend ends, then put rigid elements (whose total length is the length of a flange pair) at the bend ends where the flange pairs exist. As a guideline, British Standard 806 recommends stiffening the bends whenever a component that significantly stiffens the pipe cross section is found within two diameters of either bend end. The flanges in the figures below are modeled only to the extent that the y affect the stiffness and the stress intensification for the bends. Flanges and Bends:


10

Bends

180º Return Fitting-to-Fitting 90º Bends Separate two 90º bends by twice the bend radius. The far-point of the first bend is the same as the near-point of the second, or following, bend. Intergraph CAS recommends that you place nodes at the midpoint of each bend that comprise the 180º return. 180º Bend:


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Bends

DX = 2R

Mitered Bends Evenly spaced mitered bends, whether closely or widely spaced, are defined b y two parameters: 

The number of cuts (changes in direction)



The equivalent radius or miter spacing

For closely spaced miters, the equivalent radius is equal to the code defined as R1 for B31.3 and R for B31.1. The equation for the equivalent radius to the spacing for evenly spaced miters is: Req = S / [ 2 tan(θ) ]

Where: Req

=

equivalent miter bend radius

S

=

spacing of the miter cuts along the centerline

θ

=

code-defined half-angle between adjacent miter cuts: θ = α / 2N

α

=

total bend angle

N

=

number of cuts

Where:


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Bends When using B31.1, an additional parameter, B (length of miter segment at crotch), is examined for closely spaced miters. The following equation is used to compute B for evenly spaced miters: B

=

S [ 1 - r o / Req ]

=

outside radius of pipe cross-section

Where: r o

Closely-Spaced Mitered Bend Miter bends are closely spaced if: S < r [1 + tan (θ)]

Where: S = miter spacing r = average pipe cross section radius: (ri +ro)/2 θ = one-half the angle between adjacent miter cuts

B31.1 has additional requirements: B > 6 tn θ ≤ 22.5 deg.

B = length of the miter segment at the crotch. tn = nominal wall thickness of pipe.

Closely spaced miters, regardless of the number of m iter cuts, can be defined as a single bend. CAESAR II always calculates the spacing from the bend radius. If you have the miter spacing but not the bend radius, the radius must be calculated as shown in the following example. The mitered bend shown below has four cuts through 90º and a spacing of 15.913 inches. Req

=

S / [ 2 tan (θ)]

θ

=

α / 2N

= 90 / [2(4)] = 11.25º Req

= 15.913 / [2 tan (11.25º)] = 40


13

Bends

Closely Spaced Miter Bend:

Widely-Spaced Mitered Bend Mitered bends are widely spaced if: S³ r * [1 + tan ( θ)]

Where: S

=

spacing between miter points along the miter segment centerline

r

=

average cross section radius (r i+r ο)/2

θ

=

one-half angle between adjacent miter cuts

B31.1 has the following additional requirement: θ ≤ 22.5º

In CAESAR II, you must enter widel y spaced miters as individual, singl e-cut miters, each having a bend radius equal to: R = r [1 + cot (θ)] / 2

Where: R = reduced bend radius for widely spaced miters.


14

Bends During error checking, CAESAR II produces a warning message for each mitered component that does not pass the test for a closely spaced miter. The following components should be re-entered as a group of single cut joints. Widely Spaced Miter: Pipe O.D. = 10.375 in. Pipe Thickness = 0.500 in. Bend Angle = 90º Cuts = 2 Req = 45 inches

As su mi ng cl os ely sp aced :

Calculate the ∆ coordinates to get from the tangent intersection point of the sing le cut miter bend at node 10 to the single cut miter bend at node 15. The straight pipe section coming into and going out of the bend must be ≥ Reqsin (θ).

Enter widely spaced miters as indi vidual straight pipe elements, with bends specifi ed, having one miter cut.


15

Bends

Figure 2: Between the First and Second Cuts Figure 1: Coming Up to the First Cut

Figure 3: Coming Out of the Second Cut


16

Bends

Elbow s - Different Wall Thickness When you define the fitting thickness for the bend, CAESAR II changes the thick ness only of the curved portion of the bend element. The thick ness of any preceding or following st raight pipe is unaffected. The specified fitting thickness applies only for the current elbow and is not persisted to any subsequent elbows in the job. Stresses at the elbow are calculated based on the section modulus of the matching pipe as specified in the B31 codes. However, stress intensification factors and flexibility factors for the bend are based on the elbow wall thickness. Thick Elbow:

The elbow at node 10 has a thickness larger than the matching pipe wall. The matching pipe has a thickness of 0.5.


17

Bends

Bend Flexibility Factor Usually, bend flexibility factors are calculated according to code requirements. However, you can override the code calculation by typing a value for the K-factor . For example, if you type 1.500, then the bend is 1.5 times as flexible as a straight pipe of the same length.


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SECTION 3

Restraints This chapter provides guidelines and recommended best practices for modeling various types of restraints in CAESAR II.

In This Section Anchors .......................................................................................... 20 Flexible Nozzle - WRC Bulletin 297 ............................................... 24 Double-Acting Restraints ............................................................... 30 Single-Directional Restraints ......................................................... 31 Guides............................................................................................ 32 Limit Stops ..................................................................................... 34 Windows ........................................................................................ 36 Rotational Directional Restraints with Gaps .................................. 37 Single-Directional Restraint with Predefined Displacement .......... 38 Single-Directional Restraint and Guide with Gap and Predefined Displacement ................................................................................. 39 Restraint Settlement ...................................................................... 40 Skewed Double-Acting Restraint with Gap.................................... 41 Skewed Single-Directional Restraint ............................................. 42 Restraint between Two Pipes Using CNodes ................................ 43 Restraint between Vessel and Pipe Models .................................. 44 Restraints on a Bend at 45º ........................................................... 45 Restraints on a Bend at 30º and 60º ............................................. 46 Vertical Dummy Leg on Bends ...................................................... 46 Vertical Leg Attachment Angle ...................................................... 50 Horizontal Dummy Leg on Bends .................................................. 50 Large Rotation Rods - Basic Model ............................................... 51 Large Rotation Rods - Chain Supports .......................................... 53 Bi-Linear Restraints ....................................................................... 54 Static Snubbers ............................................................................. 56 Plastic Hinges ................................................................................ 56 Sway Brace Assemblies ................................................................ 57


19

Restraints

An chors Use the following guidelines when modeling anchors in CAESAR II: 









The anchor default stiffness for translational and rotational degrees-of-freedom is defined in the CAESAR II configuration file. You can use connecting nodes with anchors to rigidly fix one point in the piping system to any other point in the piping system. The value that you define for the Stif property applies to all six anchor degrees-of-freedom. Do not specify displacements at an anchor. If you know the displacements of a particular point, specify them without any additional restraints or anchors. Accurate definition of piping boundary conditions (restraints) is considered the single most important part of system modeling. It requires experience with piping fabrication and erection and with CAESAR II.

The example below shows a nozzle connection modeled as an anchor, along with the corresponding rigid anchor input on the Restraints Auxil iary Data tab in the Classic Piping Input dialog box. Rigid Anch or at Node 5


20

Restraints

Anchors w ith Displacements Follow these guidelines to model anchors with displacements in CAESAR II: 

Enter only displacements for the node.



Do not specify restraints or anchors at the node to be displaced.



For anchors with displacements, ensure that all six degrees-of-freedom at the node are defined.

Degrees-of-freedom not defined (left blank) in any displacement vector are presum ed to be free in all load cases. You can define up to nine different displacement vectors. Non-zero displacements are usually part of the thermal expansio n effects. They should be added into any analysis case containing the corresponding thermal, such as W+P1+T1+D1. We recommend load cases that do this automatically. The translations and/or rotations for any nodal degree of freedom that has displacements specified in any displacement vector are zero for the following load cases: 



Those that do not contain a vector as part of the load case identification Those where the specified non-zero value for load cases containing the vector is part of the load case identification

For example, defined displacements are used if the load case is W +P1+T1+D1 (OPE), and those displacements are held to zero if the load case is W+P1 (SUS). After a degree-of-freedom is fixed in one displacement vector, it cannot be free in another displacement vector at the same node. The default value for a blank displacement box is zero. The figure below shows an anchor with predefined displacements. An ch or s w it h Pr edef in ed Di sp lac emen ts


21

Restraints

Flexible Anchors Use the following guidelines to model flexible anchors in CAESAR II: 



Use six flexible restraints. Refer to the following flexible nozzle examples to improve modeling methods for intersections of this type. Flexible Restraints for Nozzle and Shell

For more information about defining restraints, see "Restraints" in the CAESAR II User's Guide.


22

Restraints

Flexible Ancho rs with Predefined Displacements To model flexible anchors with predefined displacements, use the following guidelines: 





Use six flexible restraints. Define a unique connecting node (CNode) at each of the six restraints. All six restraints should have the same connecting node. Specify the displacements at the connecting node.

Flexible Anchors wi th Predefined Displacements

The CNode in the example below is 1005. CNode numbers must be unique.


23

Restraints

Flexible Nozzle - WRC Bulletin 297 When modeling flexible nozzles, adhere to the following requirements: 

Frame only one pipe element into the nozzle node.



Do not place restraints at the nozzle node.



Do not place anchors at the nozzle node.



Do not specify displacements for the nozzle node. Refer to the displacements at flexible nozzles below.

CAESAR II automatically performs the following functions: 

Calculates nozzle flexibilities for user-defined nozzle/vessel data.



Calculates and inserts restraints to simulate nozzle flexibilities.



Calculates flexibilities for the axial translations, circumferential, and longitudinal bending.

You must complete the error check process to view these calculated values. CAESAR II uses the following criteria for its calculations: 





Shear and torsional stiffnesses are assumed rigid. Nozzle configurations outside of the WRC 297 curve limits are considered rigid. It is not unusual for one stiffness value to be rigid because of curve lim its and for the others to be suitably flexible. You can use Vessel Temperature and Material in the WRC 297 auxiliary data area to optionally compute a reduced modulus of elasticit y for the local stiffness calculations.


24

Restraints Schematic of Nozzle and Vessel to b e Modeled Using WRC 297

WRC 297 Input Examp le


25

Restraints Example WRC 297 output is shown below.

WRC 297 Nozzl e Cal cul at i ons

Flexib le Nozzle with Predefined Displacements To model flexible nozzles with predefined displacements (WRC-297), use the following guidelines: 



Define a unique vessel node in the nozzle auxiliary data area (Classic Piping Input dialog box). Apply the predefined displacements to the vessel node.

These displacements can be specified on any element. The displacement node does not need to be on an element that defines it. The CAESAR II-generated nozzle/vessel flexibilities are inserted in restraints that act between the nozzle node and the vessel node. Flexible Nozzle With Predefined Displ acements on Vessel Node


26

Restraints

Flexib le Nozzle with Complete Vessel Model To model a flexible nozzle that includes a complete vessel, use the following guidelines: 

Define a unique vessel node on the nozzle auxiliary data area. The nozzle auxiliary data area displays when you select Nozzle Flex on the Classic Piping Input dialog box. For more information, see "Nozzle Flex" in the CAESAR II User's Guide.







Run a rigid element between the vessel node defined in the nozzle auxiliary data area and the centerline of the vessel. The outside diameter of the rigid element should be approximately equal to the outside diameter of the vessel. The weight of the r igid element should be zero. Model the actual vessel length using pipe elements. Model the vessel diameter and wall thicknesses as accurately as possible Use an anchor to model the vessel anchorage point.

The CAESAR II-generated nozzle/vessel flexibilities are inserted b etween the nozzle node and the vessel node. The graphics below show a full WRC 297 model schematic. Full WRC 297 Model Schemati c


27

Restraints Full WRC 297 and Vessel Model

Figure 4: Pipe Entering Nozzle

Figure 5: WRC 297 Auxiliary Input

Figure 7: Rigid Weight is Blank (0,0)

Figure 6: Vessel Radius Element


28

Restraints

Figure 8: Vessel Element

Figure 9: Vessel Skirt Element At Ambient Temperature

Nozzle Calculations

An example of the WRC 297 results found at the end of error checking is shown below.


29

Restraints

Double-Acting Restraints Double-acting restraints are those that act in both directions a long the line of action. T he most commonly used restraints are double-acting. A CNode is the connecting node. If you do not enter a value in the CNode box, the restrained node is connected using the restraint stiffness to a rigid point in space. If you do enter a value for CNode, the restrained node is connected using the restraint stiffness to the connecting node. If you specify a gap, it is the amount of free movement along the positive or negative line of action of the restraint before resistance to movement occurs. A gap is a length, so it is always positive.

Translational Restraints Restraint acts along both the positive and negative directions. Friction at double-acting restraints acts orthogonally to the line of action of the restraint. Double-Acting Restraint at Node 55 in the Z-Direction


30

Restraints Rotational Restraints The behavior of rotation restraints is similar to double-acting translational restraints. Friction is not defined for rotational restraints. Hinged-End ro d Free to Rotate About th e Z-Axis

Restrain all except Rz

Four restraints on one element containing node 105 and remaining restraint on the next element.

Single-Directional Restraints The following are some im portant facts pertaining to single-directional restraints: 





The plus or minus sign (+/-) on the single-directional restraint gives the direction of free movement. For example, a positive Y restraint (+Y) can move freely in the positive Y-direction and is restrained against movement in the negative Y-direction. Single-directional restraints can define restraint along positive, negative, or skewed axes. Any number of single-directional restraints can act along the same line of action. If more than one single-directional restraint acts along the sa me line of action, then there are usual ly two in opposite directions and are used to model unequal leg gaps.


31

Restraints 



A CNode is the connecting node. If you do not define a value for CNode, then the restrained node is connected using the restraint stiffness to a rigid point in space. If you enter a value for CNode, then the restrained node is connected using the restraint stiffness to the connecting node. You can specify friction and gaps with single-directional restraints. The gap is in the opposite direction of the free travel. Rigid Single - Directional Restraint in Y at Node 20

The plus or minus sign (+/-) on the restraint gives the direction of free movement. Because the stiffness is omitted, the restraint is rigid.

Guides Use the following guidelines when modeling guides in CAESAR II: 

Guides are double-acting restraints with or without a specified gap.



You can use connecting nodes (CNodes) with guides.





You can define guides using the global system coordinates or using the applicable options in the restraints auxiliary data area. The restraints auxiliary data area displays when you select Restraints on the Classic Piping Input dialog box.

A guided pipe in the horizontal or skewed direction has a single restraint, acting in the horizontal plane, orthogonal to the axis of the pipe. A guided vertical pipe has X- and Z-direction supports. CAESAR II computes direction cosines for guides. T he software ignores user-defined guide direction cosines.


32

Restraints Guide on Horizontal Pipe with Singl e Direction al Restraint

Node 25 is guided in Z with a gap of 2.5 inches. A single-directional restraint in the Y-direction also exists. Both restraints are rigid. In this example, replacing the guide restraint with a Z restraint is equivalent.

Guided Pipe in Bot h Horizontal and Vertical Directi ons


33

Restraints

Limit Stops Use the following guidelines when working with limit stops in CAESAR II: 







Limit stops are single- or double-acting restraints whose line of action is along the axis of the pipe. The plus or minus sign (+/-) on the single-directional restraint gives the direction of unlimited free movement. A limit stop and a single-directional restraint can have gaps. The gap is the distance of permitted free movement along the restraining line of action. A gap is a length and is always positive. The plus or minus sign (+/-) on the restraint determines the orientation of the gap along the line of action. The gap direction is opposite to the direction of free travel.



You can use connecting nodes (CNode) with any limit stop model.



A limit stop can be defined using the restraint type LIM .





Limit stops provide double- or single-acting support parallel to the pipe axis. Limit stops can have gaps and friction. The positive line of action of the limit stop is defined by the From and To node on the element. CAESAR II computes direction cosines for orthogonal or skewed limit stops. The software ignores user-defined limit stop direction cosines.


34

Restraints Directional Limit Stop with a Gap

Two Limit Stops Acting in Opposite Directions

The stop at node 45 permits unlimited free movement in the positive X-direction and 1.0 inches of free movement in the negative X-direction before the limit stop becomes active.

The stop at node 195 p ermits unlimited free movement in the negative X-direction and 1.0 inches of free movement in the positive X-direction.


35

Restraints

Windows Use the following guidelines when modeling windows in CAES AR II. 





Equal leg windows are modeled using two double-acting restraints with gaps orthogonal to the pipe axis. Unequal leg windows are modeled using four single-acting restraints with gaps orthogonal to the pipe axis, as shown in the example below. The gap is always positive. The plus or minus sign (+/-) on the restraint determines the direction of movement before the gap closes. If there is no plus or minus sign ( +/-) on the restraint, then the restraint is double-acting, and the gap exists on both sides of the line of action of the restraint. If there is a plus or minus sign (+/-) on the restraint, then the gap exists on the restrained line of action of the restraint. For example, a positive Y (+Y) restraint is restrained against movement in the negative Y-direction. Any gap associated with a positive Y restraint is the free movement in the negative Y-direction before the restraint begins acting.


36

Restraints Window Modeled with Four Single-Directional Restraints w ith Gaps

Rotational Directional Restraints with Gaps Rotation directional restraints with gaps are considered specialty items. Usually, they used only in sophisticated expansion joint or hinge models. Rotational Directional Restraints

The figure on the left shows a bi-directional rotational restraint with gap. Allowable rotation of 5º in either direction about the Z-axis before resistance to rotation is encountered.


37

Restraints

The figure on the left shows a hinge assembly with directional rotational restraint. Hinge assembly at node 50 can rotate relative to assembly at node 55 only in the positive direction about the Z-axis.

Sing inglele-D Directio irectional nal Restr Restra aint wi with th Pre Predefined defined Displacement Define the one-directional restraint as usual, and enter a unique node number in the CNode box. Specify the predefined displacements for the CNode. Single-Direction Single-Direction al Restraint wit h Predefined Displacement Displacement

Piping at node 55 rests on top of the restraint that is displaced in the Y-direction, simulated by node 1055.


38

Restraints

Sing inglele-D Directio irectional nal Restr Restra aint and Guid uide e with Ga Gap p and Predefined Displacement Define the single-directional restraint and guide as usual. Enter a unique node number in the CNode boxes for the single-directional restraint and th e guide. Use the same node number for both. Specify the predefined displacements for the CNode. Guide Plus Single-Directional Restraint wit h Predefined Displacement:

Guided piping at node 70 rests on a structural member node 1070. The structure undergoes a predefined displacement.

In the example above, ignoring other loads that may be acting on the pipe, the 0.250-inch gap will be closed in the negative X-direction, and the pipe will have a displacement of 0.50-inches in the positive X-direction.


39

Restraints

Restraint Settlement Use the following guidelines when modeling restraint settlement in CAESAR II: 







Model restraint settlement using a single-directional restraint with predefined displacements. The magnitude of the predefined displacement is the amount of anticipated settlement in the negative Y-direction. The operating load case is used to include the the effect effect of settlement. The settlement displacements are prescribed for the connecting node at the single directional restraint. For more information, see Single-Directional Restraint with Predefined Displacement (on page 38) page 38).. Settlement is considered to be an EXP load with a half-cycle. Restraint Restraint Settlement Settlement

The weight of this pipe at node 95 exerts a sufficient load on the foundation node 1095 to cause a calculated 0.325-inch settlement.


40

Restraints

Skewed Doub Doublele-Acti Acting ng Re Restr stra aint w ith Gap Use the following guidelines when modeling skewed restraints in CAESAR II: 











You can use direction vectors or direction cosines cosines to define the line of action of the restraint. restraint. If direction vectors are used, CAESAR II converts them to direction cosines. Direction cosines can be quickly checked in the graphics processor. Any translational axis can be used in the restraint description. The redefinition of the axis does not affect any other restraint description for the el ement. Use caution caution when entering skewed skewed direction input data. A common common mistake is to to specify an axial instead of transverse restraint when modeling a skewed guide. Plotted section views of the restrained nodes are a useful check of the skewed direction specification. The sense of the the direction direction or cosine unit vector is unimportant. unimportant. In In the definition of double-acting restraints, the direction vector and cosines are used only to define the restraint line of action and are not concerned with a direction along that line. You can use a simple rule for finding perpendicular, skewed, or direction vectors. vectors. The restraint is to be perpendicular to the pipe. If the pipe has skewed delta dimensions DX and DZ, the perpendicular restraint directions vector are (DX, 0, -DZ). Skewed Skewed Double-Acting Restraint w ith Gap

Figure 10: Double Acting Restraint Y, Guided @ 45 Deg.


41

Restraints

Figure 11: Input Using Unit Direction Vectors

Figure 12: Input Using Direction Cosines

Figure 13: Input Using Perpendicular Vector

Figure 14: Input Using Guide Restraint Type

Skewed Single-Directional Restraint Use the following when modeling skewed single-directional restraints in CAESAR II: 





Skewed restraints can be nonlinear. Direction vectors vectors or direction cosines can be used to define the line of action of the restraint. restraint. If direction vectors are used, CAESAR II converts them to direction cosines. The direction of the cosines or the direction vector is along the positive line of action of the positive (+) restraint. Refer to the example below



Direction cosines can be quickly checked in the graphics processor.



You can use connecting nodes (CNode) with any skewed single-directional restraint.


42

Restraints Skewed Singl e-Direction al Restraint

Restraint between Two Pipes Using CNodes For the following two examples, make sure Connect Geometry Through CNodes is not selected to avoid plotting and geometry errors. For more information, see "Geometry Definitions" in the CAESAR II User's Guide. Nonlinear or linear restraints can act between two different pipe nodes. The CNode property effectively represents to what the other end of the restraint is attached. Nonlinear Restraint Between Two Pipes


43

Restraints Rigid Rod Hanger Betw een Two Pipes

Restraint between Vessel and Pipe Models Use the following guidelines when modeling restraints that between vessel and pipe in CAESAR II: 





Use a restraint with a connecting node to link the pipe to the rigid element that extends from the vessel shell. You can specify any number of restraints between the restrained node and the connecting node. Restraints can be linear or nonlinear with gaps and/or friction. Restraint Between Vessel and Piping


44

Restraints

Restraints on a Bend at 45º Linear and/or non-linear restraints can act at an y point on the bend curvature. Points on the bend curvature are like any other point in the piping system. The figure below shows a bend supported vertically by a rigid rod. The rod is allowed to take tensile loads only, so it is modeled as a single-directional restraint that can move freely in the positive Y-direction. If the positions of nodes 19 an d 20 are unclear, see Bends (on page 9). The line of action of the rod is shifted away from node 19. A downward force at node 15 produces a positive Z-moment about node 20 in the system as modeled, but a negative Z-moment about node 20 in real-time. The magnitude of this moment is a function of the load and the moment arm (the amount of the shift). If this is considered significant, then a ri gid element with zero weight can be placed between node 19 and the a ctual point of rod attachment. The restraint is then pl aced at the actual point of rod attachment. 90º Bend Restrained at Midpoi nt


45

Restraints

Restraints on a Bend at 30º and 60º You can define up to three nodes at any angle on the bend curvature, presuming the points are more than 5º apart. Restraints can be m odeled on any of these nodes. If necessar y, one of the points can be at the 0º point on the bend, which is the bend near point. The To node of the bend is placed at the tangent intersection point for geometric construction, but it is placed at the bend far point for analysis purposes. Consequently, specifying a node at the bend far-weld point generates an error. Nodes and angles on the bend curvature can be specified in any order. Restraints o n Intermediate Points Along a Bend

Vertical Dummy Leg on Bends You can model dummy legs on bends in several wa ys. The three most common methods used to model dummy legs are outlined below.

Near/Far Point Method 

Easy input



Dummy leg acts along centerline of vertical run



Dummy leg does not act at the proper place on the bend curvature


46

Restraints On Curvature Method 

Easy input



Dummy leg acts at the proper place on the bend curvature



Dummy leg does not act along the centerline of the vertical run

Offset Element Method 

Difficult input



Dummy leg acts at the proper place on the bend curvature



Dummy leg acts along centerline of vertical run

The element immediately after the bend must define the downstream side of the bend. Do not define dummy legs on the element immediately following a bend. Dummy legs and/or any other elements attached to the bend curvature should be coded to the bend tangent intersection point. The length of the dummy leg is taken directly from the DX, DY, and DZ properties defined for the dummy leg of the pipe. There is no automatic alteration of the dummy leg length due to the difference bet ween the bend tangent intersection point and the actual point on the bend cu rvature where the dummy leg acts. Enter the true length of the dummy leg in the DX, DY, and DZ boxes in the Classic Piping Input dialog box. Input and output plots of the dummy leg always show it going to the bend tangent intersection point. During error checking, the software generates a warning message for each dummy leg/bend model. Verify that the warning message description of the bend is accurate.


47

Restraints

Verti cal Dummy Leg on Bend

The bend shown on the left enters from the top left corner of the control station nodes 80 to 85 a nd exits horizontally to the right (nodes 85 to 90). The dummy leg is attached near the 45º point on the bend. The centerline of the dummy leg should line up with the centerline of the vertical run of pipe.

Coding the bend is the same for all three methods except alpha is not required for the Near/Far Point Method.


48

Restraints Figure 15: Near Point Method

Dummy leg length is equal to the distance from its base to the bend tangent intersection point plus the bend radius. Figure 16: On Curvature Method



Calculate α from the relation shown on the following page.



Enter α on the bend specification instead of the midpoint.





Calculate the offset distance y = R(1-sinα). Add the distance y to the dummy leg length.

Figure 17: Offset Element Method








Calculate α and the offset distance x and y from the figure on the following page.

Add x and y to the dummy leg length. Activate the offsets to subtract them from the dummy leg.

49

Restraints

Vertical Leg Attachment Angle Dummy Leg Attachment Angle Calculatio n

Horizontal Dummy Leg on Bends The element leaving the bend m ust define the downstream side of the bend. Do not define dummy legs on the element immediately following a bend. Enter the true length of the dummy leg in the DX, DY, and DZ boxes in the Classic Piping Input dialog box. Input and output plots of the dumm y leg always show the dummy leg going to the bend tangent intersection point. During error checking, the software generates a warning message for each dummy leg/bend model. Verify that the warning message description is accurate. Horizontal Dummy Leg on Midpoint of Bend

The dummy leg is defined as a zero weight rigid supported on one end by a spring can.


