COADE Pipe Stress Analysis Seminar Notes

ENGINEERING

SOFTWARE

PIPE STRESS ANALYSIS SEMINAR NOTES

Notice: Unless otherwise noted herein, the information contained in these course notes is proprietary and may not be translated or duplicated in whole or in part without the expressed written consent of COADE Engineering Software, 12777 Jones Rd., Suite 480, Houston, Texas 77070. Copyright {c} 1985 - 1998 COADE, Inc.

1

COADE Pipe Stress Analysis Seminar Notes Section 1 Table of Contents

1.0 Introduction to Pipe Stress Analysis ........................................................................ 1 1.1 Theory and Development of Pipe Stress Requirements ........................................... 8 1.1.1 Basic Stress Concepts ............................................................................... 8-14 1.1.2 3-D State of Stress in the Pipe Wall ....................................................... 14-15 1.1.3 Failure Theories ........................................................................................... 16 1.1.4 Maximum Stress Intensity Criterion ..................................................... 18-19 1.2 Fatigue Failure ....................................................................................................... 20 1.2.1 Fatigue Basics .............................................................................................. 20 1.2.2 Fatigue Curves ............................................................................................. 22 1.2.3 Effect of Fatigue on Piping ..................................................................... 24-25 1.2.4 Cyclic Reduction Factor ............................................................................... 25 1.2.5 Effect of Sustained Loads on Fatigue Strength .......................................... 26 1.3 Stress Intensification Factors ............................................................................ 28-33 1.4 Welding Research Council Bulletin 330 ................................................................. 34 1.5 Code Compliance ..................................................................................................... 43 1.5.1 Primary vs. Secondary Loads ................................................................. 43-45 1.5.2 Code Stress Equations ............................................................................ 45-46 1.5.3 B31.1 Power Piping ..................................................................................... 46 1.5.4 B31.3 Chemical Plant and Petroleum Refmery Piping .............................. 47 1.5.5 ASME Section III, Subsections NC & ND (Nuclear Class 2 & 3) .......... 49-50 1.5.6 B31.4 Fuel Gas Piping ................................................................................. 51 1.5.7 B31.8 Gas Transmission and Distribution Piping Code ............................. 52 1.5.8 Canadian Z183/Z184 Oil/Gas Pipeline Systems ......................................... 54 1.5.9 RCC-M C ...................................................................................................... 55 1.5.10 Stoomwezen ................................................................................................. 56 1.5.11 Special Considerations of Code Compliance ........................................... 56-59 1.5.12 Evaluation of Multiple Expansion Range Cases ......................................... 59

COADE Pipe Stress Analysis Seminar Notes

1.0 Introduction to Pipe Stress Analysis In order to properly design a piping system, the engineer must understand both a system's behavior under potentialloadings, as weIl as the regulatory requirements imposed upon it by the governing codes. A system's behavior can be quantified through the aggregate values of numerous physical parameters, such as accelerations, velocities, displacements, internaI forces and moments, stresses, and external reactions developed under applied loads. Allowable values for each of the se parameters are set after review of the appropriate failure criteria for the system. System response and failure criteria are dependent on the type of loadings, which can be classified by various distinctions, such as primary vs. secondary, sustained vs. occasional, or static vs. dynamic. The ASME/ANSI B31 piping codes are the result of approximately 8 decades ofwork by the American Society ofMechanical Engineers and the American National Standards Institute (formerly American Standards Association) aimed at the codification ofdesign and engineering standards for piping systems. The B31 pressure piping codes (and their successors, such as the ASME Boiler and Pressure Vessel Section III nuclearpiping codes) prescribe minimum design, materials, fabrication, assembly, erection, test, and inspection requirements for piping systems intended for use in power, petrochemical/refinery, fuel gas, gas transmission, and nuclear applications. Due to the extensive calculations required during the analysis of a piping system, this field of engineering provides a natural application for computerized calculations, especially during the last two to three decades. The proliferation of easy-to-use pipe stress software has had a two-fold effect: first, it has taken pipe stress analysis out ofthe hands ofthe highlypaid specialists and made it accessible to the engineering generalist, but likewise it has made everyone, even those with inadequate piping backgrounds, capable of turning out officiallooking results. The intention ofthis course is to provide the appropriate background for engineers entering the world of pipe stress analysis. The course concentrates on the design requirements (particularly from a stress analysis point ofview) of the codes, as weIl as the techniques to be applied in order to satisfy those requirements. Although the course is taught using the CAESAR II Pipe Stress Analysis Software, the skills learned here are directly applicable to any means of pipe stress analysis, whether the engineer uses a competing software program or even manual calculational methods.

Why do we Perform Pipe Stress Analysis? There are a number ofreasons for performing stress analysis on a piping system. A few of these foIlow: In order to keep stresses in the pipe and fittings within code allowable levels.

1 2

-

In order to keep nozzle loadings on attached equipment within allowables of manufacturers or recognized standards (NEMA SM23, API 610, API 617, etc.).