50

Restraints

Large Rotation Rods - Basic Model Large rotation rods are used to model relati vely short rods where large orthogonal m ovement of the pipe causes shortening of the restraint along the original line of action. Large rotation rods can be entered in any direction. Use the Type list on the Classic Piping Input dialog box, to select the rotation rod: XROD, YROD, or ZROD. After the rotation rod is selected, the restraint options change as follows: 



Gap changes to Len , which is the length of large rotation swing. Mu changes to Fi , which is the initial load on the restraint if used to model variable support spring hanger. (Imagine the large rotation rod as providing a bowl in which the pipe node is free to move.)

Enter large rotation rods only where the y are needed. Repeatedly using large rotation rods when they are not necessary can cause the system to become unstable during the nonlinear iteration. First, analyze the system without the large rotation rods, and then add large rotation rods where horizontal movement at support points is greatest. Usually, you should add only one rod in an area at a time. The rod angle tolerance is currently set at 1º. Large rotation is generally considered to become significant when the angle of swing becomes greater than 5º degrees. As for any other support, connecting nodes can be used for large rotation rods. Graphically, the connecting nodes and the restraint node do n ot have to be at the same point in space. There is no plot connectivity forced between large rotation rod nodes and connecting nodes. The plus or minus signs ( +/-) on the large rotation rod determine the ori entation of the swing axis. A positive YROD (+YROD) is equivalent to an YROD and indicates that the concave side of the curvature is in the positi ve Y-direction.


51

Restraints

In the example below, the rod p ivots about the structural steel support. There is a very short swing arm, so even a sm all amount of horizontal movement produces a relatively large s wing. The output report for this restraint shows X- and Y-direction loads. Large Rotation Rods :


52

Restraints

Large Rotation Rods - Chain Supports In the model below, imagine that you want the large rotation to swing only in the plane of the chain support (the Y-Z plane). The two pipes should move freely relative to each other in the axial direction (the X-Y plane). T o do this, use three restraints with connecting no des. The first restraint is the large rotation rod with its connecting nodes. In turn, the first restraint is connected to the second and third line ar restraints, allowing only Y-Z interaction b etween the large rotation rod connecting node and the top pipe node. Chain Supports


53

Restraints

Bi-Linear Restraints Bi-linear restraints are used most often to model soil support where some soil ultimate load bearing capacity can be calculated. Bi-lin ear restraints are designated by typing a 2 immediately after the direction in the restraint Type box.

After a bi-linear spring is entered, the restraint options change as follows: 

Stif changes to K1 , which is the initial stiffness.



Gap changes to K2 , which is the yield stiffness.



Mu changes to Fy , which is the yield load.

Both the yield stiffness ( K2 ) and the yield load ( Fy ) are required. If necessary, the yield stiffness can be defined as a negative value. The initial stiffness ( K1 ) is optional. If it is left blank, the software presumes a rigid initial stiffness. Some sub-sea pipeline resistance tests have shown that load carrying capacity drops after the ultimate load is reached, and displacement continues. For more information about the use of the spring types used to model underground piping systems, "Buried Pipe Modeler" in the CAESAR II User Guide. Characteristics of Bi-Linear Supports


54

Restraints Pipe in a Trench Bi-Linear Restraint Modeling

Estimated Stiffness Models

Figure 18: Estimated Lateral Stiffness: 60,000 lb/in.


Figure 19: Estimated Vertical Stiffness: 533,333 lb/in.

55

Restraints

Static Snubbers Static snubbers are translational restraints designed to provide resistance to d ynamic loads such as seismic and wind. Snubbers are activated by checking the Snubbers Active? check box for the appropriate load case (the one that contains the occasional load) on the Load Case Options tab in the Load Case Edito r . The figure below shows that snubbers will be active in load case L2 to restrain the pipe against the uniform load.

Figure 20: Load Case Options Tab

Static snubbers, or static analysis snubbers, have SNB following the translational direction that displays in the restraint Type box (located on the Classic Piping Input dialog box). After you enter a snubber, Gap and Mu are no longer available. Static snubbers can be directional. That is, they can be preceded by a plus (+) or minus (-) sign. To model static snubbers, follow the steps below. 1.

Run the operating case without defining a snubber.

2.

Note the displacements, in all six degrees of freedom, at each snubber location.

3.

In the Classic Piping Input dialog box, add each snubber with a distinct CNode.

4.

Place the displacements from Step 1 on all snubber CNodes.

5. Modify the load cases by including D1 everywhere that T1 displays. 6. Activate snubbers in the Load Case Edito r for all load cases containing occasional loads.

Plastic Hinges The steps in setting up a plastic hinge are illustrated below. The leg from A to B is overheated, causing bending of the B-D support leg. This example models the p lastic deformation at cross-section E-E. The plastic hinge is formed between nodes 10 and 15. The expansion joint is used to provide translational and torsional rigidity at the plastic hinge junction. Two bi-linear supports are used to model rigid resistance to bending until a breakaway force ( yield force) is exceeded, at which point bending is essentially free.


56

Restraints 1

Plastic Hinge in a Support L eg

The yield force is computed using the following equation: Fy = SyZ(SF)

Where Sy is the yield stress Z is the section modulus SF is the safety factor

1

The plastic hinge is modeled as a zero length expansion joint with rotational bi-linear restraints.

Sway Brace Assemblies The sway brace is commonly used to allow unrestrained thermal movements while tuning the system dynamically to eliminate vibration. A sway brace resembles a spring in that it can be pre-loaded in the cold (installed) position. After pipe thermal growth, it reaches the neutral position and the load on the system in the operating condition is zero or negligible. The sway brace is composed of a single compression spring enclosed between two movable plates. The spring is precompressed a full inch, which provides an initial force that instantaneously opposes vibration. Any movement from the sway brace neutral position is opposed by a load equal to the pre-load plus travel from the neutral position multiplied by the sway brace spring constant. After maximum allowed travel (usuall y 3-inches in either direction) is reached, the sway brace locks, preventing additional movement. Manufacturers typically recommend a specific size sway brace for a given pipe nominal diameter.


57

Restraints A more specific sway brace selection is possible when the exact restraining force required to control the piping vibration is known. The energy necessary to control the pi ping is proportional to the mass, amplitude of movement, and the force causing the vibration. From this relation, the exact restraining force required to control the piping vibration can be calculated and an appropriate sway brace size selected. After it is selected, the sway brace can be modeled in CAESAR II using a combination of a bi-linear restraint and a translational restraint: In the event that the sway brace is to be installed in the operating condition (or the neutral position is to be adjusted in the operating position), the modeling in CAESAR II is a little more complex. In this case, before modeling the swa y brace, you must analyze the piping system without the sway brace to obtain d isplacements from the cold to neutral operating position: Run an analysis on the system without the sway brace to obtain the displacements from cold to operating condition. In the example illustrated below, the assumed CAESAR II-calculated displacement from cold to operating position is 0.5 inches In the SUS case, the displacement D2 (vector 2) represents the pre-load in cold position. Under shutdown conditions, the pipe returns to its cold position and the brace exerts a force as previously described. Sustained case restraint loads on sway brace = Pre-Load + Hot Deflection * Spring Rate

In OPE, the displacement allows thermal expansion, and the sway assumes neutral position exerting zero or negligible load on the pipe. Operating case restraint loads on sway brace =~ 0.0 (does not restrain thermal expansion) Sway Brace Installed in Operating Conditi on

Sway brace opposing compression force (movement occurs after pre-load is overcome). 

Spring Rate: 150 lb../in.



Initial Loading: 150 lb.






Allowed Movement: 3 in. Calculated Displacement: .5 in.

58

Restraints Be sure to include D2 in the sustained and operating cases.


59

SECTION 4

Hangers This chapter illustrates methods for incorporating spring hanger design into CAESAR II models. For detailed information about using the software to define hangers, see Hangers in the Boundary Conditions section of the CAESAR II User's Guide.

Hanger Design Basics Select Model > Hanger Design Cont rol Data from the Classic Piping Input dialog box to enter values that affect hanger design throughout the model. T he hanger control options, with default values, are shown below. Complete descriptions of each item can be found in the CAESAR II User's Guide. To set the default value displayed in the Hanger Table box, specify the Default Spring Hanger Table setting in Configuration Editor > Database Definitions .

Whenever CAESAR II detects a proposed spring hanger location supporting zero load, the software displays a warning and does not select a hanger for this location. You can remove this proposed hanger location from the model to eliminate the warning.

Hanger Design Recommendation s There are instances where the stiffness of the adjacent piping and the hanger location restraints in the restrained weight case in teract unfavorably. This can produce an unwanted distribution of


60

Hangers loads. Often, reducing the stiffness used to compute the hanger loads in the restrained weight run eliminates these load distribution problems. T he default for this stiffness is 1.0E12. Valu es on the order of 50,000 or 75,000 have been used successfully to somewhat relax the system and redistribute these piping loads. You can define the value the software uses for calculating hanger restrained weight loads in the Hanger Default Restraint Stiff ness configuration setting in Tools > Configur e/Setup > Computation al Control. . The operating case for hanger travel (free thermal case) can be analyzed either with no spring stiffness at the hanger locations, or with the s tiffness of the selected springs inserted at those locations. In the second case, the software selects the springs through an iterative process. You can specify the manner in which the software handles spring hangers in the Include Spring Hanger Stiffness in Hanger OPE Travel Cases configuration setting in Tools > Configure/Setup > Computational Control . Inserting the actual hanger stiffness into the Operating Case for Hanger Travel may give a more accurate result, but may also introduce convergence problems. Therefore, in the latter case, it is very important that you adjust the hanger load in the cold case (in the physical system) to match the reported hanger cold load.

In This Section Hanger Design Basics ................................................................... 61 Simple Hanger Design ................................................................... 63 Single Can Design ......................................................................... 64 Constant Effort Support Design ..................................................... 64 Constant Effort Supports - No Design ........................................... 65 Existing Springs - No Design ......................................................... 67 Multiple Can Design ....................................................................... 68 Old Spring Redesign...................................................................... 68 Pipe and Hanger Supported From Vessel ..................................... 70 Hanger Design with Support Thermal Movement.......................... 70 Hanger Between Two Pipes .......................................................... 71 Hanger Design with Anchors in the Vicinity ................................... 72 Hanger Design with User-Specified Operating Load ..................... 74 Simple Bottomed Out Spring ......................................................... 75 Lift Off Spring Can ......................................................................... 76 Modeling Spring Cans with Friction ............................................... 77

Hanger Design Basics Select Model > Hanger Design Cont rol Data from the Classic Piping Input dialog box to enter values that affect hanger design throughout the model. T he hanger control options, with default values, are shown below. Complete descriptions of each item can be found in the CAESAR II User's Guide.


61

Hangers To set the default value displa yed in the Hanger Table box, specify the Default Spring Hanger Table setting in Configuratio n Editor > Database Definitio ns.

Whenever CAESAR II detects a proposed spring hanger location supporting zero load, the software displays a warning and does not select a hanger for this location. You can remove this proposed hanger location from the model to eliminate the warning.

Hanger Design Recommendation s There are instances where the stiffness of the adjacent piping and the hanger location restraints in the restrained weight case in teract unfavorably. This can produce an unwanted distribution of loads. Often, reducing the stiffness used to compute the hanger loads in the restrained weight run eliminates these load distribution problems. T he default for this stiffness is 1.0E12. Values on the order of 50,000 or 75,000 have been used successfully to somewhat relax the system and redistribute these piping loads. You can define the value the software uses for calculating hanger restrained weight loads in the Hanger Default Restraint Stiffness configuration setting in Tools > Configur e/Setup > Computation al Control. . The operating case for hanger travel (free thermal case) can be analyzed either with no spring stiffness at the hanger locations, or with the s tiffness of the selected springs inserted at those locations. In the second case, the software selects the springs through an iterative process. You can specify the manner in which the software handles spring hangers in the Include Spring Hanger Stiffness in Hanger OPE Travel Cases configuration setting in Tools > Configure/Setup > Computational Control . Inserting the actual hanger stiffness into the Operating Case for Hanger Travel may give a more accurate result, but may also introduce convergence problems. Therefore, in the latter case, it is very important that you adjust the hanger load in the cold case (in the physical system) to match the reported hanger cold load.


62

Hangers

Simple Hanger Desig n Double-click the Hanger check box on the Classic Piping Input dialog box to enter the spring hanger data for a particular node. For a simple hanger, no additional input is required. A number of the parameters from the Hanger Design Cont rol Data dialog box also display on the individual hanger auxiliary data fields. You can set these parameters globall y on the Hanger Design Contro l Data dialog box. You can override them locally on the auxiliary data area for each hanger. Simple Hanger Design


63

Hangers

Single Can Design To indicate that the pipe is supported from below, type a negative number in the Av ail abl e Space box on the Hangers auxiliary data tab. This value represents the distance between the pipe support and the concrete foundation, or baseplate. S ee the CAESAR II User's Guide for each of the manufacturer's definitions of available space. If the available space is not a criteria in the hanger design, type a large n egative value (such as -1000). CAESAR II input plots use a different symbol for these base supports. Design of Sing le Can at One Node


64

Hangers

Constant Effort Suppor t Design To design a constant effort support, type a ver y small value (such as 0.001) in the Max. Al lo wed Travel Li mi t box. Design of Constant Effor t Support Design


65

Hangers

Constant Effort Supports - No Design Entering Constant Effort Support Data: 1.

Enter the constant effort support load (per hanger) in the Predefined Hanger Data box.

2.

Enter the number of constant support hangers at the location. Do not enter the spring rate or theoretical cold load.

The hanger design algorithm does not design hangers that are completely predefined. Multipl e Predefined Const ant Effor t Supports

The two constant effort supports at node 377 should carry 10484 pounds each.

Any other data entered on this Hanger dialog box is ignored.


66

Hangers

Existing Springs - No Design Entering Existing Spring Data: 1.

Enter the Spring Rate and the Theoretical Cold Load (installation load, on a per hanger basis) values.

2.

Enter the number of Variable Suppor t Hangers at the location.

The hanger design algorithm does not desi gn hangers that are completely predefined. Other data can exist for the spring location, but this data is not used. E ntered spring rates and theoretical cold loads are multiplied by the number of hangers at this location. CAESAR II requires the Theoretical Cold (Installation) Lo ad to pre-define the spring.

Theor et i cal Col d Load = Hot Load + Tr avel x Spr i ng Rat e where upward travel is positive. Predefined Spri ng Hanger

Hot Load = 2000 pounds Cold Load = 2000 pounds + abs(1.375*590) = 2811 pounds


67

Hangers

Multiple Can Design Enter the number of hangers or cans as a positive number in the No. of Hangers at Lo cation box. Placing a negative number in that field allows CAESAR II to design up to that number of hangers at the location. All other hanger design parameters are still active. Trapeze Hanger Assembly

Power Piping Springs Allowable Load Variation:15%

The software designs up to three cans at the support if the load is too high for a single or double can configuration.

Rigid Support Displacement Criteria: 0.05 inches


68

Hangers

Old Spring Redesign Use this option to determine if the old spring can still be used. If the old spring can be used, then the new preset (initial cold load) is determined. If the old spring cannot be used, then a new spring design is recommended. The old spring is always left in the problem for subsequent load case analysis. The old hanger information needed for the redesign is: 

The hanger table



The number of springs at the location



The old spring rate

The old spring rate is entered in the Spring Rate box under Predefined Hanger Data. The Theoretical Cold (Installation) Lo ad must not be specified. Old Spring Design

Three springs at node 97. Each h ave a spring rate of 1105 pounds/inch.


69

Hangers

Pipe and Hanger Suppo rted From Vessel Connecting nodes associated with hangers and cans function in the same way as connecting nodes with restraints. Connecting node displacements are incorporated in the hanger design algorithm. Pipe Supported by Hanger From Vessel

You must first disable Connect Geometry Through CNodes in the Geometry Directives category of Tools > Config ure/Setup . For more information, see "Configuration and Environment" in the CAESAR II User's Guide.

Hanger Design with Support Thermal Movement Type unique connecting node numbers that do not exist on any pipe element on the Hanger Au xi li ary Data tab in the Hanger Connecting Node box. The hanger is designed to act with one end at the hanger node and with one end at the hanger-connecting node. Thermal growth of the hanger-connecting node can be specified on any pipe element. In the following example, the hanger at node 9 is supported from a structural steel extension (node 1009) from a large vertical vessel. T he vessel at the point where the hanger is attached grows thermally in the plus Y direction by 3.5 inches.


70

Hangers Hanger wit h Support Thermal Movement

Hanger Between Two Pipes A pipe crossing overhead supports part of the weight of the lower pipe. The node on the pipe passing overhead is typed into the Hanger Aux iliary Data tab as the CNode. When using hangers with connecting nodes to design springs, be particularly careful that CAESAR II’s design hot load is accurate. To find the hot load, CAESAR II puts a rigid element between the pipe node and the support node (which may be another pipe node as in the example below), and runs a weight case. If both nodes are expected to deflect in the weight run, then the hanger weight loads are distributed to other parts of the piping s ystem and not to the hanger. In this case, you may have t o estimate the loads on the hanger in an independent run, and then manually type the operating load on the particular spring hanger with the connecting node. If zero-load constant-effort supports are designed for a spring location with a connecting node, switch the hanger node and the connecting node. In this situation, the pipe node tends to deflect downward in the weight run less than the connecting node. To CAESAR II, this looks like the connecting node is pushing down on the hanger node, thus “holding the pipe down.” Switching the hanger node and the hanger-connecting node eliminates this problem.


71

Hangers You must first disable the Connect Geometry Through CNodes configuration option to avoid plot and geometry errors. For more information, see "Configuration and Environment" in the CAESAR II User's Guide. Hanger Between Two Pipes

The pipe at 65 is supported by a spring hanger by the pipe at 470.

Hanger Design w ith Anchors in the Vicinity Hangers are designed to support a given weight load through a specified travel with a minimum of load variation. Most often, the weight load is that of the pipe between an anchor and the hanger. The travel is the displacement of the hanger node as it thermally expands away from the anchor. When weight sensitive anchors (such as equipment nozzles) are relatively close to the hangers (that is, less than four or five pipe diameters in the horizontal plane), the anchors should probably be freed during the hanger restrained weight run. When the anchors are freed, the weight of the pipe between the anchor and the hanger should fall almost in its entirety on the hanger. Anchor nodes to be released are typed on the specific Hanger Auxi liary Data tab. The anchor degrees of freedom are released according to the spe cified Free Code value. Anchor degrees of freedom are released for the hanger design restrained weight run only. If the Free Code value is not specified for an anchor or restraint to be freed, all degrees of freedom associated with the anchor or restraint are released for the restrained weight solution. Only linear restraints and anchors can be freed to cause additional wei ght to be carried by the hanger. See the following example for more information.


72

Hangers Hanger Design in the Vicinit y at Equipment o r Vessel Nozzle

The anchor at 5 is freed in the Y-direction; the anchor at 105 is freed in all directions.


73

Hangers

Hanger Design with User-Specified Operating Load In certain situations around equipment nozzles, usually where the piping leaving the nozzle is very complex or very rigid, the hanger design algorithm selects operating loads that are too small. In these cases you can o verride CAESAR II’s calculated operating (hot) lo ads. The design algorithm proceeds normally, except that the user-entered hot load is substituted for CAESAR II’s calculated value for both the hanger design and all post hanger design analysis load cases. Hanger Design w ith User-Specified Operating Load

In this configuration, freeing the anchors at 5 and 60 did not help the thermal case nozzle loads. It was postulated that, due to the stiffness of the overhead branches, the hanger calculated hot load was not sufficient. The calculated hot load was 2376 pounds. A new hot load of 4500 pounds. is tried here.


74

Hangers

Simple Bottomed Out Spring Spring supports that may bottom out have SPR f ollowing a translation direction in the restraint Type box (such as, YSPR for a vertical bottomed-out spring). When a bottom out spring is entered, the restraint auxiliary screen changes as follo ws: The Gap value changes to " x" (the permitted travel), and the Mu value changes to F (the initial spring load). The direction of permitted travel is assumed to be opposite to the initial load on the pipe. These definitions are setup to handle vertical springs. Because of this, the " x " and F inputs are always entered as positive as shown in the following example. Input for Bot tom Out Spring Can Model

Simple bottomed-out spring restraints are used most often to conveniently enter predefined springs into the piping system m odel. These spring restraints provide a bottoming out capability that occurs when the spring has exceeded its maximum travel limit. Always enter the stiffness Stif , the allowed travel " x" , and the initial load on the spring F, to use the bottomed out spring model. If the travel " x " is not entered, it defaults to zero. If the initial load is not entered it also defaults to zero, and its sign is positive. No hanger should be entered at the same position as a bottomed out spring. Known information: 

Theoretical Installed Load = 1023 lb.



Spring Rate = 260 lb./in.


75

Hangers 

Largest Load in Hanger Table = 1690 lb.

Lift Off Spring Can Known information: 

Theoretical Installed Load = 1023 pounds.



Spring Rate = 260 pounds/inch.



Smallest Load in Hanger Table = 910 pounds.

To get from the installed condition to the initiate lift-off condition the can m ust displace in the positive Y direction. Input for L ift Off Sprin g Can Model

K1 = Spring Rate FY = 1023 pounds - 910 pounds = 113 pounds

FY = Theoretical Installed Load


76

Hangers

Modeling Spring Cans w ith Frictio n In many systems, portions of the pipe are supported by s pring cans. These spring cans perform the same function as spring hangers, except that they are below the pipe, pushing up. In some models, these spring cans are allowed to slide on their foundation, subjecting the system to friction forces. Each support of this type needs the following: 





A rigid element from the pipe center to the top of the can. Length equals pipe radius + insulation thickness + cladding thickness + shoe height + any trunnion height. A CNode to connect to the spring. Except for the vertical spring stiffness, all other displays of freedoms are rigidly connected. A rigid element representing the height of the spring can.

These points are illustrated in the example below. Model of Spring Can with Friction

Alternatively, element 15-20 can be omitted with the +Y restraint (with friction) placed directly on node 15. This modeling technique can also be used in situations where the shoe or trunnion slides on top of a bolted spring can.


77

SECTION 5

Expa xpansio nsion n Joints Joi nts This section provides guidelines and recommended best practices for modeling expansion joints in CAESAR II.

In This Section Universal Expansion Joints - Simple Model .................................. 78 Tied Bellows Expansion Joint - Simple Model ............................... 82 Universal Joint with Lateral Controls Stops - Comprehensive Tie Rod Model ............................................................................................. 85 Hinged Joint ................................................................................... 86 Slotted Hinge Joint - Simple Model ............................................... 87 Slotted Hinge Joint - Comprehensive Model ................................. 89 Tied Bellows - Simple vs. Complex Model .................................... 90 Slip Joint ........................................................................................ 91 Gimbal Joints ................................................................................. ................................................................................. 92 Tied Bellows Expansion Joint - Complex Model............................ Model............................ 93 Dual Gimbal ................................................................................... 95 Simple Bellows with Pressure Thrust ............................................ 97 Pressure-Balanced Tees and Elbows ........................................... 100 Universal Joint - Comprehensive Tie Rod ..................................... ..................................... 101

Uni nive versal rsal Expansion Join Joints ts - Simple Model Model 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, assum ing three or more tie rods exist. Instead of lateral spring rates, enter bending spring rat es from the manufacturer’s catalog. For more information about bellows stiffnesses, see Simple Bellows with Pressure Thrust (on page 97). 97). Manufacturers publish a wide variety of data for u niversal expansion joints. In most cases, the published spring rates are for the universal joint as an 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 assem bly displacement limit so that the piping designer can verif y that neither of the bellows are over-extended. Many universal expansion joint assemblies have s tops 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, you must check the results to verify that the stops are not engaged. Stops should t ypically 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. Check the displacement limits for each of the expansion joints after the protected equipment An aly si s > Exp ans io n J oi nt Ratin Rat in g command loads are within the allowables. You can use the Analy to calculate relative bellows movements for evaluating the strength of the convolution. The


78

Expansion Joints Expansion Joint Rating analysis module works only on single bellows, which requires that you first model and then check each bellows in the universal assembly.

Some manufacturers believe that friction at the tie bar ends, plus other effects, serve to limit the overall lateral flexibility of this joint. A 10% increase in overall lateral stiffness is sometimes used to compensate for these frictional effects. The complex models are built b y running pipe elements, whose diameters are equal to the diameter of the tie bars and w hose wall thicknesses are equal to half of the tie bar diameter, between rigid elements that extend normal to the p ipe axis and from the c enterline and to their intersection with the tie bar centerline. The weights of the bellows and associated hardware are 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. Field situations, such as loose nuts on tie bars, can be modeled using the complex expansion joint model. Descriptions of various universal models are shown below. Each model also includes example inputs. Only use simple models when you know that both ends of the tie bars are fixed to the flanges, that is, 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. Model this configuration with a complex joint model unless you are certain that all tie bars will remain in tension. The top model is used when you are given global assembly data for the universal, such as the assembly lateral stiffness. The second model is us ed when you are given angular spring rates for each of the two bellows used in the model.


79

Expansion Joints When provided equivalent single bellows lateral stiffness for the whole assembly: Universal Expansion Expansion Jo ints - Simple Models

The model below does not show the addition of any extra hardware or bellows weights, which could affect load distribution and spring hanger design in the area.


80

Expansion Joints When provided individual bellows angular stiffness:

Universal Expansion Expansion Joi nts - Simple Models Indivi Indivi dual Bellows

Tie rods and center spool pieces should normally be modeled at ambient temperature.

Pressure thrust is contained by double-nutted tie rods. Effective ID and axial restraints can be eliminated.


81

Expansion Joints Universal Expansion Expansion Joi nts - Simple Models Indivi Indivi dual Bellows

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.

Tied Tie d Bellow s Expansi Expansion on Join Jointt - Simpl Simple e Model Model Calculate the lateral stiffness for the bellows. T he flexible length, or convolution l ength, of the bellows is not listed in most expansion joint catalogs. The listed lengths include the rigid end pieces, such as flanges or pip e ends. Because 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 the following example, the total length of a four-convolution joint is 8-inches, and the total length of an 8-convolution joint is 12-inches. This means that the extra four convolutions a dd 4-inches, making the length of all twelve convolutions 12-inches. This also indicates that the rigid end pieces on this joint of four, eight, or twelve convolutions are 4-inches. 1/2

Deff = (4Aeff /π) = 10.0 in. 2

KTR = (3/2) (K AX) (Deff /L)

L = Flexible Convolution Length = 12 in. KTR = (3/2) (850) (10.0/12.0)

2

= 885.4 lb/in.


82

Expansion Joints Tied Bellows - Simple Model

Zero-weight rigid element (tie rod) Build the CAESAR II model of the flexible portion of the expansion joint. The rotational restraints between nodes 29 and 30 keep the two flanges parallel, which assumes three or more tie rods. In the field, the tie bars at four points around the expansion joint keep the flanges parallel. The flanges and the tie b ars form a parallelogram upon lateral deflection.


83

Expansion Joints Tied Bellows - Simple Model, continued


84

Expansion Joints

Universal Joint with Lateral Controls Stops Comprehensive Tie Rod Model Double-acting restraints with connecting nodes and gaps are used to model stop gaps along the tie bars. Stops along the tie bars are installed to restrict lateral motion at each end of the universal joint.

The following groups illustrate the m ethod used to construct the universal joint with lateral stops shown above. Only the right-side tie rod elements are shown below.

— St andar d pi pe el ement s — 34- 36 / 36- 38 — Ri gi d f l ange el ement s — 30- 32 / 40- 42 — Bel l ows el ement s — 32- 34 / 38- 40 — Ri gi d el ement s f r om t he pi pe t o the t i e bar cent er l i ne — ( Normal t o t he pi pe axi s) 30- 1030 / 36- 1036 / 42- 1042 — Ti e bar el ement s — 1003- 1002 / 1002- 1001 — Rest r ai nt s wi t h connect i ng nodes — RESTR NODE=1001 CNODE = 1042 TYPE = +Y , X , Z RESTR NODE=1002 CNODE = 1036 TYPE = Y w/ gap=1. 5 , X , Z


85

Expansion Joints

Hinged Joint Hinged joints use a zero-length expansion joint with rigid axial, transverse, and torsional stiffnesses. The bending stiffness equals the bending s tiffness of the hinge. You define the hinge directions using restraints and connecting nodes. The restraint line of action is always normal to the hinge axis. Manufacturers define hinged joints to take pressure thrust. Verify that the joi nt 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. As the axial load that the hinge carries becomes large, the hinge friction effect increases. You can make approximations to this increase in bending stiffness by increasing the stiffness of the bello ws proportionally to the axial load on the hinge. The expansion joint manufacturer can provide assistance in determining this. Typical geometries for hinged expansion joints are shown in the following figures:

In the next example, the hinged joint is zero length and is defined between nodes 45 and 46. X is the hinge axis, meaning that all relative rotations are permitted between nodes 45 and 46 about the X-axis. Nodes 45 and 46 are fixed rotationally relative to each other in the Y-axis.

The table below shows the input data used for the hinged joint as shown above.


86

Expansion Joints Hinged Joint Model


87

Expansion Joints

Slotted Hinge Joint - Simple Model The hinged joint is modeled using a zero length expansion joint and rigid elements with zero weight to define the interaction of the h inge geometry. Hinge directions are defined using restraints with connecting nodes. The restraint line of action is always normal to the hinge axis. Slotted Hinged Joint - Simple Model

Elements from nodes 10 to 15 and from nodes 16 to 20 are weightless, 9-inch long rigids.

In the example model shown above, the software presumes the relative rotation at the hinge about the Y-axis to be zero. The slots on either side 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. 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.


88

Expansion Joints

Slotted Hinge Joint - Comprehensi ve Model The next model is somewhat different from the previous m odel 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. Slotted Hinge Joint - Compr ehensive

Zero weight rigid elements that define the hinge assembly are listed below:

10 10 55 55 15 35 50 30

-

15 35 30 50 20 40 45 25

Normal t o pi pe axi s t o cent erl i ne of hi nge assy. " " " Par al l el t o pi pe axi s t o cent er l i ne of hi nge axi s. " " "

The finite length bellows must be defined accurately between nodes 10 and 55. Typically, this means entering the correct flexible length and using the manufacturer’s axial and lateral spring rates.


89

Expansion Joints The manufacturer’s published angular spring rates may not be appropriate for use in finite length expansion joint models.

Tied Bellows - Simple vs. Comp lex Model Although there are no strict rules that dictate when to use a simple model or when to use a complex model, there are some basic guidelines that you can follow. Complex Model Guidelines

Use a complex model when any the following conditions exist: 











You are investigating a failure.

Simple Model Guidelines

Use a simple model when any of the following conditions exist: 

The pipe diameter and number of convolutions become large. The nuts are only on the outside of the flange, allowing the tie bars to only carry tension.



The tie bars are either guaranteed to be carrying tension or have nuts on either side of the flange and ca n carry compression, if needed. There is no relative rotation of the ends.

You have good values for the load distribution in the tie bars. Simple models give no indication of the l oad distribution. In cases where the tie bars combine to resist relative bending of the joint ends, o ne pair of tie bars can be in compression while the other pair is in tension. T his effective redistribution of load in the tie bars is never 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. For bellows with only two tie rods, there can only be rotation in one direction.

Because of the uncertainty of the application, enter the lateral instead of the bending spring rate from the manufacturer’s catalog. Add the weights of the bellow and associated hardware to the flange weights on either side of the bellow. This is particularly true if the expansion joint is between a hanger to be sized and an anchor.


90

Expansion Joints When using expansion joints, verify that the displacement limits for the expansion joint after the protected equipment loads are within the allowables. In CAESAR II, you can use the Analysis > Expansion Joint Rating command to help compute relative bel low movements for evaluating the bellow distortion. You can build simple models of single tied bellows by entering a large axial stiffness. This axial stiffness simulates the tie bars, preventing relative axial movement of the bellows. You can model tie rods with a single rigid element along the centerline of the bellow. With zero weight and rotational restraints, this prevents the ends of the j oint from rotating relative to one another. In reality, the tie bars being offset from the centerline prevent this rotation. You can build complex models of tied bellows by running pipe elements whose diameters are equal to the diameter of the tie bars, and whose wall thicknesses are 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. For more information on building complex models, see the Tied Bellows Expansion Joint - Complex Model (on page 93). Some manufacturers believe that friction at the tie bar ends, plus other effects, serve to limit the overall lateral flexibility of this joint. 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.

Slip Joint 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. They are usually categorized as having two annular packing glands that are separated axially along the joint by a dead air space or by a small bellows sleeve. 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.

Slip Joint


91

Expansion Joints



Typical delta dimensions are: 







5 - 10. The distance from the closest guide or support to the end of the joint. The same values are also used for 25 - 30. 10 - 15. The effective length of the joint, if known, or the travel expected plus 4-inches, or a 12-inch estimate, if nothing else is known.

K1 is the spring stiffness for forces below the yield force, FY. 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 an d supports given by the vendor. 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 100feet.



FY is the joint friction thrust from the vendor catalog. Typical values are given as 400 pounds multiplied by the nominal pipe size.

Gimbal Joint s Gimbaled joints are designed to resist pressure thrust. CAESAR II recommends that you verify that the joint manufacturer is aware of the des ign loads on the gim bals. External loads are not always given. If axial loads are given, the bellows designer must know whether it includes or excludes pressure thrust. You can model the angular-only gimbal as a zero length expansion joint with rigid axial, transverse, and torsional stiffnesses. The bending stiffness is set equal to the rotational stiffness specified in the manufacturer's catalog. CAESAR II recommends that you thoroughly model angular and offset gimbals, as shown in the following figures. Angular and offset gimbaled joints are usually installed in large diameter lines where lumped-property assumptions for the bellows may not be within reasonable engineering accuracy.


92

Expansion Joints An gu lar -Onl y Gi mb aled Jo in t

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

Tied Bellow s Expansion Joint - Complex Model In the example shown below, the flexible joint is between nodes 30 and 35. The flanged ends of the joint are modeled as the rigid elements in nodes 20 to 30 and nodes 35 to 45. Additional rigid elements, perpendicular to the pipe axis, extend from each flange. The tie rods are 1-inch in diameter. The following nodal layout and corresponding data input is used to build a comprehensive model of the tied bellows.


93

Expansion Joints Tied Bellows Complex Model


94

Expansion Joints Weightless rigid elements extend from the flange centerline to the outside edge of the flanges where the tie rods are attached. Only two of eight element data inputs are sh own in these examples.

Tie rod is usually at ambient temperature, but it is important to specify this correctly.

Dual Gimb al Dual gimbal joints are two, usually angular-only, gimbaled joints in series in the pipeline. Putting two (or three) angular-only gimbaled joints together provides t he ability to absorb lateral and possibly axial deformation. Pipe flexibility software is unable to model the axial-only component of the possible deformation because it requires large rotation of the e xpansion joint components. Use the single angular deformation only gimbals in series with at least one other gimbaled joint. It is only in series that the angular deformation only gimbal provides for any lateral movement. Gimbaled joints are designed to take pressure thrust. Verify that the joint manufacturer is aware of the design loads on the gimbal. Model each individual angular-only gimbal joint as a zero length expansion joint with rigid axial, transverse, and torsional stiffnesses. Ensure that the bending stiffness is equal to the manufacturer's published rotational stiffness term. The minimum required distance, or L , between adjacent single gimbaled joints (shown as 8-7 in the following example) is principally a function of the angular and rotational deformation to be absorbed, the diameter, and the number of convolutions per j oint. The following figure shows a dual gimbal comprised of two angular-onl y gimbals. The bending stiffness for each gimbaled joint is 490.0-inches lb./deg.


95

Expansion Joints Dual Gimbal (Angul ar-Only)

Because both expansion joints use the same stiffness values, it is shown only once in this example.


96

Expansion Joints

Simple Bellow s with Pressure Thrust The numbers used for the expansion joint ex amples in this topic are for illustrative purposes only. CAESAR II recommends that you verify all values used from a qualified technical source. You can model bellows expansion joints using either a zero or a finite length. When finite length bellows are used, leave either the bending or the transverse stiffness blank. CAESAR II calculates the exact stiffness coefficient for the term left blank. If all stiffnesses are defined, CAESAR II uses the defined data and skips the stiffness calculations. For f inite length expansion joints, leave the Bending Stif box empty and enter the la teral stiffness given by the manufacturer into the Trans Stif box on the Expansion Joints Auxiliary Data tab in the Classic Piping Input dialog box. CAESAR II uses the Expansion Joint Manufacturer's Association (EJMA) equations to calculate the bending (angular) or transverse (lateral) stiffness of metallic bello ws. EJMA defines these stiffness values as: 2

Ktr = (1.5K ax)(Deff /L)

(1)

2

Kb = (Kax) (Deff ) /8

(2)

Where Kb = Bending angular stiffness of the bellows Ktr = Transverse lateral stiffness of the bello ws Kax = Axial stiffness of the bellows Deff = Effective diameter mean diameter of the bellows L = Length of the bellows for all convolutions Equation (2) above is based on modeling bellows as zero-length where the expansion joint acts as a hinge. Because it is possible to leave the Effective ID box blank (zero), CAESAR II use s derived formulas for the calculation of the bending or transverse stiffness in terms of the other stiffness, thereby eliminating the effective diameter from the equations. 2

Kb = (Ktr L /3)(π/180)

(3)

Equation (3) has been increased b y a factor of 4 from that derived from equations (1) and (2) above. or 2

Ktr = (3Kb/L )(180/π)

(4)

Equation (4) has been decreased b y a factor of 4 from that derived from equations (1) and (2) above. Bellows were originally modeled as zero-length elements. Finite-length bellows in bending cause a translation while zero-length bellows do not. For zero-length bellows, hold the translation of a finite-length bellows to zero while in bending. This requires a bending stiffness of four times that of the zero-length expansion joint, regardless of the length of the finite-length bellows. This is similar to the guided cantilever methods.


97

Expansion Joints Most expansion joint manufacturer's catalog values for bending stiffness are still based on zero-length bellows, while the transverse stiffness is normally based on the finite-length of bellows. For more information refer to equations (1) and (2) above. For this reason, enter the transverse stiffness and leave the Bending Stif box blank. CAESAR II then calculates the proper bending stiffness according to equation (3) and uses this value for the flexibility analysis. If you enter a value for K TR that does not conform to equation (1), equation (3) uses the entered value to calculate Kb. If a catalog value of K b is used and that value is based on zero-length values, CAESAR II uses equation (4), which results in one-quarter of the value of K TR.

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For a zero length expansion joint, CAESAR II uses 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. Accurate torsional stiffnesses and allowable torsional rotations should be obtained from the vendor. Systems using untied bellows should either be of very low pressure or adequately anchored and guided to withstand the possibl y 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 a dded as concentrated forces. This is particularly true when the bellow is part of a hanger sizing weight calculation.

A zero or blank Effective ID results in a zero pressure thrust. The Effective ID is the mean diameter of the bellows and is used to find the area f or 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 if the pressure is positive. The magnitude of the thrust load is P × Aeff , where P is the pressure in the pipe above atmospheric, and A eff is the area, found from Aeff = π/4 × (DEff )

2

Many manufacturers specify the effective area of the bellows. The Effective ID for CAESAR II input can be calculated using the following equation: ½

Deff = [4Aeff /π]

In the example below, the untied bellows runs between nodes 8 and 9. The elbow at node 11 is anchored to take the thrust load developed in the bellows. The manufacturer's specification for the axial stiffness of the joint is 5131 lbs/in. with a transverse stiffness of 27058 lbs/in. The bending stiffness is left blank and is calculated by CAESAR II because the bellows has a finite length. The pump and the baseplate at node 5 must be able to withstand the large axial force that develops due to pressure thrust in the bellows.


98

Expansion Joints Bellows with Pressure Thrust

Aeff = 69 in

2

P = 175 psi Thrust = 69(175)=12075 lbs. (automatically applied by CAESAR II)

* The bending stiffness ( Bending Stif) calculated by CAESAR II: 2

KBend = 1/3Ktr L (π/180) = 984 in-lbs/degree


99

Expansion Joints

Pressu re-Balanced Tees and Elbows Pressure-balanced tees and elbows are used primaril y 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 the coding of a pressure-balanced tee in a turbine exhaust line. The bottom side of the tee is bla nked off. The tee is a standard, unreinforced fabricated tee. The tie bars only act in tension.


100

Expansion Joints

Universal Joint - Compr ehensive Tie Rod The comprehensive universal joint model involves defining all tie rods and connections between tie rods and end plates.

The following groups illustrate the m ethod used to construct the universal expansion joi nt model shown above.

—Ri gi d El ement s ( Fl anges) — 15- 17 / 31- 33 —Ri gi d El ement s nor mal t o t he pi pe axi s and bet ween t he pi pe and t i e bar c ent er l i nes . Used 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 —Ri gi d El ement s nor mal t o t he pi pe axi s, and bet ween t he pi pe and t i e bar c ent er l i nes . Used at the end where there are nuts only on the backside of the flange:

15- 1015 / 15- 2015 / 15- 3015 ——I nt er medi at e l at er al t ee suppor t s ( Ri gi d) — 23- 1023 / 23- 2023 / 23- 3023 25- 1025 / 25- 2025 / 25- 3025 ——Ti e bar s — 1033- 1034- 1035- 1036 2033- 2034- 2035- 2036 3033- 3034- 3035- 3036 — Rest r ai nt s wi t h connect i ng nodes at t he t ensi on- onl y f l ange end. —— RESTR NODE = 1036 CNODE = 1015 TYPE = -X , Y , Z RESTR NODE = 2036 CNODE = 2015 TYPE = -X , Y , Z RESTR NODE = 3036 CNODE = 3015 TYPE = -X , Y , Z — Rest r ai nt s wi t h connect i ng nodes at t he i nt er medi at e support poi nt s. RESTR NODE = 1035 CNODE = 1023 TYPE = Y , Z RESTR NODE = 2035 CNODE = 2023 TYPE = Y , Z RESTR NODE = 3035 CNODE = 3023 TYPE = Y , Z


101

Expansion Joints RESTR NODE RESTR NODE RESTR NODE

= = =

1034 2034 3034


CNODE = CNODE = CNODE =

1025 2025 3025

TYPE TYPE TYPE

= = =

Y , Z Y , Z Y , Z

102

SECTION 6

Miscellaneous Models This section discusses modeling techniques for the components not explicitly covered in e arlier sections.

In This Section Reducers........................................................................................ Ball Joints....................................................................................... Jacketed Pipe ................................................................................ Cold Spring .................................................................................... Connecting Equipment ..................................................................

103 104 106 108 108

Reducers 1.

Define the reducer length as you would any pipe element. For eccentric reducers, skew the element so that the To node matches the centerline of the following pipe elements.

2.

Double-click the Reducer check box on the Classic Piping Input dialog box. If the elements preceding and following the reducer are already defined (such as if you are inserting this element) then CAESAR II automatically calculates all of the reducer input data. In that case, leave the Reducer Auxi liary Data tab fields blank.

3.

Type the diameter and wall thickness of the pipe that follows the reducer.





You can type nominal diameter and wall thickness and CAESAR II will convert these to actual diameter and wall thickness if this option is activated in the units file. (If the Diameter and Wt/Sch boxes on the Classic Piping Input dialog box convert nominal to actual, then the Reducer Aux iliary Data tab boxes will also make this conversion.) Al ph a is the slope of the (concentric) reducer transition in degrees. If unspecified, CAESAR II calculates alpha using 60 percent of the entered reducer length.

Alpha = atan[ (D1-D2) / (2*length of the sloped portion of the reducer*0.6) ] IGE /TD -12 requires entry of the reducer Al ph a as well as R1 and R2 (the reducer transition radii of the large and sm all ends).


103

Miscellaneous Models Reducer Example

Ball Joints Ball joints can be modeled with zero-length expansion joints, or with restraints and connecting nodes. When using expansion joints, each ball and socket is defined with one zero-length expansion joint having rigid axial and transverse stiffnesses and zero bending and torsional stiffnesses. If torsional stiffness is small, use a value of 1.0. Results are invalid for large rotations.


104

Miscellaneous Models Two Methods of B all Joint Modeling Method 1

Modeling a ball joint between nodes 20 and 21 using a zero-length expansion joint.

Input Specifics

Ball Joint Diagram

Node-CNode Model


105

Miscellaneous Models Method 2

Modeling a ball joint between nodes 20 and 21 using axial, translational restraints with CNodes. Input Specifics The next element starts at node 21 (the From node).

Ball Joint Diagram

Node-CNode Model

Jacketed Pipe Jacketed piping systems are modeled by running the jacket elements directly on top of the core elements so that the two are concentric. You can generate a jacketed pipe model by modeling the entire core and then duplicating the core piping using an appropriate node increment (such as 1000). This produces a second run of pipe which is modified to build the jacket model. For the jacket, change the pipe size, temperature, bend radii, and so on, to finish the model. Then, attach the jacket and core by changing the node numbers and adding restraints. Typically, the end caps connecting the c ore to the jacket pipe are much stiffer than either the core or the jacket. For this reason, node p airs like (10 and 1010), (25 and 1025), (35 and 1035), and (40 and 1040) are often j oined by using the sam e node for each. For example, the


106

Miscellaneous Models displacements and rotations at the end of the c ore pipe are assumed to be the same as the displacements and rotations at the end of the j acket pipe. Internal spiders offer negligible resistance to bending and axial relative deformation. Node 15 might be connected to node 1015 by a restraint with a connecting node. For an X-run of pipe, rigid restraints would exist between the two nodes for the Y- and Z-degrees of freedom. The +Y support acting on the jacket at node 1020 does not cause any stiffnesses to be inserted between 20 and 1020. Node 20 is included in the model so that the outside diameter interference can be checked at the 20-1020 cross sections. If there is any concern about interference or interference-related stresses at the 20-1020 nodes, then you can use restraints with connecting nodes and gaps to approximate the pipe-inside-a-pipe with clearance geometry. Because CAESAR II constructs the jacketed piping model by associating nodal degrees of freedom, the software does not know one pipe is inside of another. Therefore, consider the following items: 





If both the jacket and the core are fluid-filled, the fluid density of the jacket must be reduced to avoid excess (incorrect) weight. If wind or wave loads are specified, the wind or wave loading must be deactivated for the core so that the core does not pick up the load. The core pipe should probably have its insulation thickness set to zero.


107

Miscellaneous Models

Cold Spring See the "Technical Discussions" section of the CAESAR II User's Guide for a detailed discussion of the method for analyzing cold springs. Cut Short

Material 18 is used for cut short .

Connecting Equipment Verti cal Vessels Average diameter of the skirt = 102 + 78/ 2 = 90 inches. Average temperature of the skirt = 87º F. Temperature of the vessel = 325º F. Nozzle N1: 

OD = 10.750.

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Wall = 0.5.

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Length of Flange = 4.0 inches.


108

Miscellaneous Models 

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Weight of single flange = 112.

Element 20 to 125 is rigid. The associated diameter and wall thickness is that of the vessel. The element from 20 to 125 is stiff relative to the vessel. The element from 15 to 215 is stiff relative to the vessel. The rigid element from 135 to 140 models the flange at the end of the nozzle and is rigid relative to the nozzle diameter. The rigid element 225 to 230 models the flange at the end of the nozzle and is rigid relative to the nozzle diameter. Local shell flexibilities are defined between 130 and 135 and 220 and 215. The example considers the cantilever bending, the shear in the vessel from the skirt, and the local flexibility of the vessel shell in the vicinity of the nozzle.


109


Vessel Skir t The following partial dialog boxes illustrate the modeling techniques used to define the vertical vessel.


110


Vessel

Rigid Elements These rigid elements are weightless, but the y do undergo thermal expansion.


111


N1 Nozzle

Flange


112


N2 Nozzle


113


Horizontal Vessels Horizontal vessel models are built using combinations of straight pipe and nozzle flexibility simulations (WRC 297). The following example illustrates the most accurate way to define horizontal vessel flexibility.

NOZZLE N3: 

OD = 12.750.

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Wall= 0.687.

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Flange length = 5.0 inches.

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Flange weight = 250 lb.

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Elements 5 to 6, 6 to 10, 15 to 16, 16 to 20, and 20 to 22 are rigid and the associated diameter and wall thickness is that of the vessel. These rigid elem ents are stiff relative to the vessel. The rigid element from 26 to 30 models the flange at the end of the nozzle and are rigid relative to the nozzle diameter.


114

Miscellaneous Models 

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Local shell flexibilities are defined between the nodes 22 and 24. The above model considers the flexibility of the horizontal vessel section, the free translation horizontal restraint at 15, and the local flexibility of the vessel shell in the vicinity of the nozzle.

The following partial dialog boxes illustrate the modeling techniques used to define the horizontal vessel. Vessel Legs - portion not in contact with vessel T=100Fº, P=0.0psi Vessel Legs Not i n Contact w ith Vessel

Element 15-16 is identical except for the restraints as shown above.


115

Miscellaneous Models Vessel Legs in Contact wit h Vessel

DY = Vessel Radius Element 16-20 is identical.

Vessel Modeled As Pipe Vessel modeled as pipe (not rigid) P≠0, Fluid Density ≠0: Vessel Modeled As Pipe

Zero-Weight Rigid To Nozzle Interface Location

Pipe element (not rigid) P≠0, Fluid Density ≠0:


116

Miscellaneous Models Nozzle Element With WRC 297 Connection Defined

The vessel flexibilities are calculated and reported during error checking. These values are used automatically in the analysis. This WRC297 data is available to output in the Miscellaneous Data report.


117

SECTION 7

Tutorial A This tutorial provides a step-by-step instructions for defining a typical piping system, analyzing flexibility and stresses, and reporting results. This process includes: 

Creation and entry of the pipe stress model.



Analysis and evaluation of the results.



Redesign of the system (as covered in Tutorial B (on page 158)).

The piping system you will model defines part of a refining process that moves crude from the bottom pump to a steam stripper unit. The end suction top discharge pump has a 10-inch suction nozzle and an 8-inch discharge nozzle. The 8-inch line runs through a check valve with a 6-inch bypass to a spring hanger support. The 8-inch line then runs over a hard support before entering the vertical vessel.

The boundaries of the system are the pump discharge nozzle (on the right) and the vessel nozzle (on the left). The pump nozzle is a satisfactory boundary because the movement of that point (as the pump heats up in operation) is known and easil y calculated from the thermal strain between the pump nozzle and the base point. The vessel nozzle is an adequate boundary because of the known thermal growth of the vessel and the greater stiffness of the vessel with respect to the 8-inch pipe.


118

Tutorial A

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You can take an opposite approach by modeling the pipe ends to immovable points, such as the vessel foundation and the pump support (or base) point. When you require a more accurate model for supporting structures, you can include structural steel in the model.

The check valve sits on top of the welding tee for the 6-inch bypass piping. The 6-inch line runs through a gate valve before reentering the 8-inch line through a second welding tee above the check valve. The total weight and length of this valve assembly is unknown. Because of this, the valve lengths and weights are pulled from the CAESAR II generic database. The spring hanger above this val ve assembly carries the deadweight and absorbs the thermal growth of the vertical pipe run. The hanger attaches to the elbow in line with the vertical pipe at the near end of the elbow. The hanger is quite sensitive to the weights used. The difference between the actual installed valve weights and modeled weights should be used to adjust the spring preload. In Tutorial B you will verify that the hot load on the spring is toward the center of the manufacturer's recommended spring working range to allow errors in load estimation. An appreciable change in these weights requ ires reanalysis of the system.