1-1


3 4

In order to keep vessel stresses at piping connections within ASME Section VIII allowable levels. -

5

In order to calculate design loads for sizing supports and restraints. In order to determine piping displacements for interference checks.

6

-

In order to solve dynamic problems in piping, such as those due to mechanical vibration, acoustic vibration, fluid hammer, pulsation, transient flow, and relief valve discharge.

7

-

In order to help optimize piping design.

Typical Pipe Stress Documentation

Documentation typically associated with stress analysis problems consists of the stress isometric, the stress analysis input echo, and the stress analysis results output. Examples ofthese documents are shown in Figures 1-1 through 1-5 on subsequent pages. The stress isometric (Figure 1-1) is a sketch, drawn in an isometric coordinate system, which gives the viewer a rough 3-D idea of the piping system. The stress isometric often summarizes the piping design data, as gathered from other documents, such as the line list, piping specification, piping drawing, Appendix A (Figure 1-2) of the applicable piping code, etc. Design data typically required in order to do pipe stress analysis consists of pipe materials and sizes; operating parameters, such as temperature, pressure, and fluid contents; code stress allowables; and loading parameters, such as insulation weight, external equipment movements, and wind and earthquake criteria. Points of interest on the stress isometric are identified by node points. Node points are required at any location where it is necessary to provide information to, or obtain information from, the pipe stress software. Typically, node points are located as required in order to: define geometry (system start, end, direction changes, intersection, etc.)

1 2

-

define element stiffness parameters (changes in pipe cross section or material, rigid elements, or expansion joints)

3 4

note changes in operating conditions (system start, isolation or pressure reduction valves, etc.)

-

5

designate boundary conditions (restraints and imposed displacements) specify mass points (for refinement of dynamic model)

6

-

note loading conditions (insulation weight, imposed forces, response spectra, earthquake g-factors, wind exposure, snow, etc.)

7

-

retrieve information from the stress analysis (stresses at piping mid spans, displacements at wall penetrations, etc.)

1-2

COADE Pipe Stress Analysis Seminar Notes The input echo (Figure 1-3) provides more detailed information on the system, and is meant to be used by the engineer in conjunction with the stress isometric. The analysis output provides results, such as displacements, internal forces and moments, stresses, and restraint loadings at each node point of the pipe, acting under the specified loading conditions. CAESAR II provides results in either graphic or text format; Figures 1-4 and 1-5 present stress and dis placement results graphically. The output also provides a code check calculation for the appropriate piping code, from which the analyst can determine which locations are over stressed.

SSEMl tUI

tower-:'~[

Haterial A186 Gr.B SH @ 788 deg. = 16.588 psi SC @ 78 deg. = 28.888 psi t = 788 deg. F. Flue Gas P = 125 psi Dia = 28" Std.Wall Insul = 2" Calciul!I Silicate

,~~.y..

SUpport ......

rD_

~3S ~..~145 j;

COl!lputed therl!lal expansion of the vessel is 17.268E-6 in/in/deg.F. at a telllp of 828 deg.F. Node 188 is 28.88 ft. above vessel skirt 'i

A

Z

Disp. @ 188 = (828-7B)deg.F(17.268E-6)in/in/deg* (28.88)(12)ft.in/rt. = 3.121 in. X

Disp. D 128 = (B28-78) (17.268E-6)(28.88+6.5-15)(12) = 1.8 in.

Figure 1-1

1-3

Exchanger 0

0

0 ANSI/ASME 831.3-1984 ROmON TABLEA·I

ASME CODJ! FOI. PIUlSSUIUI PlPINO CHEMICAL PLANT AND PBTROLBUM Rl!FlNBIlY 'IPINO

ASMJ! COD! fOI. 'IU!SSUIUI'IPINO CIIEMICAL PLANT AND PBfR.OLEUM lEFINI!IlY.IPINO

TABLE A-1 (CONT'OI BASIC ALLOWABLE STRESSES IN TENSION FOR METALS lU Nurilers ln '-thKes Rtftr to Notes ,., ~ A TMIes; Specifications AIt ASTM Uilleu DIIIIIwIIt lIIdItaIId

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COADE Pipe Stress Analysis Seminar Notes CAESAR II VERS 3.18 JOBNAME:SSEM1 PIPE DATA

DEC 10, 1992

3:05 am

Page 1

From 100 To 105 DY= 3.500 ft. PIPE Dia= 20.000 in. Wall= .375 in. Insul= 2.000 in. GENERAL T1= 700 F P1= 125.0000 lb./sq.in. Mat= (l)LOW CARBON STEEL E= 27,900,000 lb./sq.in. v = .292 Density= .2899 lb./eu.in. RIGID Weight= 3,290.00 lb. DISPLACEMENTS Node 100 DX= .000 in. DY= 3.121 in. DZ= .000 in. RX= .000 RY= .000 RZ= .000 ALLOWABLE STRESSES B31.3 (1990) Se= 20,000 lb./sq.in. Sh1= 16,500 lb./sq.in. From 105 To 110 DY= 3.000 ft. BEND at "TO" end Radius= 30.000 in. (LONG) Bend Angle= 90.000 Angle/Node @2= .00 108 From 110 To 115 DX= 12.000 ft. BEND at "TO" end Bend Angle= 90.000 Radius= 30.000 in. (LONG) Angle/Node @2= .00 113 From 115 To DISPLACEMENTS Node 120 RZ= FREE

120

Angle/Node @1= 45.00 114

DY= -15.000 ft.