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The weld point on the vertical run of the elbow is the near end and the horizontal run weld point is the far end. The other end of the hanger attaches to available structure above the model. Because of the vertical thermal growth of the hanger attachment point, you cannot use a simple rod hanger. The horizontal piping rests on an unspecified support at the far end of the next elbow. This support, modeled as a rigid nonlinear restraint acting on the pipe centerline, allo ws the piping to move upward but prevents downward motion.

Steps in this tutorial 1.

Review the piping system drawing (on page 120)

2.

Configure the software (on page 121)

3.

Define the first pipe element (on page 122)

4.

Define the vertical pipe run (on page 127)

5.

Define the horizontal pipe run (on page 131)

6.

Define the 6-inch bypass (on page 136)

7.

Review the piping input (on page 141)

8.

Check for errors (on page 144)

9.

Check the static load cases (on page 144)

10. Run the static analysis (on page 146) 11. Review the analysis results (on page 146) 12. Conclusions (on page 157)


119

Tutorial A

Review the piping system drawing The following drawing contains the detail you need to construct the model.

You will assign node numbers where there is: 

A change in geometry, such as a pipe diameter or wall thickness.



A change in direction, materials, temperature, or pressure operating conditions.



The application of boundary conditions such as restraints, point loads, and displacements.



Any other location for which you want output.

Node numbering increments by fives, starting with node 5 at the pump nozzle. T he 6-inch bypass piping uses the same progression, but starts with node 600. The software uses the nodes to analyze the piping stress. Output for each elbow is available for nodes at the near, mid, and far points of a bend (at 0, 45, and 90-degrees).


120

Tutorial A

Configure the software 1.

Open CAESAR II from the Windows Start menu. The CAESAR II main window displays.

2.

In the main window ribbon, click Home > Setup > Configure

.

The CAESAR II Config uration Editor displays.

3.

Set the numeric increment between nodes, using an increment of 5 between node numbers. The default increment is 10, so you must change it. Click the Geometry Directives category, and then select 5 in the Au to Nod e Num ber Increment list.

4.

Click Save and Exi t

.

The software saves the change, closes the CAESAR II Config uration Edit or , and returns to the main CAESAR II window.

5.

In the main window ribbon, click Home > New

.

The New Jo b Name Specifi cation dialog box displays.


121

Tutorial A 6. Type Tutor as the file name. Note the folder in which to store the file. The default folder is C:\ProgramData\Intergraph CAS\CAESAR II\\Examples. You can click Browse to navigate to another folder.

C:\ProgramData is a hidden folder. 7.

Click OK . The software saves the job file, closes the New Jo b Name Specifi cation dialog box, and displays the Review Current Unit s dialog box with the English units used for all piping element fields.

8.

Click OK to close the Review Current Units dialog box and return to the main CAESAR II window.

Define the firs t pipe element Periodically click Save or File > Save as you create elements. CAESAR II also interrupts your session and prompts you to save your work 30 minutes after the last save.

Node 5-10 Pipe Element The first pipe element runs from the pump discharge nozzle (node 5) to the centerline intersection of the 8-inch main line with the 6-inch bypass (node 10). The length of this element is 2 ft. in the Y direction.


122

Tutorial A

1.

In the main window ribbon, click Home > Input > Piping Input

.

Because this is a new job, the Review Curr ent Units dialog box again displays. Click OK to close it. A new window with command toolbars, the Classic Piping Input dialog box, and a graphic view displays.

CAESAR II automatically generates the From and To nodes when you start a new piping element. The From box displays 5 and the To box displays 10 because you set the node increment to 5 in the CAESAR II Config uration Editor dialog box. The default -Mill Tol % value for the piping code also displays. The graphics view displays and updates your piping system as you define element properties.


123

Tutorial A 2.

In the DY box, type 2- to specify the element length of 2 ft. The dash ( - ) symbol indicates feet. If you type a length without the dash, the units are inches.

3.

In the Diameter box, type 8 to specify the nominal pipe size of 8 in. The software replaces the nominal value with the actual outer diameter (OD) of 8.6250 in. The pipe element updates in the graphic view.

4.

In the Wt/Sch box, type S to specify standard wall thickness in inches. The software replaces S with the actual wall thickness of 0.3220 in.

5.

6.

Type the following values: Corrosion

0.0313 (the corrosion allowance in inches)

Temp 1

600 (the maximum operating temperature in degrees F)

Pressure 1

30 (the maximum incidental pressure in lb./sq.in.)

In the Fluid Density box, type 0.8SG to specify the specific gravity of 80 percent of the deadweight of water. The software replaces the specific gravity with the fluid density of 0.02889 lb./cu.in.

7.

In the Material list, select (1)LOW CARBON as the pipe material. The software queries the material database and adds values for Elastic Modulus, Poisson’s Ratio, and Pipe Den. The software also references the material number to add the coefficient of expansion for the specified temperatures.


124

Tutorial A 8.

In the Insul Thk box, type 3 to specify the insulation thickness in inches. In the Insulation Density list, select Calcium Silicate. The software replaces the insulation material type with the insulation density of 0.00666 lb./cu.in.

Optionally, type 11.5/1728 in the Insulation Density box. 11.5 is the density in 3 lb./cu.ft. 1728 is the factor (12 ) to convert the density to lb./cu.in. Numeric fields allow simple math. 9.

Double-click the Displacements check box to display the Displacement tab on the right. 5 displays as the value for Node 1.

10. Define displacements for the thermal growth of the pump discharge nozzle from the base support (anchor) point. For anchor displacement DY in the Vector 1 column, type 0.077. For anchor displacement DZ in the Vector 1 column, type 0.046. The displacement vector displays in the graphic view.

11. Type 0 for the other four degrees of freedom ( DX, RX, RY, and RZ).

If you do not assign values for DX, RX, RY, and RZ, then node 5 is free to move in these directions. 12. Double-click the Al lo wab le St res s check box to display the Allow able Stress tab on the right. 13. In the Code list, select B31.3 if it is not the default selection.


125

Tutorial A 14. Type the following values: SC

20000 (the cold allowable stress in psi)

SH1

17300 (the hot allowable stress in psi)

(1)LOW CARBON is a generic material without associated allowable stress values in the material database. You must m anually define allowable stresses.

Do not use commas when typing allowable stress values. You can use an exponential format, such as 20e3. You now have the following material property and allowable stress values defined:

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If a job is new, CAESAR II first opens the Review Current Units dialog box. For an existing job, the Classic Piping Input dialog box opens directly with the first piping element active. If the Units File Label box on the Review Current Units dialog box does not show Imperial units, click Cancel , and then select Tools > Configur e Setup. Click the Database Definitions category and select English in the Units File Name list.


126

Tutorial A

Define the vertical pip e run Node 10 Welding Tee Intersectio n Node 10 defines the bottom intersection of the 8-inch main and 6-inch bypass lines. This intersection is constructed using an 8 x 6 welding tee. Piping codes recognize the reduced strength of this piping component by increasing the calculated stress at this point i n the system. For CAESAR II to include this stress intensification factor in the stress calculation, the node must be identified as a welding tee. 1.

Double-click the SIFs & Tees check box to display the SIFs/Tees tab on the right.

2.

In the Node box, specify 10 as the intersection node.

3. In the Type list, select 3 - Welding . CAESAR II calculates the SIFs at this intersection according to the selected B31.3 piping code.

Node 10-15 Pipe Element for Welding Tee This element runs from the intersection point (node 10) to the beginning of the check valve (node 15). This short run finishes out the welding tee. The length of the element is 7 inches in the Y direction.

1.

Click Continue

on the Navigation Tools toolbar to define the next length of pipe.

You can also press ALT-C or click Edit > Continue. The To node 10 of the previous element now displays as the From node. The To node is 15. All pipe material properties carry forward from the previous element. Allowable stress data also carries forward, even though the Allow able Stress box is cleared.

2. In the DY box, type 7 to specify the element length of 7 in. Press TAB.


127

Tutorial A

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Do not select Al lo wabl e Str ess unless you have a change in material, code, or temperature. Uniform Loads and Wind also carry forward without selecting the check box. No other component information, boundary conditions, or loading conditions carry forward.

Node 15-20 Check Valve The next element is the flanged c heck valve and the mating flanges. The software considers a valve to be a rigid element and includes the mating flanges because flange piping components are stiffer than the attached pipe. The software accesses the valve/flange database to automatically generate the valve and its data.

1.

Click Continue

on the Navigation Tools toolbar.

The To node 15 of the previous element now displays as the From node. The To node is 20.

2.

Click Model > Valve, or click Valve/Flange Database

on the Input Tools toolbar.

The Valve and Flange Database dialog box displays.

3. Select CHECK from the Rigid Type list and FLG from the End Type list. Click OK .


128

Tutorial A A 150 psi class flanged check valve displays between nodes 15 and 20.

The software adds valve data to the Classic Piping Input dialog box. The element length DY is 2 ft. 3.75 in., Rigid is selected, and the weight of the valve and flanges of 470.000 displays in the Rigids tab.





If you know the length and weight of the valve, you can directly specify the DY length and the Rigid Weight in the Rigids tab. If the Valve and Flange Database dialog box does not display, see Configuration Editor in the CAESAR II User's Guide.

Node 20-25 Pipe Element for Welding Tee This element runs from the end of the c heck valve (node 20) to the top intersection point (node 25) This short run finishes out the top welding tee. The length of this element is 7 inches in the Y direction (half of the total length of the 8-inch x 6-inch welding tee).

1.

Click Continue




129

Tutorial A 2.

In the DY box, type 7 to specify the element length of 7 in.

Node 25 Welding Tee Intersectio n Node 25 defines the top intersection of the 8-inch main and 6-inch bypass lines. This intersection is constructed using a second 8-inch x 6-inch welding tee that sits on top of the check valve.

1. Double-click the SIFs & Tees check box to display the SIFs/Tees tab on the right. 2. In the Node box, specify 25 as the intersection node. 3. In the Type list, select 3 - Welding . CAESAR II calculates the SIFs at this intersection according to the selected B31.3 piping code.

Node 25-30 Pipe Element The next element runs from the top intersection of the 8-inch main and 6-inch bypass lines (node 25) to the intersection of the vertical pipe centerline and the horizontal pipe centerline (node 30). The length of this element is 10 ft. 2 in. in the Y direction. Node 30 does not remain at this location on the piping system. Any additional input specified at 30 and all output for node 30 is located at the far weld point of the elbow, which connects the vertical and horizontal runs.

1.

Click Continue




130

Tutorial A 2.

In the DY box, type 10- 2 to specify the element length of 10 ft. 2 in. Press TAB.

Define the horizontal pip e run Node 30 Elbow and Hanger Node 30 defines the 90-degree elbow and the hanger as the pipe run turns from vertical to horizontal. When you define the el bow, the software adds nodes at the elbow near point and far point.

1.

Double-click the Bend check box to display the Bends tab on the right. The software creates a long radius elbow (1-1/2 times the nominal pipe diameter), node 28 at the elbow near point ( An gl e 2 of 0.000), and node 29 at the midpoint ( An gl e 1 of M). Node 30 moves to the elbow far point.

The elbow is part of the vertical pipe run, but does not display until you create the horizontal pipe run.

2.

Double-click the Hangers check box to display the Hangers tab on the right.


131

Tutorial A 3. Type 28 in the Node box to place the hanger at the elbow near point. 4.







Clear the Al lo w Sh or t Ran ge Sp ri ng s box. Use the default value of 1 - ANVIL for Hanger Table.

You can change the Radius value on the Bends tab. For the first iteration of the analysis, you will use the default hanger settings with no additional design data. Click in a hanger setting box on the Hangers tab and press F1 to open the help for more information. A short-range spring is not used at this point because a mid-range spring is usually less expensive.


132

Tutorial A Node 35-40 Pipe Element The next element starts the horizontal pipe run in the X direction from the far point of the vertical elbow (node 30) to the intersection of the horizontal pipe centerlines at the horizontal bend (node 35).

1.

Click Continue



2. In the DX box, type -12 ft., and then press TAB. The elbow on the vertical pipe element and the new horizontal pipe element display.

Node 35 Elbow and Restraint Node 35 defines the horizontal 90-degree el bow and the rigid restraint. When you define the elbow, the software adds nodes at the elbow near point and far point. The length is 12 ft. The restraint is non-linear, meaning its stiffness is not constant but is a function of load or displacement. It does not a llow the pipe to move downward, but allows the pipe to move upward. 1.

Double-click the Bend check box to display the Bends tab on the right. The software creates a long radius elbow (1-1/2 times the nominal pipe diameter), node 33 at the elbow near point ( An gl e 2 of 0.000), and node 34 at the midpoint ( An gl e 1 of M). Node 35 moves to the elbow far point. .


133

Tutorial A

Similar to the first elbow, the elbow is part of the horizontal pipe run, but does not display until you create the next pipe element.

2.

Double-click the Restraints check box to display the Restraints tab on the right.

3. Type +Y in the Type box to place the support at the elbow far point.

With no stiffness specified with the restraint, CAESAR II sets the restraint as very stiff (rigid). This means that under any practical load, the pipe does not push the restraint downward. The restraint displays as a vector.



The software represents all restraints, except anchors, as vectors.



You can specify up to four restraints on the Restraints tab.



To restrain lateral motion at node 35, use an X value to define a second guide restraint.


134

Tutorial A Node 35-40 Pipe Element The next element finished the horizontal pipe run in the Z direction from the far point of the horizontal elbow (node 35) to the termination at the intersection with the vessel wall (node 40). The length is 1 8 ft.

1.

Click Continue



2. In the DZ box, type 18-, and then press TAB . The horizontal elbow and the new horizontal pipe element display.

3.

Double-click the Displacements check box to display the Displacement tab on the right. 40 displays as the value for Node 1.

4.

Define displacements for the thermal growth of the pump discharge nozzle from the base support (anchor) point. For anchor displacement DY in the Vector 1 column, type 0.077. For anchor displacement DZ in the Vector 1 column, type 0.046.

5.

Define displacements for the thermal growth of the vessel nozzle. In the Vector 1 column, type the following values: DX

0.0

DY

0.28

DZ

-0.1

RX, RY, & RZ

0.0


135

Tutorial A The displacement vector displays in the graphic view.

Define the 6-inch bypass You now return to the 6-inch bypass piping around the 8-inch check valve above the pump. The welding tee nodes of 10 and 25 are defined as reducing tees when you model the 6-inch piping elements. Start the bypass from node 10 and create elements with 600-series nodes.

Node 10-605 Pipe Element The first element of the bypass runs from the reducing tee at node 10 to the intersection of the horizontal pipe centerline and the vertical pipe centerline (node 605). The length of the element is 2 ft.

1.

Click Continue


The To node displays 40 and the From node displays 45. The software assumes that you want to continue modeling from the last-created node.

2. Type 10 in the From node box and type 605 in the To node box. 3.

In the DX box, type -2-. The software measures the element length from the 8-inch centerline to the centerline of the vertical 6-inch line.

4.

In the Diameter box, type 6 to specify the nominal pipe size of 6 in. The software replaces the nominal value with the actual outer diameter (OD) of 6.6250 in.

5.

In the Wt/Sch box, type S to specify standard wall thickness in inches.


136

Tutorial A The software replaces S with the actual wall thickness of 0.2800 in.

Node 605 Elbow Node 605 defines the first 90-degree elbow of the bypass. The software adds nodes at the elbow near point and far point.

1.

Double-click the Bend check box to display the Bends tab on the right. The software creates a long radius elbow (1-1/2 times the nominal pipe diameter) for the 6-inch line, node 603 at the elbow near point ( An gl e 2 of 0.000), and node 604 at the midpoint ( An gl e 1 of M). Node 605 moves to the elbow far point.

2.

In the Type list, select 1 -Single Flange.

Similar to the previous elbows, the elbow does not display until you create the next pipe element.

Each elbow on the bypass is flanged on the end closest to the gate valve. The flange acts like a stiffening ring, reducing the bending flexibility of the elbow. This characteristic of flanged elbows is addressed by the piping codes through a modification of the flexibility factor and stress intensification for the elbow.

Node 605-610 Pipe Element This short element runs from the far point of the elbow (node 605) to the beginning of the bypass gate valve (node 610).


137

Tutorial A 1.

Click Continue


The To node displays 610 and the From node displays 605.

2.

In the DY box, type 9 to specify the element length of 9 in. The horizontal elbow and the new horizontal pipe element display. The 9-inch length puts node 610 at the far end of the bend. The locations of nodes 605 and 610 are coincident. To prevent a zero-length element, the software attaches an element length of 0.45 in., equal to 1 percent of the bend radius.

You can change the default 1percent attachment length b y specifying a new value for Bend Length Att achment Percent in Tools > Configure/Setup .

Node 610-615 Gate Valve This element is the 6-inch 150-psi class flanged gate valve.

1.

Click Continue



2.

Click Valve/Flange Database


The Valve and Flange Database dialog box displays.

3. Select GATE from the Rigid Type list and FLG from the End Type list. Click OK .


138

Tutorial A A 150 psi class flanged gate valve displays.

The software adds valve data to the Classic Piping Input dialog box. The element length DY is 1 ft. 5.750 in., Rigid is selected, and the weight of the valve and flanges of 225.000 displays in the Rigids tab. The software includes these values in the analysis.

Select NOFLG in the End Type list of the Valve and Flange Database dialog box if you do not want to include the deadweight and length of the attached flanges in the analysis.

Node 615-620 Pipe Element and Node 620 Elbow This element runs from the end of the gate valve (node 615) to the top horizontal pipe centerline and the vertical pipe centerline (node 620). You need to find the Y distance between nodes 615 and 620, where 620 is at the same height as the top welding tee node 25.

1.

Click Distance


The Distance dialog box displays.

2. Select Between Element Nodes and To and From Nodes, type 615 and 25 as the nodes, and then click Calculate. 1' 3" displays as the needed DY distance.

3.

Click Continue




139

Tutorial A 4.

In the DY box, type 15 to specify the element length of 1 ft. 3 in.

5.

Double-click the Bend check box to display the Bends tab on the right. The software creates a long radius elbow (1-1/2 times the nominal pipe diameter) for the 6-inch line, node 618 at the elbow near point ( An ( An gl e 2 of 2 of 0.000), 0.000), and node 619 at the midpoint ( An An gl e 1 of 1 of M). Node 620 moves to the elbow far point.

6. In the Type list, select 1 -Single Flange.

Similar to the previous elbows, the elbow does not display until you create the next pipe element.

Node 620-2 620-255 Pipe Element Element Clos ing the Byp ass Loop This element runs from the far point of the top bypass elbow to the reducing tee at node 25, returning the 6-inch pipe to the 8-inch pipe and closing the bypass loop.

1.

Click Continue


The To The To node node displays 625 and the From node From node displays 620. 620.

2. Type 25 in the To node box.


140

Tutorial A 3.

Click Close Loop


The software creates the element and displays the length of 2 ft. in ft. in the DX box.

4.

To save the model click Save

or File > Save.

Your model is c omplete!


141

Tutorial A

Rev evie iew w the piping input You can review the values in your model graphically or by viewing a list of pi ping input data.

Review Graphically You built your model using the default piping input layout, with a graphic view displaying to the right of the Classic Piping Input dialog box. The size of the graphic view can be increased by clicking Au to Hid e in the upper right corner of the Classic Piping Input . To display the node numbers, click Node Numbers on the Plot Tools toolbar, or press N.

Standard Operators Operators toolbar, and then use the arrow keys to rotate the Click Orbit on the Standard plot. You can also use the arrow keys or mouse to pan the plot after clicking Pan . Scrolling the mouse zooms the model and pressing the center mouse button pans the plot. Click ing the right mouse button, and then clicking Operator s > Pan from the pop-up menu provides an alternative method of panning the plot. The m odel then follows the cursor. The p lus sign (+) zooms in and the minus sig n (-) zooms out. There are toolbar buttons and menu items to alter the pan view and to d isplay element and restraint information on the plot. Use these different items to become familiar with them. To reset the plot to the default, click Reset on the Reset Prin t or click Print toolbar or click View > Reset . To print a copy of the display click File > Prin on the Standard toolbar.

Because the graphics are included in the input processor, the graphic must be clicked to set the focus before printing.


142

Tutorial A The V key toggles different views. The volume plot shown below is especially useful for larger models because it uses less of the computer's resources.

The illustration below shows a view down the Z-axis with a zoom and pan to show the pipe valves. This volume plot shows the nodes and identifies the tees. To see the displacements specified in the model, click Displacements or Options > Displacements .

Review Review using L ist Click List , or Edit > List to quickly review and edit different categories of data in the job. Clicking on the row number to the left of a line of data highlights the entire row. Hold the Shift key while clicking on a second row of data to highlight all rows in b etween the two. Different types of data sets are available b y choosing the appropriate tab along the bo ttom of the dialog


143

Tutorial A box. Use the scroll bar alon g the bottom of the list to vi ew more element data such as temperatures and pressures. The Element list is shown in the following figure.

Check for erro rrors rs 1.

Click Start Run

Checker . to run the CAESAR II Error Checker

CAESAR II checks the job for errors and lists a variety of notes and warnings. Error checking generates three notes and one warning for the tutorial model. Two notes address the hanger in the model, and the third one is the center of gravity report. The warning addresses insulation thickness. The notes indicate that the software must analyze and size the hanger.

The software builds the intermediate (scratch) files for the static analysis and saves binary data for this model with the file extension ._a. With the scratch files created, the i nput process is complete and control is returned to the CAE SAR II piping input. The analysis may proceed with notes and warnings, but fatal errors must be corrected before continuing.

Check the th e static lo ad ca c ases After error checking the the model, review the load cases. 1.

Click Edit Static Load Cases Static Load Cases .

on the CAESAR II Tool s toolbar, or click Edit > Edit

The Static Analysi s - Load Case Editor Editor dialog dialog box displays. CAESAR II begins with a standard set of load cases based upon the piping code selected and the loads defined in input. For this tutorial, the software includes load cases to size the hanger before performing the standard structural and stress analyses. The hanger sizing algorithm requires two analyses before analyzing the standard load cases:


144

Tutorial A The operating condition for this anal ysis consists of the deadweight of the pipe, its contents and insulation, the design temperature and pressure, and the preload o n the hanger at node 28. The installed condition includes the d eadweight and hanger preload. In addition t o these structural analyses, certain stress conditions must be addressed. For the piping code, the sus tained and expansion stresses must be calculated. Sustained stresses include deadweight, preloads, and pressure. Sustained stresses can be taken from the installed condition analysis if the pressure loads are included. CAESAR II includes the pressure term in the installed case because pressure, in most cases, has no impact on t he structural loads on the piping. With the installed case structural analysis also serving as the sustained case stress analysis, no additional load case must be added to calculate the sustained stresses. Expansion stresses reflect the change in s ystem position from its installed position to its operating position. Because of system non-linearity, this change i n position cannot be determined by analyzing thermal loads alone. By default, CAESAR II constructs a third load case to calculate the expansi on stress (range). This case is not a third, complete analysis of the system. Instead, it is a product of the operating and installed structural analyses already performed. The difference in system displacements between these two cases is the displacements stress range from which the expansion stresses are calculated. The third class of stress in piping – occasional stresses (as opposed to expansion and sustained) – is not included in the recommended analyses a nd must be specified by you. Likewise, Fatigue (FAT) stress cases are provided only when specificall y required by the active piping code (TD/12, for example). For most systems, the recommended load cases are exactly what you want to analyze. L1 (W) - Weight f or hanger l oads

Calculates the deadweight carried by the pr oposed spring at node 28. L2 (W+D1+T1+P1) - Operating for hanger t ravel

Calculates the vertical travel of the proposed spr ing. All load categories, which compose the operating load case, are used for this analysis. These are deadweight, displacements, thermal set 1, and pressure set 1. With these two numbers—the load carried by the hanger and the amount of travel it must accommodate—The software selects the appropriate spring from the Anvil catalog. This spring and its proper preload are installed in the model for the remaining analyses. L3 (W+D1+T1+P1+H) - Operatin g case co ndi tio n 1

Defines the operating hanger load case. L3 is identical to L2 but includes the sized hanger preload (H). This analysis produces the operating forces and moments on the supports, and the deflections of all points in the system.L3 is a structural analysis case and not a B31.3 stress analysis case. The refining piping code does not recognize pipe stress in the operating condition as a test for system failure and does not establish a limit for this state of stress. L4 (W+P1+H) Alternate sustained stress b ased on operating c ondit ion 1

Provides an alternate method for the sustained case of L5, using the restraint status from the L3 operating load case to evaluate the stresses induced by prim ary loads. L5 (W+P1+H) - Sustained case condition 1

Defines a structural and stress case of a cold system. The load case eliminates the (assumed) thermal effects (D1+T1). By including pressure (P1), this case also has the necessary components to be used to report the s ystem’s sustained stresses.


145

Tutorial A L6 (L3-L5) - Expansion case conditi on 1

Defines an algebraic combination of two bas ic load cases. The displacem ents of L5 are subtracted from the displacements of L3 to produce these results. This c ase develops the displacement range of the system in its growth from the installed position to the operating position. This displacement range is used for the calculation of the system’s expansion stresses.

Run the static analysis 1.

To proceed with the analysis, click File > Analyze, or click Run the Analysis

.

The software continues with the data processing by building, sorting, and storing the equation (matrix) data for the system and the basic load cases. When complete, the Static Output Processor displays.

You can stop the analysis at any time by clicking Cancel . CAESAR II analyzes the basic loads (hanger design, operating, and installed). The displacement results of cases 3 and 4 are used with the el ement stiffness matrices to calculate the forces, moments, and stresses throughout the system. The difference between the two sets of displacements is used to establish the displacement range of the piping system as defined in L6. This new displacement set is similarly used to calculate forces, moments, and stresses.





All the results are loaded into the binary data file Tutor._p. The ._p file can only be examined through the Static Output Processor . If you want to review analysis results at a later time, you do not need to rerun the static analysis . Click Output > Reports > Static display the output.

from the ribbon on the main window to

Review the analysis results The Static Output Proc essor dialog box displays directhly after static analysis or by later clicking Output > Reports > Static

from the main CAESAR II window.

Usually, the first look at output is to verify that the piping model is responding as e xpected. Checking deflections and restraint loads in the operating and installed cases should quickly uncover any major problems with the s ystem layout or input. If there are unusual results, re-examine the input for correctness. If the output verifies the model, the r esults can be used to collect pipe stresses, support and equipment loads, and any other useful data found in the output. This information is useful in documenting a good piping design or troubleshooting an inadequate one.


146

Tutorial A 1.

Selecting the (OPE) load case, and then click Graphical Output

.

A view of the operating displacements of t his piping system displays.

As in other CAESAR II dialog boxes, both the toolbar buttons and menu items may be used to select display options. 2.

From the menu, select Show > Displacement > Deflected Shape . The plot shows the centerline plot along with a normalized deflected shape of the system in the operating condition.

3.

When you are finished viewing the plotted output for the operating case, change the case to Sustained in the drop list on the left of the second toolbar.

4.

Click Show > Stress > Overstress and verify that there are no o ver-stressed points in the system.


147

Tutorial A 5.

Reset the plot, and click Show > Stress > Stress > Code to display the code-defined stresses throughout the system. The stress symbols appear on the screen and locate the highest stress points in the system.

6. Click Show > Stress > Maximum to list the stress values on the plot. 7. Press Enter to list the stresses one at a time starting with the highest. The node number is shown in parenthesis following the stress value. The element containing this node is highlighted. The highest (first) sustained stress listed is at node 40 (nozzle to vessel connection) with a value of 1591 psi.

For a quick review of the stresses as well as the displacements and restraint loads, the on the toolbar. This information displays Element Viewer can be displayed by clicking in the next figure. 8.

Return to the output processor menu by clicking Window > .

9.

For a quick look at the selected hanger data, click Hanger Table with Text from the General Computed Results column in the main output processor. Then, click View Reports . The software reports the Anvil Fig. B-268 Size 10 spring selected at node 28.


148

Tutorial A This selection is based on the values found in the first two analyses. Both analyses provide no load case reports in the output processor. 

The expected hot load for the proposed support at node 28 (1209 lb.).



The thermal growth of node 28 (0.750 in.).

10. Return to the Output Menu , and select only the operating l oad case (OPE) Displacements and Restraint Summary by holding down the Ctrl key. The restraint loads at nodes 5 and 40 are compared to the pump and vessel load limits. Note the different output tabs at the bottom of the screen.

11. Return to the Output Menu , and select the sustained case (SUS) to examine the installed condition of the piping system. Turn off 3 and turn on 4. Both the operating and sustained cases can be reviewed together by having both 3 and 4 highlighted at the same time. 12. Return to the Output Menu , and highlight the sustained and expansion cases (4 and 5) and stresses. Each stress report begins with a summary stating that the code stresses are below their allowable stress. In the table that follows the summary, the stresses display for each node in the system. These nodes are listed in pairs with their associated element. The last column lists the ratio of actual stress to allowable stress in terms of percentage.

These results can be sent to the printer or to a file rather then sent to the screen. Before creating the report, a title line for the hardcopy can be generated through Options-Title Lines on the Output Menu . 13. Type the following two lines for the report header:

CAESAR I I TUTORI AL BOTTOMS PUMP TO STEAM STRI PPER 14. To send a specific output to the printer, click File > Print . Use the output wizard to create a b ook of reports in a specific order and then send them to an output device. Click More>> in the Static Output Processor to access the wizard. Start the report with the hanger table by selecting it and clicking [Add] . 15. Select the operating and sustained load cases and displacements and restraint summary reports. 16. Click Ad d again. 17. Add the sustained and expansion stress reports by having only load cases 4, 5, and Stresses highlighted. 18. Click Ad d to service this request. This completes a typical output report after reviewing the reports order.

19. Select the output device, and then click Generate TOC, if needed. 20. Click Finish . Segments of the output reports are included at the end of this section.

An input echo is available through the output processor. A complete input listing can start the printed report or output file created by this processor.


149

Tutorial A To archive the static analysis el ectronically, the report can be sent to a data file rather than to the printer. Use the above instructions substituting Save for Print , or use the appropriate output choice on the wizard screen. The first time you click Save, the software prompts you for a filename. T he resulting data file, Tutor.out, can be copied with the CA ESAR II input and output files Tutor._a and Tutor._p to a CD. These files, along with the configuration file (Caesar.cfg), and the time sequencing file (Tutor.otl) provide a complete record of the analysis an d should be stored with the drawing and any listings.

Static Analysi s Output Li sting The following figure is an example CAESAR II tutorial output report:



The output listed in the example includes significant output only.



Notes, which discuss the results, are included with each report.



The following reports are included in this output: 

Complete Hanger Report



Operating Case Displacement Report



Installed (Sustained) Case Displacement Report



Operating & Installed Restraint Summary



Sustained Stress Summary and Stress Report



Expansion Stress Summary and Stress Report. (Stresses in the operating condition are not used in B31.3 analyses)

The hot load of 1209 lbf. was calculated in the initial weight run (load case 1) with a rigid Y restraint installed at node 28. The load on the restraint was 1209 lbf. A 1209 lbf. +Y load replaced the rigid Y restraint at 28 and then an operating case was analyzed (load case 2). Node 28 moved 0.750 in. in the +Y direction in this analysis. CAESAR II entered the Anvil hanger table with these two values and selected an appropriate mid-range spring. The size 10 spring has the hot load of 1209 lbf. in its working range. This mid-range spring (short range springs were excluded) has a spring rate of 260 lbf./in. Assuming


150

Tutorial A that node 28 moves 0.750 inches between the cold to hot position, this increases the spring load by (.750)(260) or 195 lbf. The cold load on the size 10 spring is 1222+195 or 1404 lbf. This cold load is also within the working range of the size 10 spring so CAESAR II selects it.



The deflections of nodes 5 and 40 were entered as input.



Node 28 again moves up 0.750 in. in the Y direction with the spring installed.


151

Tutorial A Look at the zero position of nodes 5 and 40. When the imposed displacements are not included in the analysis, the node is fixed with zero movement in each of the defined directions.

This restraint report lists the piping forces and moments on the restraint. It does not list the restraint loads on the piping. The loads at node 5 are the nozzle loads and can be used without sign change to check the API 610 allowable loads. Loads for node 40 can be used to check the vessel stresses due to the nozzle loads. The loads at 28 show the operating load and the actual installation load (with contents) for the selected spring. The spring carries the designed load of 1209 pounds in the operation condition.


152

Tutorial A The +Y restraint at node 35 shows it is nonlinear nature. In the cold condition, the restraint is active. As the piping moves to the hot position, it disengages from the support. Refer back to the displacement reports to confirm that the Y displacement is 0 .0 in the installed (sustained) condition and +Y in the operating condition.


153

Tutorial A The summary shows that the sustained stresses throughout the system are below their allowable values. The sustained stress closest to its allowable limit is at the vessel node 40.


154

Tutorial A For the previous stress detail report, note the appl ication of the tee and bend stress intensification factors. The tee at 25 has SIFs other than 1.00 for all three listings: 25 to 28, 20 to 25, and 25 to 620. Bend SIFs are applied only on the bend side of the node compare node 28 on 25-28 and 28-29. No stresses are listed for rigid elements as no valid moment of inertia is provided for these elements.


155

Tutorial A The summary shows that the expansion stresses throughout the system are below their allowable values. The expansion stress closest to its allowable limit occurs along the header at the node 10 tee.

For the previous stress detail report, compare the bend side of 30 with the straigh t side of 30. The SIF doubles the calculated stress. Also note the changing allowable stress. This is the result of applying an allo wable stress, which takes credit for "unused" stress in the sustained case.


156

Tutorial A

Conclusions The review of piping stresses shows that the piping has adequate wall thickness and support to keep within the sustained allowable stress, as well as enough flexibility to remain below the expansion allowable stress limit. A quick review of the s ystem displacements does not reveal any interference problems from pipe expansion. Equipment loads must still be checked to ensure a safe and effective design. T he pump loads at node 5 may be compared to the API (American Petroleum Institute) Standard 610 (Seventh Edition, February 1989), Centrifugal Pumps for General Refinery Service. T he nozzle loads, too, can be compared to the allowed maximum limits. The nozzle loads can be translated into local stresses using Welding Research Council Bulletins 107 or 297 - Local Stresses in Cylindrical Shells Due to External Loadings on Nozzles (WRC 107) or it's Supplement (WRC 297). These local stresses can then be compared to allowable stress values established in "ASME Section VIII Division 2 Appendix 4, Mandatory Design Based on Stress Analysis". Because the loads on these boundary co nditions are related to the piping system layout, the piping system cannot be properly approved until these load limits are also verified. These verifications are performed in Tutorial B (on page 158).


157

SECTION 8

Tutorial B This section continues the tutorial by guiding you through equipment checks and the redesign of the system analyzed previously.

In This Section Evaluating Pump Discharge Loads ............................................... 158 Creating Accurate Models ............................................................. 166 WRC 297 Calculations Completed at the End of Error Checking . 169 Checking Nozzle Loads................................................................. 175 System Redesign ........................................................................... 177 Conclusion ..................................................................................... 185

Evaluating Pump Disch arge Loads Collecting pump and load information is the first step in reviewing the pump loads. API 610 (10th Edition) examines pump loads at two levels: 

Individual nozzle loads



Combined nozzle loads on the pump housing

The suction and discharge nozzles have a set of allowable load limits based on nozzle orientation and nozzle size. Both the individual X, Y, and Z components and the resultant forces and moments are checked. Additionally, to assure maintenance of proper pump/motor alignment, all loads on the pump are resolved about a base point and compared to their allowable values. The CAESAR II API 610 processor requires the suction and discharge size, position, and orientation and the loads on these no zzles. The processor provides the load l imits. For this evaluation only, the discharge nozzle loads have been calculated. Therefore, only the discharge nozzle is checked and neither the suction limits nor the resolution to the base point are evaluated.


158

Tutorial B Even though all the loads are not known, the entire description of the pump is collected for the API 610 processor in CAESAR II. The dimensioned isometric shown in the next figure defines the orientation of this pump with its end suction nozzle and top discharge nozzle. Both nozzles are dimensioned back to the base point which is the i ntersection of the shaft axis and the support line for the pump. This pump’s drive shaft is along the X-axis.


159

Tutorial B The discharge nozzle loads are found in the static analysis output from Tutorial A. Because the discharge nozzle served as a boundary condition for this analysis, the nozzle loads are conveniently listed in the restraint reports. These forces and moments on the restraint at node 5 are the piping loads acting on the discharge nozzle. No sign change is required. The operating loads and installation loads must both fall below the d efined limits. Examination of the restraint summary for the operating and sustained (installed) cases reveals that the operating loads are the controlling case. The terminal output showing these numbers is found in the following figure. The operating case loads are used for the discharge nozzle analysis.

1.

Before starting Tutorial B, create a new folder called Tutorial B.

2.

Open the Tutorial A file and save it in the Tutorial B folder.

3.

Rename the file to Tutorial B.

Evaluating Pump Discharge Loads 1. Select Analy si s-API 610 in the CAESAR II main menu to enter the API 610 processor. The software displays the Open dialog box.

2. Type in Tutorpump , and then click Open . 3.

Click Yes to create the new file named Tutorpump .


160

Tutorial B The API The API 610 dialog 610 dialog box displays.

4.

Type in the Equipment Description related to the analysis as shown in the example figure above.

5.

Click the Input Data tab, and enter the information as described below. Arbitrary node numbers are assigned for the pump base point and for the pump suction nozzle (1 & 105 respectively).

6. Use the data shown in the following figure to add the remaining values. It It is best to enter as much data as is currently availabl e so that when the rem aining (suction) data is determined, recollection of data is minimal. The factors for the Table 4 load multipliers are left blank. CAESAR II uses the default values established in API 610. If the pump manufacturer defines pump load limits that are different from those defined in API 610, enter the modified limit here (This value must be between 1.0 and 2.0).


161

Tutorial B 7.

Define the pump shaft centerline direction, the nozzle types, types, node numbers, and nominal diameters on the Input Data tab.

8.

Nozzle tab, and enter the distance f rom the base point to the suction Click the Suction Nozzle nozzle (not from the nozzle to the b ase point) and the nozzle loads.

Because the nozzle loads are unknown at this time, no forces and moments are entered.

The next figure shows the Discharge Nozzle tab with the Nozzle location from the center of the pump. The nozzle location is t aken from the piping isometric described at the beginning of the chapter. The Nozzle locations are in global coordinates.

9.

Select Loads Jo b and Lo ad Case Case . Click Select


162

Tutorial B 10. Navigate to and choose the name of the output file that contains the restraint loads for this pump. In this case, select TUTOR-B.C2 from the list, and then click Open .

The next dialog box allows you to choose the appropriate load case for inclusion in the API-610 analysis.

11. Select the operating case, and and click OK .


163

Tutorial B The loads from the restraint report at node 5 are read automatically. This is the end of the input for the API Standard 610 pump load evaluation.

An aly ze 12. Click Analy

to perform the API-610 equipment check.

An aly ze from the menu or from the EQP toolbar. You can select Analy The results are available on the Equipment Report tab. Report tab.

With no suction nozzle data entered, the suction nozzle cannot be evaluated. This report has some value in that the individual load component limits for the suction nozzle are listed. The discharge nozzle report is complete in its comparison of the operating loads on the nozzle and the defined limits. If the nozzle load components are less than the Table 4 limits, no additional checks must be made. If the nozzle load components are greater than the Table 4 values but


164

Tutorial B less than two times the Table 4 values, the pump may still pass if other checks are within their allowable values. The CAESAR II report first compares these loads to the Table 4 l imits. If the ratios in the report (see the following figure) are all less than 1.0, the pump is adequate. If all the ratios are less than 2.0, the pump must pass additional ch ecks. Two times the Table 4 limits were not used in this analysis because the loading on the suction side of the pump is unknown. The moments about the X-, Y-, and Z-axes are greater than the API 610 standards and additional checks are not valid. T he moment about the X-axis is 10,174 ft-lbf and the (conditional) limit is 2600 ft-lbf. The m oment about the Y-axis is 5905 ft-lbf and t he limit is 1300 ft-lbf. The discharge nozzle loads must be reduced. The pump loading shown below is in the pump local coordinate system.

If the analysis was using two times the Table 4 values, the checks shown in the next part of the report would be used to qualif y the pump loads. Here, the resultant of the applied nozzle forces and moments on each nozzle are compared to their related Table 4 lim its (Condition F.1.2.2). Both the suction and discharge loads are also resolved to the pump base point and again compared with a Table 4 l imit (Condition F.1.2.3). For this analysis, these data have no significance as the components of the discharge l oads are greater than the Table 4 values. After the output has been reviewed, you can review the reports reports again or send the report to a file or to the system printer ( File > Print ). For this tutorial, the limits on the discharge nozzle are noted for quick checks on future redesign anal ysis. After this piping system is redesigned s o that the discharge nozzle is not overloaded, the existing data in the equipment file TUTOR can be updated for the final pump verification report.


165

Tutorial B

Creating Accurate Models The operating moments (X, Y, and Z) on the pump nozzle are too large. The system appears to be modeled correctly, so it must be m odified to reduce these loads. You first need to determine the causes of these large loads to make the most effective change to the s ystem. Return to the static output for the operating load case. There are two major clues as to the cause of these excessive loads: 1.

Compare the operating loads on the pump to the installed loads on the pump. If they are vastly different, the thermal effects are the cause of the overload. If they are similar, the sustained effects cause the high loads. In this case, only the operating loads are high. Therefore, this system has a thermal expansion problem. For a given amount of thermal growth, the thermal forces and moments are reduced by adding flexibility to the system. F = KX, so for a given X the thermal growth between the end points F or M can be reduced by reducing K. If the system was overloading the pump due to sustained effects, the system pressure or deadweight is causing the problem. Systems with pressure problems usually include untied expansion joints. Deadweight problems can be traced back to im proper system support. Typically, this is either spring pre-loads or support locations.

2.

Go back to the displaced shapes plot of the operating load case to examine the source of the high moments. Most engineers and analysts find it easier to understand system response to loads in terms of system displacements rather than internal forces and moments. The displacement plot is useful in identifying which runs of pipe are generating the thermal strain and which runs of pipe are turning that thermal strain into the large forces and moments on the pump.

The next figure makes it clear that the large moment about the Z-axis at the p ump is caused by the thermal growth of B working against the stiffness of legs A and C. The large moment about the X-axis is due to the thermal growth of A working against the stiffness of legs B and C. The thermal growth of the vessel connection may also contribute to these high loads.

How can these excessive loads be reduced? How can additional flexibility be added to the system so that these loads drop? Two possible solutions are: 

Addition of an expansion loop to the piping.


166

Tutorial B 

Installation of an expansion joint.

Before either of these choices is made, a much simpler and cheaper solution is examined—improving the model to incorporate the inherent flexibilities found in the vessel/nozzle intersection. The pump loads due to expansion would drop if the thermal growth of the three legs A, B , and C could deflect the vessel nozzle. Such nozzle flexibilities are defined in Welding Research Council (WRC) Bulletin 297, "Local Stresses in Cylindrical Shells Due to External Loadings on Nozzles" (Supplement to WRC Bulletin No. 107). WRC Bulletin 297 supplies curves by which the outer diameters and thicknesses of the vessel and nozzle are used to define local nozzle flexibilities. These curves are limited to certain ratios of nozzle and vessel terms, such as the following: d/D < 0.5 d/t > 20 20 < D/T < 2500 d/T > 5 Where: d = nozzle OD (= 8.625 in.) t = nozzle thickness (= 0.322 in.) D = vessel OD (= 60 in.) T = vessel thickness (= 7/16 in.) In this system where the vessel is vertical and the nozzle is in the Z direction, flexibilities are defined at node 40 for translation in the Z direction and rotation about the X- and Y-axes. The other three degrees-of-freedom (the three local shear terms) remain rigid as in the original model where this nozzle was modeled as a rigid connection with its thermal deflections. The vessel wall thickness is 3/16 in. but the nozzle has a 1/4 in. pad reinforcing the connection; this produces an effective vessel wall of 7/16 in.


167

Tutorial B

Before any costly system modification is made, the m odel is refined to incorporate these WRC 297 nozzle flexibilities. It is possible that a more thorough and accurate m odel of the system shows that redesign is not needed. To assist in this model update, CAESAR II provides a processor which calculates and inserts these flexibil ities into the system. This change constitutes the second analysis of this la yout. Save the TUTOR-B.c2 file as T UTOR-B2.c2. Open TUTOR-B2.c2. Return to the input processor for the job TUTOR-B2. Go to the nozzle node (40) and double-click the Nozzle Flex check box. Type the correct data in the Au xi li ary Dat a. The nozzle pipe size is imported from the spreadsheet. If this nozzle connection had no associated thermal growth, the vessel node number need not be entered. Because this vessel has thermal growth, the vessel node number must be identified and the thermal displacements previously assigned to node 40 m ust be reassigned to this new n ode number. 1. Type the vessel node number as node 6000. The calculated nozzle flexibilities are applied between nodes 40 and 6000. The vessel dimensions are entered here in terms of outer diameter, wall thickness, and reinforcing pad thickness. WRC 297 flexibilities are also sensitive to the proximity of stiffeners to the nozzle. Here, a tray in the vessel is closest to the nozzle and is 4 feet above the nozzle. On the other side of the nozzle, the bottom head tangent and skirt connection is 6 feet below.

The vessel orientation, based on a direction vector, is entered next. 2. Type 1 in the Y direction to indicate a vertical vessel.


168

Tutorial B This Z nozzle and Y vessel define the orientation of the local stiffnesses assigned through WRC 297. This completes the definition of the nozzle. There is no piping element defined between nodes 40 and 6000.

The displacements provided at node 40 must be moved to node 6000. 3.

Click Displacements and change node 40 to 6000.

WRC 297 Calculation s Completed at the End of Error Checking With the nozzle specification and the no de number change for the vessel deflections, the jo b is ready for analysis. 1.

Click File > Error Check to start the error checker. The error checker again produces the two notes regarding the hanger sizing. Additionally there is a warning generated regarding the specification of a vessel node number in the WRC 297 input when this node number is not included on any piping element. This warning


169

Tutorial B message (75E) is shown in the following figure. There is no trouble with this job because the displacements of the vessel node (node 6000) are defined.

The figure below shows the nozzle flexibility calculations performed by the CAESAR II error processor.

The previous report lists the flexibilities extracted from WRC 297: 

Axial stiffness of 319, 226 lb./in.



Longitudinal bending stiffness of 291,036 in.lb./deg.



Circumferential bending stiffness of 58,665 in.lb./deg.

These three numbers are much less than the magnitude of the default rigid stiffness, which is 10E12. The local coordinate system is defined by the nozzle/vessel orientation. With the nozzle in the Z-direction and the vessel in the Y-direction, this new axial stiffness is in the global Z direction (the nozzle centerline), longitudinal bending is about the global X-axis (bending into the vessel centerline or long axis), and circumferential bending is about the global Y-axis (about the vessel centerline). After the display of the WRC 297 calculations, CAESAR II shows the error processor is completed by summarizing the type and number of messages. 2. With no fatal errors encountered, close the piping input processor and return to the CAESAR II main menu.


170

Tutorial B The model is now ready for a second static analysis.

3.

Click Analy si s > St ati cs to proceed. There are five analyses - two for the hanger sizing followed by the operating case, the installed or sustained case, and the expansion case.

4.

Click Analy ze

.

After the analysis is completed, t he Output Processor dialog box displays for output review.

5.

With only a minor change to the input, a rigorous, error-checking review of the results should not be necessary. Instead: a. Check the sustained and expansion stresses to confirm that they are still below their allowable limits. b. Check the hanger selection. c.

When the operating and sustained restraint summary is displayed, check the loads on the pump nozzle and node 5.


171

Tutorial B

The highest sustained and expansion stresses are 1282 psi and 14,103 psi, respe ctively. This is well below the allowable limits. The software selected a lighter spring for installation at node 28. Previously, a size 10 spring was selected, now a size 9 is recommended. In the first analysis, the spring carried 1209 lb. in the hot position, now it carries only 904 lb. The system should still weigh the same, so why is the spring load smaller? The reduced longitudinal bending stiffness at the nozzle may explain this change.


172

Tutorial B Something can be said about each of these restraints: 













The pump discharge nozzle loads at node 5 reveal the impact of the change in flexibility at node 40. The operating moment about the Z-axis shows the greatest change dropping to 747 ft.lb. from 5905 ft.lb. The shear force in the X-direction has also been reduced by 50%. The axial force in the Y direction, however, has risen from 1556 lb. to 1809 lb. This higher pump load is tied directly to the hanger selection, which was also affected by the WRC 297 nozzle flexibilities. The spring support at node 28 is shown next. While the previous analysis had the spring carrying 1209 lb. in the operating position, it now carries only 904 lb. This 300 lb. reduction in the spring load returns as an additional 300 lb. load on the pump nozzle. With the spring installed directly above the pump nozzle, simply increasing the load carried by the spring reduces the load on the nozzle. If another anal ysis is required, the hanger sizing procedure is adjusted so that the hanger carries more load so that the pump load drops. The +Y support at node 35 reveals why the hanger load has changed so much. In the first analysis, the support at node 35 was not active in the operating case; the pipe rested on the support in its installed position but lifted off the support as it went into operation. The hanger sizing algorithm readjusted the spring load so that it would carry its portion of the system no longer resting at 35. In this second anal ysis, the restraint at 35 remains active in the operating p osition, therefore the hanger at 28 does not c arry any additional load from 35. The added longitudinal bending flexibility at node 40 is what allows the pipe to rest at node 35. The support definition at node 40 shows the changes inherent in the WRC 297 nozzle flexibility calculations. Flexibilities are added in the axial and bending directions (Z, RX, and RY) while the shear terms remain rigid (X, Y and RZ). This added flexibility greatly reduces the bending moments about the X and Y-axes at node 40. Again, these reduced loads are not a result of design modifications but modeling refinements. If the vessel nozzle connection meets the requirements of Welding Research Council Bulletin 297, there is much to gain in nozzle flexibility.


173

Tutorial B 6.

To further investigate the effect of the nozzle flexibilities, show the displaced position of the piping system in its operating condition.

The final report from this analysis shows the displacements of node 40. The imposed thermal growth of the nozzle was removed from node 40 and redefined at node 6000. This output shows the operating position of node 6000 as (0, 0.28, -0.10; 0, 0, 0) [defined as (X, Y, Z; RX, RY, RZ)]. Comparing these numbers with node 40 above, you can see the impact of the nozzle flexibilities. The biggest difference is due to the circumferential bending flexibility (RY) but the longitudinal bending flexibility (RX) plays a large role in the weight distribution of the system.

Do the new pump loads meet the allowable limits defined in API 610?


174

Tutorial B

Checking Nozzle Loads The operating moments (X, Y, and Z) on the pump nozzle were too large in the initial model. A quick run through the API 610 processor q uickly evaluates the refined model. In the TUTOR input, only the discharge loads need be changed. 1.

Click the Discharge Nozzle tab, and then click Refresh Loads from Current Jo b to obtain the new loads.


175

Tutorial B 2.

Click Analy ze

to run the Nozzle analysis.

The API 610 report is shown below.

The situation is better but not good enough. The Y-moment on the discharge nozzle is well below the limit. The X -moment, however, remains more than twice the allowable l oad. The Z-moment also exceeds the allowable. Exceeding twice the allowable load would be fine if Condition F.1.2.b is satisfied, but it is not. Condition F.1.2.b states that even though the individual load components can be more than twice their individual limit, the loads are satisfactory if the resolved forces divided by their resolved limits plus the resolved moments divided by their resolved limits is less than two. The sum of the ratios for the discharge nozzle is 2.822, so the pump loads are still too high. There is a quick "what if" check that may prove the pump loads can be brought within their allowable values. The discussion of the restraint loads mentioned that the set load directly controls the vertical load on the discharge nozzle on the spring at node 28. This spring pre-load could be ideally set so that when the pump is in operation, there is no pump load in the


176

Tutorial B Y-direction. At this point the hanger carries 904 lb. in the operating position while the pump carries 1809 lb. If the spring load carried 2713 lb., the load on the pump would be zero in Y. Would that satisfy Condition F.1.2.b? Rerunning the API 610 processor with the Y -load set to zero shows the Condition F.1.2.b reduced to 2.33, which still remains above the limit. Spring load adjustment is useful but s ystem redesign is indicated.

System Redesign The probable causes of the large X-moment at node 5 were discussed in earlier topics. This excessive load is caused by the thermal expansion of the leg from node 35 to 40 (the A leg) working against the stiffness of the remainder of the system (legs B and C). If the thermal strain of leg A is fixed, only the system stiffness can be changed to reduce the operating load at node 5. You can reduce this stiffness by adding an expansion loop or an expansion joint. For this tutorial, an expansion loop is chosen.

Where should the expansion loop be added? As a rule of thumb, the best location for an expansion loop is determined by the orientation of the leg which produces the thermal strain causing the problem. Here, leg A sets the orientation of the loop. The added piping to generate the expansion loop lays perpendicular to leg A. For this system, pipe can be added in either the X- or Y-direction. This added pipe effectively increases the cantilever length which is displaced by leg A . By increasing the cantilever length, the stiffness is reduced and loads drop. Therefore, add a loop on the A run of pipe (nodes 35 - 40) by adding pipe in the X-direction. How long should the loop legs be? There are several conditions which set the loop si ze: 

Available support location



Maximum distance between supports



Cost of pipe



Available space

For this tutorial, an eight-foot by eight-foot loop is used. For systems that are not analyzed, the recommended maximum spacing between supports for 8-inch water-filled pipe is 19 feet (see


177

Tutorial B ASME B31.1 121.5 or MSS SP-69). The 8-foot loop run lengthens the 30 - 35 pipe from 12 feet to 20 feet, which is close to this recommended spacing.

1.

Return to the CAESAR II main menu, and start the input processor with TUTOR-B as the current job. When testing layout modifications which may not prove successful, create a new input set with the proposed changes and leave the original model intact. If the proposed changes do not produce the required results, the original model is still available for the next attempt. That way, you do not have to delete the proposed changes from the model.

2.

Click File > Save As , and give the model a new name of TUTOR-B2. The current model is copied to TUTOR-B2. Changes can be made to this new model.

3. Press PgDn to move through the elements to d isplay the element from 30 to 35. 4.

Change the length of 30 - 35 from 12 feet to 20 feet.


178

Tutorial B 5.

Change the DX value to -20-.

6. Move the +Y support from node 35 to node 33. The recommended maximum spacing between supports for this size pi pe is 19 feet. Leaving the support at node 35 would place the support 21-feet from the hanger at node 28 so the support is moved closer to node 33. 7. Select the Restraints box. The Au xi li ary Data tab displays the current +Y restraint at node 35.

8. Select the Node box and change 35 to 33. 9. Break the element from nodes 30 - 35 by adding node 32 at the midpoint. Node 32 is added as an output point to check mid-span sag. While you are still on element 30-35, click Model > Break to add node 32 to this line 10-feet from node 30 wit h no restraints at node 32.

The maximum distance between supports as specified in ASME B31.1 and MSS SP-69 ensures a very low sustained stress in the line. Because CAESAR II calculates these sustained stresses, the output would confirm that much greater distances between supports are safe. The recommended spacing also limits the pipe sag between supports to 0.1 inch. The recommended spacing is conservative but it serves as a useful guideline. 10. Break 35 - 40 eight-feet down the line by adding node 135. Press PgDn to display the element 35 - 40. Break this element and add the new node 135, 8 feet (8-) from node 35. 11. Double-click the Bend check box to add the bend specification at node 135. 12. Insert an 8-foot element after 35 - 135. While still on element 35 - 135, click Edit > Insert to start the Insert command. Click Af ter to place this new element after the element 35 - 135. CAESAR II displays a new dialog box for the new element.


179

Tutorial B 13. Type the To node as 235. 14. Type the length in the DX box as 8-feet ( 8-). 15. Double-click the Bend check box to add the bend at node 235. 16. Press PgDn to the next element (135 - 40) and change the From node (135) to the new node 235. This change finishes the new element.

17. Add a support to the new run 135 - 235. Both ASME B31.1 and MSS SP-69 provide limits to spacing between supports. These guidelines were used to set the size of this expansion l oop (maximum support spacing for 8-inch carbon steel water line is 19-feet). These guidelines also state that the maxim um run of pipe where bends are in cluded is 3/4 of the straight run limit. Here, that limit is about 15-feet. There are over 26 feet of pipe between 35 and 40 so a new support should be added. The support is added about halfway between 35 and 40 - 13-feet from the nozzle at 40 or 3-feet back from 235. 18. With display element 135 - 235 highlighted, click Model > Break . 19. Define a single node 140, 5-feet (5-) from node 135. 20. Type 33 in the Get Support From Node box. CAESAR II duplicates the +Y support entered at node 33 at this new node 140.

A large vertical load remains on the pump nozzle after the hanger at node 28 is sized and installed by CAESAR II. The spring selected from the Anvil Hanger Table should carry more of the deadweight of the pipe and valving. The sizing algorithm can be adjusted so that the pump nozzle carries no load when the software calculates the load to b e carried by the spring. This change greatly reduces the final nozzle load by sizing a larger spring at 28. 21. Go to element 25 - 30 and click the Hanger Auxi liary Data tab. 22. Type 5 in the Free Restraint at Node box. 23. Select Y from the Free Code list.


180

Tutorial B CAESAR II disconnects the Y restraint at node 5 while it calculates the deadweight load carried by the proposed spring at 28.

24. Click File > Error Check , or click checker.

on the CAESAR II Tool s toolbar, to start the error

This data should now process without error. If any errors do occur, carefully read the error messages and return to the input processor to correct them.

25. If everything looks correct, close the piping input processor and return to the main menu. The job is again ready for static analysis.

26. Click Analy si s > St ati cs from the main menu, and run Tutor2 with the same load cases that were created for Tutor. Do this by accepting the default setting on the Static Analysis dialog box. The Output Processor dialog box displays after the analysis is complete.


181

Tutorial B As previously recommended, the sustained and expansion stresses are first checked to confirm that they remain below their allowable limits. The hanger selection and the op erating and sustained (installed) restraint summary display to examine the impact of this model modification on the pump nozzle loads at node 5. The highest sustained and expansion stresses are 2017 psi and 5419 psi, respectively; well below the allowable limits. The sustained stresses increased a small amount because of the longer spans between supports while the expansion stresses show a significant reduction. The added system flexibility caused this reduction in expansion stress. That is a good indication that the nozzle loads have dropped as well.


182

Tutorial B 27. Click Hanger Table with Text from under the General Comput ed Results column. The software selected a heavier spring for installation at node 28. In the last analysis, a size 9 spring was selected. Now, a size 12 spring is recommended. The spring now carries 2202 pounds in its hot position. This greater load is the result of the modification to the spring hanger selection criteria where the pump is disconnected when the spring’s hot load is calculated. The added load-carrying capability of the spring reduces the vertical load on the pump nozzle. The spring loads can be further manipulated if the nozzle load needs additional adjustment.

28. Click Operating and Sustained Lo ad Cases and Restraint Summary to display the restraint summary report.


183

Tutorial B 29. Display the displaced shape of the piping system in its operating condition to check the effect of the loop on the overall displacement.

The pump discharge nozzle loads at node 5 look much better; revealing the impact of the change in flexibility at node 40. The loop adds flexibility in the Z-direction. The Z-force on the pump fell from 747 pounds to 235 pounds. The large operating moment about the X-axis and the target of this redesign dropped from almost 10,000 pounds to 2,755 pounds. Another interesting effect of this added flexibility is the increase in the Z-moment from -300 foot-pounds to +1519 foot-pounds. The pump load in the Y-direction exhibits the adjustment to the hanger selection. The hot load on the pump is -206 pounds and the cold load on the pump is +337 pounds. The absolute magnitude of the pump load could not be much smaller. If necessary, the hanger load could be adjusted to bring the pump installation load to zero or the pump operating load to zero. The spring support at node 28 now shows a hot and cold load of 2,202 pounds and 2,540 pounds, respectively. By releasing the anchor in the initial weight analysis, the spring carries the riser load. This load was only 904 pounds in the previous analysis. The extra flexibility has also changed the support load at node 33. Previously, the support load dropped as the pipe became hot. Now, the load increases as the pipe heats up. The vessel nozzle loads at node 40 shows a similar pattern of change as the pump nozzle. Most loads drop, but there is one moment (in this case it is X) that increases. Are the nozzle loads adequate?


184

Tutorial B You do not need to use the API 610 processor to confi rm that the discharge nozzle loads are below their maximum allowed values. Refer back to either of the previous anal yses to quickly locate the individual limits and compare them to the new operating loads on node 5: Direction

API Limi t

Model Results

X (lb.)

1700

136

Y (lb.)

2200

-206

Z (lb.)

1400

-235

RX (ft.lb.)

5200

-2755

RY (ft.lb.)

3800

-1540

RZ (ft.lb.)

2600

1520

For this exercise, the loading on the suction side of the pump is assumed to meet allowable loading; therefore two times the API allowable loading was used. Because all six components of the discharge nozzle loads are below their limits, no additional checks (conditions F.1.2.b. and F.1.2.c.) need to be made. The discharge no zzle is no longer overloaded. The final pump evaluation cannot be made until the suction nozzle loads are compared with their API 610 l imits.

Conclusion The pump discharge loads are now within their allowable limits. The vessel loads from the nozzle at node 40 should also be checked to ensure they are not too high. These loads cannot be compared to a fixed load limit as with the pump. Instead, these loads must be converted to local stresses on the vessel and these stresses compared with their limits as defined by ASME Section VIII, Division 2. As a very rough guide for evaluating loca l vessel stresses, you can check the code defined stress on the pipe connected to the vessel. If those stresses are below about 6000 psi, the vessel stresses should be adequate. Looking at the operating, sustained, and expansion stresses at node 40, the maximum stress is less than 2500 psi. The vessel loads seem fine. If the s tresses are to be checked, the W elding Research Council Bulletin 107 (WRC 107) can be used to convert the applied forces and moments to the appropriate local stresses. CAESAR II provides a processor to convert these loads into WRC 107 stresses and a second processor to combine the different stress categories (general or local primary membrane stress intensity, primary membrane plus primary bending stress intensity, and primary plus secondary stress intensity) for comparison with their design limits. Final reports should now be made to document this design change. As shown earlier in this tutorial, the input listing could be generated from the Input Processor or from the Output Processor . Include the current status of the software’s default settings i n this input echo. A h ard copy of a few input plots would also help in defining this model and anal ysis. Structural and stress results from the Output Processor substantiate the current design. Structural output includes the system displacements and restraint loads for both the operating and installed cases. The code-defined pip e stresses are generated for the sustained and


185

Tutorial B expansion cases. The hanger report should also be generated from the Output menu. The data files from this analysis can also be archived with the h ard-copy reports. Copy the files TUTOR-B2.C2 and Caesar.cfg to a CD to archive a copy of the CAESAR II input, load case definition, CAESAR II output, and software default settings. Often upon release of a new version of CAESAR II, archived files must be converted to the new version and subsequently re-analyzed. This is primarily due to frequent format changes within CAESAR II as new features are added. To avoid this, keep the old version of the software available, and use newest version for new jobs.


186

SECTION 9

Examples This section provides examples for a variety of design challenges faced by piping engineers.

In This Section Harmonic Analysis (TABLE) .......................................................... 187 Relief Valve Loads (RELIEF) ......................................................... 192 Dynamic Analysis of Water Hammer Loads (HAMMER) .............. 201 Dynamic Analysis of Independent Support Earthquake Excitation (CRYNOS) ..................................................................................... 213 Structural Analysis (FRAME) ......................................................... 225 Dynamic Analysis (NUREG9) ........................................................ 232 Omega Loop Modelling (OMEGA)................................................. 240 Jacketed Piping (JACKET) ............................................................ 245 WRC 107 ....................................................................................... 255 NEMA SM23 .................................................................................. 264

Harmonic Analysis (TABLE) 

The problem in this example is taken from the following source: I. S. Tuba and W. B. W right, Pressure Vessel and Piping 1972 Computer Programs Verification An Aid To Developers and Users, The American Society of Mechanical Engineers, New York, 1972. Problems 6 and 2.



Only the input that is germane to the dynamic analysis is discussed.


187

Examples This example first analyzes the following model for natural frequencies and then for harmonic loads imposed on the top o f the structure at nodes 8 and 13.

Enter the model as shown and set the material density on the p ipe spreadsheet to be zero. Enter all weights as concentrated masses. Do not enter bends; enter only straight elements. Member Properties Pipe Outside Diameter

2.375 in.

Pipe Wall Thickness

0.154 in.

Elastic Modulus

27.9E+06 psi

Poisson's Ratio

0.3

Run the static case, and then click Dynamic Analysis software opens the Dynamic Analysis dialog box.


on the CAESAR II toolbar. The

188

Examples On the Lump ed Masses tab, you can add additional masses or delete degrees-of-freedom. In the Eigensolution of larger systems, the deletion of un-needed degrees-of-freedom can be a very important factor in keeping run times reasonable. Usuall y, masses must neither be added nor deleted. The mass of the piping, fluid, and insulation is automatically calculated and included by CAESAR II. In the current example, the weight of the pipe is zero, and all masses are concentrated and predefined as lumped masses.

Next, use the Control Parameters tab to modify the control parameters as shown belo w:

Setting Frequency Cutoff (Hz) to zero turns it off, and setting Max. No. Eigenv alues Calculated (0 - Not Used) to 5 guarantees that the first five natural frequencies are included in the results.


189

Examples Click Run the Analysis . When the Eigensolution is completed, the calculated natural frequencies are printed as shown in the figure below.

Close the Dynamic Output Processor. Click Output > Animation > Mode Shapes on the CAESAR II main menu to view the animations of the five modes of vibration. T he first mode is back and f orth along the x-axis, the second mode is transverse along the z-axis and the third mode is a twisting about the y-axis. The next two modes are combinations of the previous three.

Harmoni c Analysis of t his System In this example, a 120 H z electric motor sits on the pipi ng structure and acts:

FX @ 8 = ( - 95 cos ω t ) l b. FX @ 13 = ( 95 cos ω t ) l b. The harmonic analysis can determine the largest stress in the small piping structure subject to these dynamic loads. The 120 Hz vibration falls between the structural resonant frequencies 115 Hz and 137 Hz. The torsional mode is most likely excited because the sign difference on the forces promotes a twisting of the structure. The model has already been built, so dynamic input only requires a slight modification. There is o nly a single harmonic frequency of excitation to investigate.


190

Examples You enter harmonic loads next. The software prompts you for harmonic forces first and then prompts you for direction. In the example piping system, harmonic forces act at nodes 8 and 13. The forces act in the X-direction with an opposite sign and with a magnitude of 95 pounds. The force acting at node 8 can be plotted as a function of time as shown in the figure below.

In the example problem, there are 120 c ycles per second. The following figure shows the h armonic force data input. If necessary, harmonic displacements can exist in the same problem with harmonic forces. This example has only harmonic forces.

The same force effect can be achieved b y entering +95.0 pounds at each node and entering a phase angle of 180.0 degrees at node 13.


191

Examples Calculations for the example problem take less than 30 seconds to complete. You can view the structure in animated motion or view standard displaced shape plots in the DynPlot window using the Display Graphical Results option (as shown below). Additionally, harmonic results, restraint loads, forces, and stresses can be calculated for the maximum displacements due to the harmonic loads.

Relief Valve Loads (RELIEF) The following example analyzes the two relief valve systems, shown below, subject to the simultaneous firing of both valves. 

Process Steam Conditi ons - 450 psi, @ 650°F



Relief Valve Orifice - JOHNSON #34A-06



Valve Opening Time - 8.0 milliseconds



Valve Closing Time - 8.0 milliseconds


2.141 in. ID.

192

Examples 

Relief Duration - 1.0 second

Relief Valve Exampl e Problem Setup You can use CAESAR II to compute the support loads, forces, and stresses in the vent piping system when the relief valves fire sim ultaneously. Venting steam stagnation properties are given. On the main menu, click An aly si s > Dyn ami cs to open the Dynamic Analysis dialog box. In the Dynamic Analysis dialog box, click Relief to compute the maximum thrust load magnitude at the vent pipe exit. This Load Synthesis dynamic load acts downward at the vent elbow nodes 65 and 100. Venting lasts for approximately one second, and the opening and closing time for the relief valve (as provided by the manufacturer) is 8.0 milliseconds. A static load case is run first to perform spring h anger sizing at nodes 20 and 22. The static load case #3 is the operating case and is used to set the nonlinear restraints for the dynamic analysis.


193

Examples CAESAR II Gas Thrust Load Calcul ations

The spectrum table name is arbitrarily selected as Relief and is defined as having a Frequency range and a Force-Multiplier ordinate. A # sign precedes the name in the spectrum definition because the shock table is to be read from an ASCII file. The spectrum definition is shown as it appears in the Dynamic Analysis dialog box is shown below:


194

Examples The DLF Spectrum Generator builds the ASCII file, Relief , which contains the relief valve spectrum table. You must define the filename, maximum table frequency, number of points, and the time-history waveform. In the example shown below, a m aximum frequency of 33 Hz and 20 data points are used to generate the tabl e.

The points in the time history waveform are entered as shown below. These p oints represent the opening of the valve, its one-second vent time, and its closing.


195

Examples The resulting DLF Spectrum is shown below.

The Frequency versus Dynamic Load Factors is written to the file Relief . Click Save to File, and then click OK to close the Spectrum Table Values dialog box. The thrust loads act at points 65 and 100. These loads are defined on the Force Sets tab and are entered as shown below.

There is only a single load case defined on the Spectrum Load Cases tab as follows:


196

Examples Only one static/dynamic combination case is defined. It is the combination of the sustained static load case with the dynamic load case. This is defined on the Static/Dynamic Combinations tab as follows:

Only one item needs to be set on the Control Parameters tab. It defines the static load case for setting the nonlinear restraints. In the following example, the setting is 3 (line 1). Alternatively, you can set the modal combination m ethod (line 11) to AB S instead of SRSS to produce conservative results.

Relief Valve Loadin g Output There are four key reports for a relief valve analysis:


197

Examples 





Mass Participatio n Report - Shows how sensitive each of the piping system’s modes is to the relief valve firing. High modal participation factors indicate that the mode is e asily excited by the applied dynamic f orces. If subsequent displacement, restraint, or stress reports indicate excessive dynamic responses, then the modes having high participation must be dampened or eliminated. After a particular mode is targeted as being a problem, it c an be viewed in tabular form using the mode shape report or graphically using the animated mode shape plots.

Displacement Report - Shows the maximum possible positive or negative displacement that can occur at some time during the reli ef valve’s firing. Values in this report are always positive.

Restraint Repor t - Shows the maximum dynamic load for which the support should be designed. The top value is t he maximum support reaction. The second value is the largest support reaction due to any one mode. T he last number on the left indicates which mode.


198

Examples





Stress Report - Shows the maximum dynamic stress due to the relief valve firing. Stresses from a dynamic shock load case should be combined with the sustained stresses from a static analysis and the result compared with the code defined occasional stress for the material. Partic ipation Factor Report - Shows which modes tend to be excited by the applied dynamic load.

The following Stress report shows element stresses due to the dynamic relief loads. T he top value is the maximum stress due to the interaction of all the s ystem modes. The second value is the largest stress due to any one m ode. The bottom number on the left indicates which mode. In the following examples, the maximum stress at node 5 is 1481 psi. The stress at node 5 due only to mode #1 is 1280 psi.

The maximum stress at node 40 on elements 40-50 is 6430 psi. T he stress at node 40 due to mode #4 is 3982 psi. Mode #4 is the largest contributor to the stress at node 40.


199

Examples

Support reactions due to the combination of the static sustained and the dynamic solutions are shown in the following two report samples:

This stress combination can be compared to the B31 code allowables for occasional stresses.


200

Examples

Dynamic Analysis of Water Hammer Loads (HAMMER) In this example, the cooling water suppl y line shown below suffers a pressure surge when the turbine driven pump drops offline due to a bearing temperature problem. The elbow at node 45 is observed to jump 6 to 8 inches in the X-direction when the turbine trip occurs. To eliminate the large field displacements associated with the turbine trip, an alternative support scheme must be designed. 

Fluid Properties - 250 psi @ 140° F



Flow Velocit y - 6 fps



Water Bulk Modul us - 313000 psi

SOLUTION The magnitude of the pump supply si de pressure wave, which emanates from the pump discharge at node 5, can be estimated from dp = ρ c dv Where: dp = the pressure rise due to the pump’s instantaneous stopping ρ = the fluid density

c = the speed of sound in the fluid dv = the change in velocity of the fluid The speed of sound in the fluid can be estimated from: 0.5

c = [Ef /(ρ+ ρ(Ef /E)(d/t))]


201

Examples Where: Ef = the bulk modulus of the fluid (313000 psi) E = the modulus of elasticity of the pipe (30E6 psi) d = the pipe mean diameter t = the pipe wall thickness ρ = the fluid density (62.4 lbm/ft3)

ρ + ρ(Ef /E)(d/t) = 62.4 lbm/ft

3

[1 + (313000/30E6)(8.62 -0.322)/0.322] = 79.1875 lbm/ft 2

3

2

3 2

2 1/2

c = (313000 lbf/in )(ft /79.1875 lbm)(32.2 lbm ft/lbf sec )(144in /ft ) = 4281 ft/sec For a more detailed discussion and evaluation of the speed of sound, see Piping Handbook, Crocker & King, Fifth Edition, McGraw-Hill pages 3-189 through 3-191 Apply the previously mentioned equation for the magnitude of the water hammer pressure wave. 3

dp = ρ c dv = (62.4 lbm/ft )(4281 ft/sec)(6.0 ft/sec) 3

2

2

2

= (62.4 lbm/ft )(4281 ft/sec)(6.0 ft/sec)(lbf sec /32.2 lbm ft)(ft /144 in ) = 345.6 psi There are two distinct pressure pulses generated when a flowing fluid is brought to a stop. One pulse originates at the supply side of the pump, and the other pulse originates at the discharge side of the pump. This example only deals with the suppl y side water hammer effect, but the magnitude and impact of the discharge side water hammer load should likewise be investigated when in a design mode. The time history waveform for both types of water hamm er pulses is shown as follo ws:

Pod - Discharge pressure Ps - Source (tank or static) pressure Pos - Suction pressure (while running) dp - Pressure fluctuation due to the instantaneous stoppage of flow through the pump Pv - Liquid vapor pressure at flow temperature There is an unbalanced load on the piping system due to the time it takes the pressure wave to pass successive elbow-elbow pairs. The magnitude of this unbalanced load can be computed from:


202

Examples F unbalanced = dp x Area The duration of the load is found from t = L/c, where L is the length of pipe between adjacent elbow-elbow pairs. For this example, the elbow-elbow pairs most likely to cause the large deflections at node 45 are nodes 45-75 and nodes 90-110. The rise time for the unbalanced dynamic loading should be obtained from the pump manufacturer or from testing, and it can be determined from graphs such as those shown above. For this example, a rise time of 5 milliseconds is assumed.

CALCULATIONS L 45-75 = 7 + 4(20) + 4 = 90 ft. L 90-110 = 3(20) + 15 = 75 ft. 2

Area = Π/4di ; di = 8.625-(2)(0.322) = 7.981 in. 2

Area = Π/4(7.981) = 50.0 in

2

F unbalanced = dp x Area = (345.6) (50.0) = 17289 lbf t duration = L/c = (90)/(4281) = 21 milliseconds, on leg from 45 to 75 = (75)/(4281) = 17.5 milliseconds, on leg from 90 to 110 t rise = 5.0 milliseconds Because the piping in this example is ductile low carbon steel, the major design variable is the large displacement. The problem is assumed to be solved when the restraint system is redesigned to limit the large displacements due to water hammer without causing an y subsequent thermal problem due to over-restraint. 1. Generate the DLF spectrum files as shown in the following examples.


203

Examples


204

Examples

2.

Define the spectrum on the Spectrum Definitions tab:

3.

Define the force sets on the Force Sets tab.

Three spectrum load cases are of interest in this example: each spectrum separately and the two of them in combination.


205

Examples

The sustained static load case is now combined with each dynamic load case for code stress checks. For operating restraint loads, the static operating case is combined with each dynamic load case, if necessary.


206

Examples

4.

Set the options on the Control Parameters tab as shown below:

Notes for Analyzing Water Hammer Loads On the pump or valve supply side, the magnitude of the pressure wave is calculated as shown in this example using the following formula: dp = ρ c dv

On the pump or valve discharge side, the maximum magnitude of the pressure wave is the difference between the fluid vapor pressure and the line pressure. On the supply side, a positi ve pressure wave moves away from the pump at the speed of sound in the fluid. The magnitude of the pressure wave is equal to the sum of the suction side pressure and dp . On the discharge side, a negative pressure wave moves away from the pump at the s peed of sound in the fluid. The m aximum magnitude of this negative pressure wave is the difference between the pump discharge pressure and the flu id vapor pressure. After the pump shuts down, the pressure at the discharge begins to drop. The momentum of the fluid in the downstream piping draws the discharge pressure down. If the fluid reaches its vapor pressure, the f luid adjacent to the pump flashes. As the negative pressure wave moves away from the p ump, these vapor bubbles collapse instantly. This local vapor implosion can cause extremely high pressure pulses. In addition, there can be a fluid backflow created due to the rapid drop in pressure. In this case, the backflow slap at the idle pump can be accentuated by the collapse of created vapor bubbles, resulting in an extremely large downstream water hammer loading. Water hammer loadings cycle to some extent. The pressure wave passes through the system once at full strength. Reflections of the wave can then cause secondary pressure transients.


207

Examples Without a transient fluid simulation or field data, the usual procedure is to assume one or two significant passes of the pressure wave. Where critical piping is concerned, or where the maximum loads on snubbers and restraints is to be computed, the independent effect of a single pass of the pressure wave should be analyzed for each elbow-elbow pair in the model. A separate force spectrum load set is defined for the elbow with the highest pressure as the wave passes between the elbow-elbow pair. The direction of the applied force is away from the elbow-elbow pair. An individual dynamic load case is run for each separate force set; com binations of different force sets are usually not run. This approach is satisfactory when applied t o large, hot steam piping systems that have very few fixed restraints and a high num ber of low modes of vibration. Extrapolation to other types of piping systems should be made at the discretion of the piping designer. CAESAR II does not check the integrity of the piping system due to the local increase in hoop stress that occurs as the fluid pressure wave passes each pipe cross-section. Slowing the mechanism that tends to reduce the flowrate can reduce the m agnitude of the water hammer loads. In the case of valve closing, it means slowly closing the valve. In the case of a pump going off line, it means slowly removing power from the pump. Slowly in each of these instances can be estimated from: T = 2L/c Where: T = Time of one wave cycle sec. L = Characteristic length of piping system. This is usually the length between the pump or valve and the source or sink. c = Speed of sound in the fluid. If the pump or valve stops in a time shorter than T, then the water hammer should be analyzed as shown in this example for instantaneous closure. Calculations for this problem are gi ven below. Of primary interest is the largest time segment that must be use d to close a valve or bring a pump flowrate to a halt such that water hammer type pressure pulses are not generated. Calculations using the lengths of several reflecting systems are made to determine the variation of the computed Ts. The longest time is for the wave to leave the supply side at node 5 and move to the tank connection at node 125. This represents a total L of about 270-feet. T = (2) (270) ft./(4281)ft/sec = 126 milliseconds The length through which the wave passes that causes the most trouble is the length between nodes 45 and 75: T = (2) (90)/(4281) = 42 milliseconds If the pump or valve can slo w down in greater than 126 milliseconds, the tendency for water hammer in the piping system is usually abated. If the pump or valve can slow down in greater than 42 milliseconds then the tendency for water hammer in the 45-75 length is abated. Water hammer excitation initially produces axial acoustic waves in the steel pipe wall that can induce locally very high, ver y short duration forces and stresses. These short duration loads are usually not a design problem in ductile steel piping systems. Where crack propagation in welds and material due to water hammer loads is a concern, use the following rules: 

A very high number of natural frequencies must usually be included in the analysis. Cutoff frequencies of 300 Hz are not unusual. T hese are the axial natural modes of the pipe between the excited elbow-elbow pairs. Higher modes must be computed until the inclusion


208

Examples of extra modes does not produce an appreciable change in the force/stress response. The maximum frequency cutoff can be estimated using SQRT (E/ρ)/L Where: E = Pipe material modulus of elasticity p = Pipe material density L = Length of a single pipe element in the primary run that is to have accurate stresses computed due to the passing of the water hammer originated acoustic stress wave. Calculation of the maximum cutoff frequency for the 45-75 e lbow-elbow pair for the 20-foot pipe lengths is given as follows: f cutoff = SQRT (E/ρ)/L = SQRT ((30E6)(32.2)(12)/(0.283))/20 = (202388 in./sec) / (20 ft. 12 in/ft) = (843.3 rad./sec) / (2 p rad./cycles) = 134.2 Hz Alternatively, including the Missing Mass Correction approximates the contribution from the omitted modes. 

The length of any element in the primary axial runs should not be greater than about ct/4, where c equals the speed of sound in the pipe and t equals the duration of the w ater hammer load. Calculation of the greatest element length for the 45-75 elbow-elbow pair is given as follows: Lmax= ct/4 = (4281) ft/sec (0.021) sec/(4) = 22.5 ft To get an accurate estimate of the stresses due to the passing of the stress wave in the pipe, individual element lengths should be smaller than about 20 feet. Shorter duration lo ads require shorter elements to monitor the passing of the stress wave.



The inclusion of the response due to the higher modes does not affect the displacement results (only the force and stress results). Displacement results, such as the 6- to 8-inches in this example, can usually be computed accurately after the inclusion of the lo w frequency modes with participation factors greater than about 0.01.


209

Examples

Water Hammer Lo ading Output Mass Participation Report This report illustrates the sensitivity between each of the modes in the pipin g system and the water hammer dynamic loading. High modal participation factors indicate that the mode is easil y excited by the applied dynam ic forces. If subsequent displacement reports indicate high dynamic responses, then the modes having high participation must be dampened or elim inated. After a particular mode is targeted as being a problem, you can view it in table format using the mode shape report. You can view it graphically using the animated mode shape plots.


210

Examples

Restr aint/Force/Stress Reports If high modes are included, as discussed in Notes for Analyzing Water Hammer Loads (on page 207), then these reports give the m aximum values of the forces and stresses in the system due to gross deformation and the propagation of an acoustic stress wave in the pipe. If the high modes are not included, then these reports give the maximum values of forces and stresses in the system due to gross deformation alone.

Displacement Report This report gives the maximum possible positive or negative displacement that can occur at some time during the event. Values in this report are always positive.


211

Examples

Combination Cases The force spectrum approach to the water hammer problem does no t include consideration of the time relationship between modal or directional maximums. Conservative results can be guaranteed by taking the absolute summation of both the modal and directional response properties. Running one load case for each main piping run and a final load case, including all of the individual load cases, can typically provide a good indication for where problems exist. In this example, the main piping run between nodes 45 and 75 added the major contribution to the system dynamic responses. The combination load case, which included the 45-75 and 90-110 contributions together, yielded little extra information.

Soluti on Summary A guide and axial limit stop at nodes 45 and 105 produces little increase in thermal stresses, which were low to begin with, and serves to attenuate the large axial displacements in the line due to the water hammer load. Loads on this support due to the low mode displacements are seen to be small. Local, very short duration loads may not be so small. The restraint should be designed with this in mind. A few simple design rules are usually sufficient: 



Flexible is optimal. The restraint should only be stiff enough to sufficiently attenuate the low frequency gross deformation. Areas of local discontinuities, such as the weld of the support to the pipe, should have extra weld or support plate area. Discontinuities at other restraints in a problem area sh ould also be reinforced to withstand the local passing of the impact stress wave.


212

Examples

Dynamic Analysis of Independent Support Earthquake Excitation (CRYNOS) This example explains how to model a cryogenic piping system in accordance with B31.3 using specific ground, building, and envelope sp ectra. The two analyses that are run in this example assume the pipe (structural steel) supports are rigid and include the flexibility of the structural steel supports by including the steel frames in the analysis. Finally, compare the results from the two analyses. The design p arameters are: 

Ambient temperature: 100°F



Operating temperature: -59°F



Pipe: 8-inches Sch 10S



Insulation: 4-inches 22.3 lb/cu feet



Insulation: 0.232 SG



Columns: W14x82



Beams: W10x12

The isometric of the complete model is shown in the following figure. This drawing shows the piping, pipe supports, and the structural steel frames

.


213

Examples The following table lists the excitation spectra to be applied to this model: Ground Response Spectra Ground Response T sec

Building Response Spectra Building Response

V in/sec

T sec

Envelope Response Spectra Envelope Response

V in/sec

T sec

V in/sec

0.05

0.787

0.05

0.787

0.05

0.787

0.2

7.874

0.2

1.3

0.2

7.874

0.5

21.653

0.5

3.4

0.5

21.653

1

39.37

1

27.3

1

39.37

2

18.89

2

30.4

2

30.4

3.5

43.7

3.5

21.12

3.5

43.7

5

11.8

5

21.3

5

21.3

10

5.9

10

5.359

10

5.9


214

Examples The necessity for the various spectra can be best understood by investigating the difference between independent support excitation and uniform support excitation. These excitation methods are shown in the following figures.


215

Examples For the analysis with steel supports, the structural steel must be included as part of the piping model. Click Environment > Include Structur al Input Files. The structural data input is shown below:

SECI D=1, W14 X 82; COLUMN CROSS SECTI ON SECI D=2, W10 X 12; BEAM CROSS SECTI ON MATI D=1, YM=29E6 POI S=0. 3 G=11E6 DENS=0. 283 DEFAULT SECI D=1 ANGLE=90 EDI M 1038 1039 DY=15- 0; DEFI NE ALL COLUMNS EDI M 1043 1044 DY=15- 0 EDI M 1048 1049 DY=15- 0 EDI M 1053 1054 DY=15- 0 DEFAULT SECI D=2 ANGLE=0 EDI M 1039 1040 DZ=- 2- 0; DEFI NE ALL BEAMS EDI M 1044 1045 DZ=- 2- 0 EDI M 1054 1055 DZ=- 2- 0 FI X 1038 ALL FI X 1043 ALL FI X 1048 ALL FI X 1053 ALL_ The dynamics input for this example is summarized in the following figure:

Details of the dynamics input are shown in the next four figures.


216

Examples

Only the results for the uniform support excitation case are shown. Using this load case, the model with and without structural steel supports is compared. The results from these two models are shown in the following tables.


217

Examples

With Structure


218

Examples

Without Structure


219

Examples

With Structure


220

Examples

Without Structure


221

Examples

With Structure


222

Examples

Without Structure


223

Examples

Results These comparison tables illustrate the differences that can exist when the structural steel models are not included in t he analysis. In some cases, the results with the structural steel included are many times higher than the resu lts computed without the structural steel. The steel models add flexibility to the piping system. More flexibility means lower natural frequencies and more modes to be excited by the sh ock. A comparison of the natural frequencies of the two models is given as follows:

With Structure


224

Examples

Without Structure

In the previous table, there are only fi ve extra mode shapes for the s ystem. The restraint moment at node 55 in the Z-direction is much larger without the steel model than it is with the steel model. Even though the piping is tied to the steel, the steel frame does not support much moment in the Z-direction. The steel fram e bends slightly about the Z-axis, and the moment is carried through from the pipe. In the piping only model, the rigid anchor at node 55 does not rotate about the Z-axis (or any other axis), and ends up carrying the entire moment load.

Structural Analysis (FRAME) The following example uses CAESAR II to analyze the braced frame shown below, subject to the given uniform load and self weight. 2

4



Column section data area = 15 in inertias = 250 in.



Column section data area = 10 in inertias = 500 in.



Brace section data area = 5 in inertias = 1 in.



Material density: 490 pcf



Material density: 200 lb/inches

2

2

4

4

This example shows how to model a structure using the CAESAR II Structural module. The figure below displays a single-bay, braced space frame. All beam and column lengths are 50-inches. This frame is subject to its own weight load as well as a uniform load of 200 pounds per inch on all of the top-level beams. You need to determine the displacements, reactions, and element forces for three load cases: self weight, uniform load, and self weight plus uniform load.


225

Examples

This example illustrates how to use most of the keyword directives in the structural preprocessor. A standard finite element modeling approach is f ollowed, where the system nodes are defined first, followed by materials and section properties, elements, and finally the loading.

Process the Input File 1.

In CAESAR II, click File > Open and open the structural file Frame.str .





2.

Example files are delivered to the [Product Directory]\Examples folder during installation. Select Structural(*.str) in the Files of type list.

Click Input > Structu ral Steel.


226

Examples The Structu ral Modeler dialog box displays the frame model.

3.

Click Save

on the Structu ral Modeler toolbar to error check and save the model.

4.

Click Yes in the Save changes to FRAME? message box.

5.

Click OK in the Model Generation Status message box.

6.

Click File > Exit to exit the structural steel input processor.

7.

Click Static Analysis


on the CAESAR II toolbar.

227

Examples CAESAR II reads the binary files created by the structural preprocessor and displays recommended load cases in the Static Analysis dialog box.

CAESAR II recommends load cases to satisfy piping code compliance and not necessarily loads that you would want to use for structural analysis. 8.

Edit the load cases as shown in the example below. Load case 2 (L2 ) consists only of U1, which is designated as an operating case (OPE). This is a construction case and is segregated here because i t may be interesting to see the loads produced by the U niform Load.


228

Examples 9.

Click Run Analysis

to begin processing.

The software displays the results for the analysis.


229

Examples


230

Examples


231

Examples

Dynamic Analysis (NUREG9) This example analyzes the piping system shown below, subjected to a series of shock spectra. The example problem, taken from NUREG/CR -1677, BNL-NUREG-51267, VOL I I, August 1985, is one of the NRC benchmark problems run to verify the dynamic capabilities of CAESAR II. The detailed input is neither shown nor discussed in this example. The required input files are located in the [Product Directory]\Examples folder.


232

Examples

NRC Example NUREG 9 NRC Example NUREG 9 The example is a three-branch system, composed of 20 pipe elements and 14 support elements. The support elements are divided into four groups that correspond to four distinct input excitation spectra sets. This example demonstrates the independent support m otion feature of CAESAR II. In modeling this problem, the 14 support elements are input as restraints with stiffnesses. All bend elements include a node at the "near" point to ensure mass and stiffness computations consistent with the NRC example. In addition to the pipe density, there is a single lumped mass applied at node 18. In this example, the contributions from the pseudo-static anchor point displacements are not included. The three solutions presented represent the following: 





Envelope spectrum; spatial then modal combinations ISM (independent support motion); directional, spatial, then modal combinations using SRSS ISM directional, spatial, then modal combinations using ABS

NRC Example Problem 2A NATURAL FREQUENCY REPORT (Hz) MODE

NRC

CAESAR II

1

9.360

9.362

2

12.71

12.708

3

15.38

15.379

4

17.80

17.800


233

Examples MODE

NRC

CAESAR II

5

21.60

21.606

6

25.10

25.102

7

32.03

32.039

8

38.07

38.075

9

40.29

40.299

10

48.90

48.905

11

57.51

57.524

12

61.50

61.510

13

62.54

62.550

14

69.35

69.359

15

77.44

77.456

16

78.88

78.893

17

101.7

101.731

18

103.6

103.598

19

108.0

107.983

20

115.1

115.116

21

135.2

135.265

22

155.2

155.244

23

160.6

160.626

24

203.8

203.820

25

209.9

209.957

NRC BULLETIN NUREG-51267 VOL.II 1980.


234

Examples TRANSLATIONS (in) DX

DY

DZ

NODE NRC

CAESAR II NRC

CAESAR II NRC

CAESAR II

2

.0105

.0105

.0

.0

.0250

.0250

4

.0431

.0431

.0049

.0049

.0907

.0907

6

.0475

.0475

.0253

.0252

.0327

.0327

8

.0280

.0280

.0379

.0379

.0491

.0491

10

.0108

.0107

.0249

.0249

.0631

.0631

12

.0285

.0285

.0186

.0186

.0633

.0633

14

.0849

.0849

.0085

.0085

.0635

.0635

16

.0476

.0476

.0001

.0001

.0402

.0401

18

.0286

.0286

.0318

.0318

.0421

.0421

20

.0131

.0131

.0095

.0095

.0001

.0001

Problem 2A NRC BULLETIN NUREG-51267 VOL.II 1980.

ROTATIONS (deg) RX

RY

RZ

NODE NRC

CAESAR II

NRC

CAESAR II NRC

CAESAR II

2

.0457

.0457

.0260

.0260

.0190

.0190

4

.0515

.0515

.0688

.0688

.0269

.0268

6

.0389

.0389

.1012

.1012

.0268

.0267

8

.0309

.0309

.0950

.0949

.0217

.0217

10

.0201

.0201

.0289

.0289

.0203

.0203

12

.0105

.0105

.0328

.0328

.0224

.0224


235

Examples 14

.0102

.0102

.0514

.0511

.0299

.0299

16

.0359

.0359

.0496

.0496

.0476

.0476

18

.0105

.0105

.0343

.0343

.0128

.0127

20

.0215

.0214

.0273

.0273

.0090

.0090

Problem 2A NRC BULLETIN NUREG-51267 VOL. II 1980.

SUPPORT FORCES (lb) FX

FY

FZ

NODE

NRC

CAESAR II NRC

CAESAR II NRC

CAESAR II

1

90

90

65

64

177

177

7

0

0

0

0

708

707

9

446

445

0

0

0

0

11

0

0

206

206

0

0

13

0

0

164

164

0

0

15

188

187

188

187

263

262

17

58

58

198

197

103

103

21

378

377

192

191

245

245

Problem 2A NRC BULLETIN NUREG-51267 VOL. II 1980.

NRC Example Problem 2B TRANSLATIONS (in) DX

DY

DZ

NODE NRC

CAESAR II

NRC

CAESAR II

NRC

CAESAR II

2

.0064

.0064

.0002

.0

0158

0158

4

.0267

.0267

.0031

.0031

.0574

.0574

6

.0295

.0295

.0162

.0162

.0207

.0207


236

Examples DX

DY

DZ

8

.0170

.0170

.0242

.0242

.0311

.0311

10

.0029

.0029

.0152

.0152

.0399

.0399

12

.0103

.0103

.0110

.0110

.0400

.0400

14

.0530

.0530

.0053

.0053

.0401

.0401

16

.0301

.0301

.0001

.0001

.0255

.0255

18

.0103

.0103

.0187

.0187

.0267

.0267

20

.0033

.0033

.0057

.0057

.0

.0

Problem 2B NRC BULLETIN NUREG-51267 VOL. II 1980.

ROTATIONS (deg) RX

RY

RZ

NODE NRC

CAESAR II NRC

CAESAR II

NRC

CAESAR II

2

.0289

.0289

.0165

.0165

.0116

.0116

4

.0326

.0326

.0435

.0435

.0172

.0171

6

.0247

.0247

.0641

.0640

.0171

.0171

8

.0199

.0199

.0599

.0598

.0132

.0132

10

.0134

.0134

.0075

.0075

.0120

.0120

12

.0071

.0071

.0204

.0204

.0134

.0134

14

.0062

.0062

.0307

.0307

.0184

.0184

16

.0228

.0228

.0276

.0276

.0301

.0301

18

.0070

.0070

.0208

.0208

.0079

.0079

20

.0128

.0128

.0074

.0074

.0053

.0053



237

Examples SUPPORT FORCES (lb) FX

FY

FZ

NODE NRC

CAESAR II NRC

CAESAR II NRC

CAESAR II

1

53

53

46

46

113

112

7

0

0

0

0

441

440

9

257

256

0

0

0

0

11

0

0

123

123

0

0

13

0

0

98

98

0

0

15

111

111

111

111

156

155

17

32

32

124

123

66

66

21

103

103

114

113

116

115


NRC Example Prob lem 2C NRC BENCHMARK SERIES NRC BULLETIN NUREG-51267 VOL.II 1980. NRC PROBLEM 2C CAESAR II JOB NUREG9

TRANSLATIONS (in) DX

DY

DZ

NODE

NRC

CAESAR II NRC

CAESAR II

NRC

CAESAR II

2

.0090

.0090

.0

.0

.0220

.0220

4

.0373

.0372

.0044

.0044

.0800

.0800

6

.0411

.0411

.0235

.0235

.0289

.0288

8

.0237

.0237

.0355

.0355

.0434

.0434

10

.0043

.0043

.0227

.0227

.0556

.0556

12

.0148

.0148

.0164

.0164

.0558

.0558


238

Examples DX

DY

DZ

14

.0741

.0740

.0074

.0074

.0560

.0560

16

.0420

.0420

.0001

.0001

.0355

.0355

18

.0148

.0148

.0281

.0372

.0372

.0372

20

.0049

.0049

.0085

.0085

.0001

.0001

Problem 2C NRC BULLETIN NUREG-51267 VOL. II 1980.

ROTATIONS (deg) RX

RY

RZ

NODE

NRC

CAESAR II

NRC

CAESAR II

NRC

CAESAR II

2

.0402

.0402

.0229

.0229

.0163

.0163

4

.0456

.0455

.0606

.0605

.0244

.0244

6

.0347

.0346

.0894

.0893

.0252

.0252

8

.0282

.0282

.0835

.0835

.0196

.0196

10

.0197

.0197

.0112

.0112

.0179

.0179

12

.0104

.0104

.0285

.0285

.0199

.0199

14

.0092

.0092

.0429

.0429

.0260

.0260

16

.0318

.0317

.0387

.0387

.0421

.0420

18

.0104

.0104

.0291

.0291

.0116

.0116

20

.0191

.0191

.0110

.0110

.0079

.0079


SUPPORT FORCES (lb) FX

FY

FZ

NODE

NRC

CAESAR II

NRC

CAESAR II

NRC

CAESAR II

1

76

76

70

69

156

155


239

Examples FX

FY

FZ

7

0

0

0

0

607

607

9

350

350

0

0

0

0

11

0

0

184

184

0

0

13

0

0

146

146

0

0

15

151

151

151

151

212

211

17

45

45

169

168

91

90

21

152

151

170

169

158

157


Omega Lo op Modelling (OMEGA) The Omega expansion loop consists of a series of back to back 135 degree bends. This example illustrates how to generate a piping model of an Omega loop according to the following sketches. 

Pipe: 3-inches, standard wall



Bend radius: 24-inches



Material: Low carbon steel



Temperature: 200°F, 300°F, 400°F


240

Examples The example uses an Omega loop as shown below. The given dimensions are the 6-ft 10-in. height, the 2-ft bend radius, and the bend angles of 135° and 2 70°. Using this information, you can derive the other dimensions shown in the following figure.

Equation 1: Figure 1 Equation 1: Figure 1

When coding a series of back-to-back bends, it is important to remember that the delta dimensions should be measured from the tangent intersection point (T IP) to the tangent intersection point. For more information about the proper coding of bends, see Bends (on page 9).


241

Examples Figure 2 shows the node points, which are coded on the Classic Piping Input dialog box to model the Omega loop. The model is anchored at nodes 1 and 35. The first bend (lower left bend) spans between nodes 5 and 10. The TIP 10 is to the far right of the bend. For analysis and output, the actual location of node 10 is at the far weld line, as shown in Figure 3.


The second bend (upper left bend) spans between nodes 10 and 15. Because you can code TIP to TIP, the delta coordinates entered in the Classic Piping Input dialog box are the X and Y distances between nodes 10 and 15 (Figure 2). The actual location of node 15 is at the far weld line, shown on Figure 3. Node 15 is the TIP for this bend and lies to the left of the pipe. The third bend (upper right bend) spans between nodes 15 and 20, where node 20 is the TIP. In coding from TIP to TIP, only a delta x is required. Figure 3 shows the actual location of node 20 on the pipe. The fourth and final bend (lower right bend) spans between nodes 20 and 25. In this case, a delta X and a delta Y are required. The actual location of node 25 is shown on Figure 3. The element from 25 to 30 is a straight element necessary to finish off the bend.


242

Examples A bend in CAESAR II requires an element beyond the far weld line to determine its orientation.


Below is an input listing for the model. The delta dimensions shown were obtained from Figure 1. Three additional, equally-spaced points are located on each bend. This example requires a change in the CAESAR II Configuration Editor to allow the error checker to accept large angle (> 95°) b ends. \


243

Examples Click Tools > Configur e / Setup to open the CAESAR II Configuration Editor. In the Categories pane, click Geometry Directives and set Maximum Allowable Bend Angle to 135.


244

Examples

Jacketed Piping (JACKET) This example guides you through the modeling techniques used in the analysis of jacketed piping systems. Where applicable, various alternatives are d iscussed that can benefit specific systems or problems. The piping system to be analyzed is shown in the following figure. It consists of an 8-inch, schedule-40 crude oil line and a 12-inch, schedule-40 steam jacket. The section of piping from the pump to the valve is completely jacketed, while the section from the valve to the vessel has only the straight sections jacketed. This variation in the jacket is used to i llustrate the two common types of jacketed systems. The core pipe is supported in the jacket using spiders. These spiders provide translational restraints in two directions, normal to the axis of the pipe. For this system, the spiders are located at each elbow weld line and in the straight runs such that the spider spacing does not exceed six feet. For this system, both the jack et and the core are low carbon steel.


245

Examples In some systems, the jacket and the core consist of different materials. This condition must be modeled very carefully because the thermal g rowth in the core is different f rom the thermal growth of the jacket. Improper axial restraints in such a system can cause extremely large loads in the pipe.

Modeling Plan The first step in modeling a ny system is to consider the m ost efficient way to create the input and, more importantly, how to best review the results. After you determine how to best review the results, you can define the input node numbering scheme. Based on the node numbering scheme, you can then decide how to generate the model to take advantage of the various rotate, duplicate, and include options. For this example system, the core piping is modeled using node n umbers from 1000 to 1999, and the jacket is modeled using node numbers starting at 2000. Additionally, similar locations on the two systems will have the same base node number, that is, nodes 1110 and 2110 describe the same point on both the core and the jacket. Setting up the n ode numbers in this manner enables one of the systems to be generated from the other, using either the duplicate or the include options of the input preprocessor. You can also view the system individually in the plot by the Range command and breaking the model at node 1999. The other advantage to this numbering scheme is that when reviewing the output, you can immediately tell from the node number whether the point in question belo ngs to the core or the j acket. Although not necessary for a small system such as this, additional node number ranges can be defined to differentiate parts of the model. To il lustrate, the following additional constraints can be placed on the node numbers. The ground level piping will have nodes in the 100-400 series, while the second level piping will have nodes in the 500-900 series. For example, node 1110 will be a core node at ground level and node 2550 will be a j acket node on the second level. To indicate locations where external supports are applied to the system, node numbers will end in 5; all other points will be multiples of 10. Similar node numbering schemes can be used to differentiate branches from headers, pipe from structural steel, and various line siz es. Starting the modeling process with a clearly defined plan can ease both input verification and output


246

Examples review. For example, consider reviewing the i nput for this system and finding a sp ring hanger at node 1530. You can quickly recognize this as an error because the 1000 series nodes make up the core piping and cannot use spring hangers. Additionally, support node numbering should end with a 5.

Node Layout The system as defined in the preceding figure consists of nine segm ents of piping. Each segment is shown in the following figure with the node numbers assigned to the various points for the core piping. Each segm ent is discussed individually.









The term segment is used solely to assist in discussing this example. CAESAR II does not require the segregation of a piping system into segments. There are no such input requirements or restrictions in CAESAR II.

Segment A - This segment runs from the pump to the first elbow. Because this section is at ground level, the 100 series nodes are used. Because the pump acts as an anchor, the start node of this segment ends in 5, so the pump is assigned node 1105. The length of the segment requires an intermediate node point for a spid er, so node 1110 is assigned 5-feet from the pump. Nodes 1120 and 1115 are assigned to the elbow. The positive Y support is not at node 1115 because 1115 is part of the core piping. The positive Y is applied at node 2115 (the jacket), so we assign the 5 to this node point. Segment B - This segment is the six-foot vertical section, beginning with the elbow at node 1120. This section can be modeled by coding to the top elbow and assigning nodes 1500 and 1510. The 500 series nodes are used here because the 2nd level piping is now being modeled.


247

Examples 













Segment C - The first horizontal run in the 2nd level requires a node at mid-span to accommodate a spring hanger (on the jacket). This m id-span node divides the segment into two 9-foot lengths, which exceed the m aximum spider spacing of 6-feet. Therefore, the eighteen-foot span is divided into four elements, each one 4-feet, 6-inches. The nodes assigned are 1520, 1525 (for the hanger location), and 1530. The segment is finis hed off with the elbow modeled by nodes 1540 and 1550. Segment D - This horizontal segment in the 2nd level is modeled using nodes 1560, 1570, and nodes 1575 and 1580 at the elbow. The nodes 1560 and 1570 are for spiders, while 1575 is a hanger location Segment E - This horizontal segment contains the valve. Nodes for this segment are 1590, 1600, 1610, and 1615. Node 1615 terminates the elbow and is also a hanger location. The element from 1590 to 1600 should be declared rigid with a weight of 452 pounds. Starting with the elbow at nodes 1610 through 1615, all of the elbows are modeled as individual elements. This eases the subsequent coding of the jacket. The elbows in this part of the model consist of two straight pieces of pipe, equal in length to the radius of the elbow. Segment F -The third horizontal leg of the expansion loop is modeled using nodes 1620, 1630, 1640, and 1650. Segment G -The last horizontal run of the 2nd level is modeled using nodes 1655, 1660, and 1670. Node 1655 is a hanger location. Segment H - The second vertical section of piping returns the system to ground level. The only additional nodes required for this section are for the elbow at nodes 1130 and 1135. Node 1135 is a positive Y location on the jacket. Segment I - This is the last segment that terminates at the vessel nozzle. The nodes used to model this segment are 1140, 1150, and 1155.


248

Examples

Core Pipin g Input During the input of the above data, make frequent use of the CAESAR II plot facility. Doing so ensures that the system is being modeled correctly an d that any input errors are detected as soon as possible. The following figure shows a volume plot of the completed core piping, with node numbers and anchors.

Figure 21: Completed Core Piping

At this phase of the input, save the input file if you have not already done so. We also recommend running the error checker at this time because the core piping model serves as the basis for generating the jacket piping model. If any errors exist in the c ore, they are duplicated in the jacket, thus doubling your correction efforts. The additional data required to finish the model, such as allowable stresses, temperatures, pressures, and so forth, are contained in the CAESAR II input file, which is delivered with the software. This data is found in the [Installation Driver]\ProgramData\Intergraph Cas\CAESAR II\5.30\Examples Jacket._a file.

Jacket Input (1st Half) There are several ways to obtain the jacket model. The most obvious m ethod is to continue using the Classic Piping Inpu t dialog box to build the jacket. A second method is to duplicate the core pipe input file, and then use the CAESAR II include feature to combine the two models. Another method is to use the List processor and duplicate the necessary elements from within the preprocessor. The latter method is the one used f or this example.


249

Examples To begin modeling the jacket, click Edit > List . The software opens the List Option dialog box, which displays all of the applicable input data. Click the Elements tab to view a list of elements and their associated delta coordinates as shown below:

For the first half of the jacket, the core piping is duplicated using the steps below. The duplicated region starts at the pump and ends at the valve. 1.

Click in the row for the element from 1105 to 1110.

2. Press Shift and click the row for the element from 1580 to 1590. This is the element just before the valve. The software highlights all rows between the two selections.

3. Right-click the highlighted list and select Duplicate. 4. In the Block Duplicate dialog box, click Identical. 5.

Click At End of Inp ut to place the duplicate block.


250

Examples 6. Specify 1000 for the node increment, and then click OK .

CAESAR II duplicates the block and increments all of the node numbers by 1000. This results in a section of pipe identical to the pipe from 1105 to 1590 with node numbers from 2105 to 2590.

Three changes must be made to the new section of pipe to obtain the jacket piping. First, the diameter and wall thickness must be changed to 12-inches, schedule 40. You can use the List Editor to find the element from 2105 to 2110, and then type new values. Next, you must also specify the following values: jack et temperature, jacket pressure, jacket insulation, and jacket fluid weight. Finally, you must change all the jacket bend radii from long to short. The most straightforward way to do this is to click the Bends tab on the List Options dialog box to open the Bends list. Then, starting with the bend at node 2120, change the radius from Long to 12.0-inches. You must make this change to all of the remaining bends. After the above changes have been made, the first half of the jacket is finished. A volume plot of the system now shows the core piping overlaid by the jacket piping. Although the two models are correctly positioned with respect to one another, they are not connected. All you have done so far is duplicate several pipes. From the standpoint of CAESAR II, there are simply two discontinuous systems in the same input file. The graphics module plots discontinuous systems such that they all start from the same point, which is why the jacket and core line up properly in this case. The next step is to correctly connect the j acket to the core and appl y any external restraints. The connection between the jacket and the core piping must model the spiders that ali gn the two in the real system. These connections can be modeled in CAESAR II using restraints with connecting nodes (CNodes).


251

Examples A CNode associates degrees-of-freedom. If a CNode connects two nodes in the Y-direction, they experience identical displacements in the Y-direction. Use CNodes to restrain two nodes to each other without restraining them to the "outside world." The modeling of the connection between the jacket and the core begi ns at the pump. In the Classic Piping Input dialog box, enter the restraint fie ld. Then add a restraint at node 1105 with a CNode at 2105 of type anchor . This associates all six degrees-of-freedom between nodes 1105 and 2105. Next, add two restraints at node 1110. Both of these restraints have a CNode at 2110, one in the Y-direction and one in the Z-direction. These two restraints model the spider between the core and the jacket. The spider was not modeled using gaps. The actual clearance between the spider and the pipes is very small, and attempting to numerically model this cl earance using restraints with gaps causes the job to be highly non-linear. Models with gaps at each spider have convergence problems and are unlikely to reach a solution. Nodes 1110 to 1120 define the first elbow. Add four restraints to this spreadsheet as follows: At node 1115, put a CNode of 2115 with Y- and Z-direction restraints At node 1120, put a CNode of 2120 with X- and Z-direction restraints. These restraints are perpendicular to the axis of the pipe. Also, at node 2115 there is a positive Y external restraint. Add this support to the system on the spreadsheet containing node 2115. Similarly, add the remaining spiders to the model. An example of the completed input file, JACKET._A, is delivered to the [CAESAR II Directory]\Examples folder. When node 1590 is reached, the CNode at 2590 is connected with an anchor. Also add the spring hangers at nodes 2525 and 2575. Aside from the two anchors at the pump and the valve, all of the spider connections between the jacket and the core are modeled using two perpendicular restraints with connecting nodes. How are the other four degrees-of-freedom restrained? What keeps this model from undergoing rigid body motion? These questions can be resolved by considering two points. First, the jacket is continuous over the core from the pump to the valve. At both of these points, you have connected all six degrees-of-freedom. Second, the translational restraints prevent motion in the three translational directions. Additionally, these restraints also prevent rotation because the jacket is continuous. Whenever a model is cons tructed, you must insure that the model, or parts of the model, cannot undergo rigid body motion. Such a model produces a singul ar stiffness matrix, and the solution cannot be attained. An example of such a poor model is a cantilever beam with a hinge at mid span. At this point in the data input process, we recommend that you click Error to run the model through interactive error checking. CAESAR II saves the input and Check reports the results in the Error and Warnings dialog box. All reported errors should be corrected before modeling the second half of the piping jacket.


252

Examples

Jacket Input (2nd Half) Modeling the second half of the j acket is more complex than the first half because the jacket only covers the straight runs of piping. For this reason, the jacket elements are co ded manually, as opposed to any form of duplication. Although duplicating portions of the model is feasible, the time required to delete the jack et from the elbows is greater than the time required to input only the straight sections. By modeling the jacket directly, the restraints for the spiders can be input as you encounter them. To start entering the necessary data in the Classic Piping Input dialog box, press Ctrl + End to go to the last spreadsheet in the model. At this point, click Continue and change the node numbers to 2600 and 2610, with a DX of 5 ft. Where is the element from 2600 to 2610? Return to the spreadsheet and temporarily change the diameter of nodes 2600 to 2610 to 24-inches and try the volume plot. The element 2600-2610 has been positioned at the plot origin because, at this time, it is not connected t o anything. Return to the dialog box and correct the diameter by changing it back to 12-inches. To properly connect the jacket to the core, you must add restraints at nodes 2600 and at 2610. At node 2610, add a CNode of 1610 with restraints in the Y- and Z-directions. At node 2600, add a CNode of 1600. Do not associate nodes 2600 and 1600 in the Y- and Z-directions. Associating the nodes in such a way allows the jacket to move freely in the X-direction and to spin about the X-axis, producing an unstable model. This problem did not exist in the first half of the model because the jacket was continuous over the elbows and the model was three-dimensional in nature. In the second half of the m odel, you must ensure that the appropriate axial and torsional restraints are applied to the jacket. At node 2600, model an anchor to 1600. T his is simpler than modeling separate X, Y, Z, and RX restraints. This causes the 8-inch line to be physically connected to the 12-inch line in all six degrees-of-freedom. The next jacket element covers the core from node 1616 (the end of the elbow) to node 1640. Node 2615 is anchored to node 1616 with a CNode. The next two elements, 2620-2630 and 2630-2640, are standard pipe elements with a DZ of -4.333 ft. Each To node is connected to the corresponding core node with a CNode associating the X- and Y-directions.


253

Examples The remaining three sections of j acket are modeled in exactly the same m anner. The final step in the modeling is to add the spring hangers at nodes 2615 and 2655 and the positive Y-restraint at 2135. The completed model is shown in the following figure.

Completed Jacketed Piping System An example of the completed input file, JACKET._A, is delivered to the [CAESAR II Directory]\Examples folder. After the input task has been completed, you must error check the job and analyze it for the specified loading conditions. Check the results to ensure that the system is modeled correctly. These checks should include the following: 







Verify the weight of the core system, the jacket system, and the combined system. You can use the Sustained Restraint report for this check. Be sure that the jacket pipe fluid density accounts for the volume lost due to the c ore. Because CAESAR II does not do this automatically; you must reduce the density of the jacket fluid accordingly. Verify that the piping system does not develop large axial loads in the core, the jacket, or the equipment anchors. This can be caused by improperly over restraining the pipe in the axial direction or by the effects of thermal growth on dissimilar metals. Check the displacements at the elbows in the operating case and verify that the core pipe does not have a tendency to move through the jacket. CAESAR II does not perform interference checking. Check the displacements at the spiders where the jacket and the core are connected. In the direction of the spiders, the displacements should be the same for both the jacket and the core.



Verify that wind and wave loads, if any are specified, are disabled on the core piping.



Ensure that the insulation thickness of the core pipe is set to zero.


254

Examples

WRC 107 The following example illustrates a comprehensive local stress analysis of a vessel/nozzle using WRC 107 and ASME Section VIII, Division 2 criteria.

To determine whether the WRC 107 Bu lletin is appropriate for the computation of the local stress state in the vessel due to external loading, review the geometry guidelines:

D = 120. 0 i n. , T = 0. 625 i n. , d = 12. 75 i n. , t = 0. 375 i n. d / D = 0. 10625 < 0. 33 Dm/ T = ( D- T) / T = 191 > 50


255

Examples In the example, both conditions are satisfied. The actual preparation of the WRC 107 ca lculation input can now begin. One of the most important steps in the WRC 107 procedure is to identify the correlation between the CAESAR II global coordinates and the WRC 107 local axes. CAESAR II performs this conversion automatically. You m ust, however, identify the vectors defining the vessel as well as the nozzle centerline. The following figure illustrates the definition of the direction vectors of the vessel and the nozzle.

Convert Forc es/Moments f rom CAESAR II Glob al Coordi nates to WRC 107 Loc al Axes In order to define a vessel direction vector, you first need to designate the output data points (A-D), as defined by the WRC 107 Bulletin. The line between data points B and A defines the vessel centerline (except for nozzles on heads, where the vessel centerline has to be defined along a direction that is perpendicular to that of the nozzle). Because, in the vessel/nozzle configuration shown, point A is assigned to the bottom of the nozzle, the vessel direction vector can be written as (0.0, -1.0, 0.0), while the nozzle direction vector is (1.0, 0.0, 0 .0). The nozzle direction vector is alwa ys defined as the vector pointing from the vessel nozzle connection to the centerline of the vessel. In the previous figure, there are two nodes occupying the same space at the nozzle/vessel surface junction: nodes 55 and 56. You can use an anchor at node 55 with a connecting node at 56 to model the local vessel flexibility as rigid. If you are not familiar with this m odeling approach, see "Piping Input" in the CAESAR II User's Guide for more details.


256

Examples You can then replace the anchor with a WRC 297 local vessel flexibility model, and re-run the job to see the range of loads and displacements that exist in the system around the vessel nozzle. In either case, the restraint loads forces and moments can be obtained from the CAESAR II Restraint report. These loads reflect the action of the piping on the vessel. The following figure displays the restraint report of the rigid anchor model.

Figure 22: CAESAR II RESTRAINT REPORT: EX107 WRC 107 EXAMPLE PROBLEM

The total sustained axial load on the nozzle may not be reflected in the restraint report. A pressure thrust load contributes an additional axial load to the nozzle. The pressure thrust force always tends to push the n ozzle away from the vessel. For example, with a pressure of 275 psi over the inside area of the 12-inch pipe, the total P load becomes:

P = - 26 - P( A) = - 26 - 275p ( 122) / 4 = - 31, 128


257

Examples If needed, the P load can be adjusted automatically for the input using the WRC 107/297 analysis module. To begin the analysis, click Analy si s > WRC 107/297 on the CAESAR II main menu. The software first prompts you to create a new WRC job and then prompts you to enter geometric data that describes the nozzle ( Nozzle Data tab) and the vessel ( Vessel Data tab), followed by loadings data ( Loads Data tab). To do a WRC 297 calculation, click dialog box.


, and a new tab appears on the left side of the

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Examples You can enter up to three sets of loadings representing Sustained (SUS), Expansion ( EXP), and Occasional (OCC) load cases. CAESAR II automatically performs the stress calculation of each of the load cases consecutively and then performs the WRC 107 stress summation routine to compare the computed stress intensities against the stress allowables as required in Appendix 4 of ASME Section VIII, Division 2. The focus in the current example is only on the sustained and thermal expansion cases. The loads are shown in the figure below. You can elect to leave any boxes blank if they are not applicable. If a static analysis has been performed on the system you are analyzing with WRC-107, CAESAR II can import the loads directly from the output file. To do this, click Import Loads from Job for each load case. CAESAR II reads in the loads for the nozzle node number that was specified on the Nozzle Data tab.


259

Examples To run the analysis, click Local Stress Analysis on the WRC 107/297 toolbar. The software opens an output dialog box and displays the processing results. You can also click View Report Using Microso ft Word on the WRC 107/297 toolbar to perform the initial WRC 107 calculation and summation and send the results directly to Microsoft™ Word . After the input echo, the parameters extracted from the WRC 107 figures are printed to this report. This step is similar to collecting the data by hand. These non-dimensional values are combined with the nozzle loads to calculate the two normal and on e shear stress. The stresses are reported on the outer and inner vessel surfaces of the four points A, B, C and D located around the nozzle. CAESAR II provides the normal and shear stresses and translates them into stress intensities, which can be used f or comparisons against material allowables. The outputs of the stress computations are shown in the following examples. As the output shows, the largest expansion stress intensity (117485 psi) occurs at the outer surface of point B (Bu).


260

Examples


261

Examples WRC 107 Stress Calculatio n for SUStained Loads


262

Examples WRC 107 Stress Calculatio n for EXPansion s Loads


263

Examples WRC 107 Stress Summation s: Vessel Stress Summati on at Nozzle Junction

Failed items display in red. Because the present nozzle loading causes stress intensities that are not acceptable to the ASME Section VIII, Division 2 criteria, the nozzle loading must be corrected. One option is to adjust the nozzle loading from its source; another option is to reinforce the nozzle connection on the vessel side either by increasing the vessel thickness or by adding a reinforcing pad. You can repeat the same analysis procedure until the fina l results are acceptable. After a reinforcing pad is selected, the CAESAR II automatically computes the stress at the edge of the pad as well.


264

Examples

NEMA SM23 This example illustrates the use of the NEMA SM-23 computations of the CAESA R II Equipment module. Two examples are given. The first example can be found in the NEMA SM-23 Standard, 7th edition Example 8A, beginning on page 47. The second example illustrates the use of metric units and the correct implementation of paragraph 8.4.6.2. Click An aly si s > NEM A SM23 on the CAESAR II main menu to enter a NEMA SM23 problem. The software displays the NEMA SM23 dialog box.

NEMA Exampl e PT69M The input data is shown in the following figures and illustrates the computations for Dc and De, the use of metric units, and the correct computation of the total moment loads resolved about the discharge nozzle.


265

Examples The output from this analysis is shown below:

Nozzle Results f or PT69M The first item of interest in this ou tput report is the variation in the units systems used. The input values are reflected in the user-defined set of units, which in this example is millimeters, newtons, and newton-meters. The computed values are reported in inches, pounds, and foot-pounds. This is necessary because of the equations used to determine code compliance. These equations combine forces and moments and then compare the sum to a dim ension. In essence, pounds plus foot-pounds must be less than inches. The results can be i nterpreted correctly only if presented in English units. For the exhaust nozzle, the input value of 254 millimeters converts to a 10-inch nominal pipe. Because this is larger than 8 inches, De is equal to (16 + 10) divided by 3, or 8.667 inches. This yields an allowable of 500 x 8.667, or 4333. The square root of the sum of the squares of the forces acting on the exhaust nozzle yields 7922 newtons, which converts to 1781 pounds. Similarly, the square root of the sum of the squares of the moments acting on the exhaust nozzle yields 3000 newton-meters, which converts to 2213 foot-pounds. Applying the 3F + M equation yields 7556. Because 7566 is larger than 4333, this nozzle fails the requirements of the SM-23 Standard. The same computations must also be performed on the inlet nozzle. The output displa yed above shows that this nozzle also fails the SM-23 Standard requirements. Also shown for the in let nozzle are the moments about the discharge nozzle caused by the inlet no zzle forces. Applying


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Examples the standard right-hand rule sign convention, a positive Y-force offset a positive Z-distance causes a negative X moment. Similarly, a positive Z-force offset a positive Y-distance causes a positive X moment. Therefore, the inlet nozzle forces cause a n MX-moment about the exhaust nozzle of -(3296*.6) + (3999*0), which yields -1978 newton-meters. The MY- and MZ-moments caused by the suction nozzle forces about the exhaust nozzle can be computed in a similar fashion. These moments are needed to correctly comply with Section 8.4.6.2. The previous report is repeated for each extraction no zzle specified. This particular example does not contain extraction nozzles, so these reports are not produced. Following the individual nozzle reports is the summation of forces and m oments about the exhaust nozzle. This report is shown in the following figure:

Nozzle Load Summ ation Report This report shows the force summations in the three global directions as well as the resultant force, computed by the SRSS method discussed in Nozzle Results for PT69M (on page 266). These forces are shown in the user-defined set of units on the l eft side of the report, followed b y the forces in pounds. The next column shows the allowable for each force, as a function of Dc, which is defined above. Following the force summation is the m oment summation. This summary reports the total moment about the three global directions and the r esultant moment, computed by the SRSS method. The total moment is the sum of the individual moments plus the contribution from the forces multiplied by their distances from the discharge nozzle. Consider, for example, the MX moment of 721 newton-meters. This value is obtained using the following calculation: 1200 + 1499 + -1978 The final line of this report combines the resultant force and resultant moment and compares the result to its allowable.


267

Index 1 180º Return Fitting-to-Fitting 90º Bends • 11

A Anchors • 20 Anchors with Displacements • 21

B Ball Joints • 104 Bend Definition • 9 Bend Flexibility Factor • 18 Bends • 9 Bi-Linear Restraints • 54

C Check for errors • 144 Check the static load cases • 144 Checking Nozzle Loads • 174 Closely-Spaced Mitered Bend • 13 Cold Spring • 108 Combination Cases • 211 Conclusion • 184 Conclusions • 156 Configure the software • 121 Connecting Equipment • 108 Constant Effort Support Design • 65 Constant Effort Supports - No Design • 66 Convert Forces/Moments from CAESAR II Global Coordinates to WRC 107 Local Axes • 255 Core Piping Input • 248 Creating Accurate Models • 165

D Define the 6-inch bypass • 136 Define the first pipe element • 122 Define the horizontal pipe run • 131 Define the vertical pipe run • 127 Displacement Report • 210 Double-Acting Restraints • 30 Dual Gimbal • 95 Dynamic Analysis (NUREG9) • 231 Dynamic Analysis of Independent Support Earthquake Excitation (CRYNOS) • 212


Dynamic Analysis of Water Hammer Loads (HAMMER) • 200

E Elbows - Different Wall Thickness • 17 Evaluating Pump Discharge Loads • 157 Examples • 186 Existing Springs - No Design • 67 Expansion Joints • 78

F Flexible Anchors • 22 Flexible Anchors with Predefined Displacements • 23 Flexible Nozzle - WRC Bulletin 297 • 24 Flexible Nozzle with Complete Vessel Model • 27 Flexible Nozzle with Predefined Displacements • 26

G Gimbal Joints • 92 Guides • 32

H Hanger Between Two Pipes • 71 Hanger Design Basics • 61 Hanger Design with Anchors in the Vicinity • 72 Hanger Design with Support Thermal Movement • 70 Hanger Design with User-Specified Operating Load • 74 Hangers • 60 Harmonic Analysis (TABLE) • 186 Harmonic Analysis of this System • 189 Hinged Joint • 86 Horizontal Dummy Leg on Bends • 50 Horizontal Vessels • 114

J Jacket Input (1st Half) • 248 Jacket Input (2nd Half) • 252 Jacketed Pipe • 106 Jacketed Piping (JACKET) • 244

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Index

L Large Rotation Rods - Basic Model • 51 Large Rotation Rods - Chain Supports • 53 Lift Off Spring Can • 76 Limit Stops • 34

M Mass Participation Report • 209 Miscellaneous Models • 103 Mitered Bends • 12 Modeling Plan • 245 Modeling Spring Cans with Friction • 77 Multiple Can Design • 68

N N1 Nozzle • 112 N2 Nozzle • 113 NEMA SM23 • 264 Node Layout • 246 Notes for Analyzing Water Hammer Loads • 206 Nozzle Load Summation Report • 266 Nozzle Results for PT69M • 265 NRC Example NUREG 9 • 232 NRC Example Problem 2A • 232 NRC Example Problem 2B • 235 NRC Example Problem 2C • 237

O Old Spring Redesign • 69 Omega Loop Modelling (OMEGA) • 239 Overview • 8

P Pipe and Hanger Supported From Vessel • 70 Plastic Hinges • 56 Pressure-Balanced Tees and Elbows • 100

R Reducers • 103 Relief Valve Example Problem Setup • 192 Relief Valve Loading Output • 196 Relief Valve Loads (RELIEF) • 191 Restraint between Two Pipes Using CNodes • 43 Restraint between Vessel and Pipe Models • 44 Restraint Settlement • 40


Restraint/Force/Stress Reports • 210 Restraints • 19 Restraints on a Bend at 30º and 60º • 46 Restraints on a Bend at 45º • 45 Results • 223 Review the analysis results • 146 Review the piping input • 142 Review the piping system drawing • 120 Rigid Elements • 111 Rotational Directional Restraints with Gaps • 37 Run the static analysis • 146

S Simple Bellows with Pressure Thrust • 97 Simple Bottomed Out Spring • 75 Simple Hanger Design • 63 Single and Double Flanged Bends or Stiffened Bends • 10 Single Can Design • 64 Single-Directional Restraint and Guide with Gap and Predefined Displacement • 39 Single-Directional Restraint with Predefined Displacement • 38 Single-Directional Restraints • 31 Skewed Double-Acting Restraint with Gap • 41 Skewed Single-Directional Restraint • 42 Slip Joint • 91 Slotted Hinge Joint - Comprehensive Model • 89 Slotted Hinge Joint - Simple Model • 88 Solution Summary • 211 Static Analysis Output Listing • 150 Static Snubbers • 56 Structural Analysis (FRAME) • 224 Sway Brace Assemblies • 57 System Redesign • 176

T Tied Bellows - Simple vs. Complex Model • 90 Tied Bellows Expansion Joint - Complex Model • 93 Tied Bellows Expansion Joint - Simple Model • 82 Tutorial A • 118 Tutorial B • 157

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Caesar II Application Guide

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