DX= FREE

DY= 1.800 in.

DZ= FREE

From 120 To 125 DY= -3.000 ft. BEND at "TO" end Bend Angle= 90.000 Radius= 30.000 in. (LONG) Angle/Node @2= .00 123 From 125 To 130 RESTRAINTS Node 130 +Y

DX= 35.000 ft.

From 130 To 135 RESTRAINTS Node 135 +Y

DX= 35.000 ft.

From 135 To 140 RESTRAINTS Node 140 +Y

DX= 35.000 ft.

From 140 To 145 DX= 20.000 ft. BEND at "TO" end Bend Angle= 90.000 Radius= 30.000 in. (LONG) Angle/Node @2= .00 143 From 145 To 150 RESTRAINTS Node 150 ANC

Angle/Node @1= 45.00 109

DY= -12.000 ft.

Figure 1-3

1-5

RX= FREE

RY= FREE

Angle/Node @1= 45.00 124

Angle/Node @1= 45.00 144


CASE 3 (EXP)D3=D1-D2

FILE:SSEI11

DEC 4.1992 12:4?am QUIT

nODES OURSTR I1AXSTR BHDlltG

TORS AXIAL STRESS S'inBOL BI'IDUIG

TORS AXIAL STRESS COLOR BI'IDItIG

TORS AXIAL STRESS

"~'" 1'I0DE= 123

OUERSTRESSED l'IODES

~"

~,

Figure 1-4 CASE 1 (OPEJW+DIS+T1+P1

F1LE:SSEnl

DEC 4.1992 12:49am

RESET QUIT

l'IODES DEFU SPECFY I1AGnIF GROW COLORS ORIGI'IL BLArtK

Iml:'I}" HRDCP'i

ItODE= 125

I1AX. DISPS.

X

Figure 1-5

1-6


What are these Stresses? The stresses calculated are not necessarily real stresses (such as could be measured by a strain gauge, for example), but are rather "code" stresses. Code stress calculations are based upon specific equations, which are the result of8 decades of compromise and simplification. The calculations reflect: 1

Inclusion or exclusion ofpiping loads, based upon convenience of calculation or selected failure. In fact the result may not even represent an absolute stress value, but rather a RANGE of values.

2

Loading type - these are segregated, and analyzed separately, as though they occur in isolation, even though they actually are present simultaneously.

3

-

Magnification, due to local fitting configuration, which may in reality reflect a decrease in fatigue strength, rather than an increase in actual stress.

4

-

Code committee tradition - every code is a result of a different set of concerns and compromises, and therefore may appear to be on a different branch of the evolutionary ladder. Because of this, every code gives different results when calculating stresses.

A summary of significant dates in the history of the development of the piping codes is presented below: 1915

Power Piping Society provides the first national code for pressure piping.

1926

The American Standards Association initiates project B31 to govern pressure piping.

1955

Markl publishes his paper ''Piping Flexibility Analysis", introducing piping analysis methods based on the "stress range".

1957

First computerized analysis ofpiping systems.

1968

Congress enacts the Natural Pipeline Safety Act, establishing CFR 192, which will in time replace B31.8 for gas pipeline transportation.

1969

Introduction of ANSI B31.7 code for Nuclear power plant piping.

1971

Introduction of ASME Section III for Nuclear power plant piping.

1974

Winter Addenda B31.1 moves away from the separation ofbending and torsional moment terms in the stress calculations and alters the intensification factor for moments on the branch leg of intersections.

1978

ANSI B31.7 is withdrawn.

1987

Welding Research Council Bulletin 330 recommends changes to the B31.1, B31.3, and ASME III Class 2 and 3 piping codes.

1-7


1.1 Theory and Development of Pipe Stress Requirements 1.1.1 Basic Stress Concepts

Normal stresses: Normal stresses are those acting in a direction normal to the face of the crystal structure ofthe material, and may he either tensile or compressive in nature. (In fact, normal stresses in piping tend more to tension due the predominant nature of internal pressure as a load case.) Normal stresses may be applied in more than one direction, and may develop from a numher of different types of loads. For a piping system, these are discussed below:

Longitudinal stress: Longitudinal, or axial, stress is the normal stress acting parallel to the longitudinal axis ofthe pipe. This may he caused by an internal force acting axially within the pipe:

- - -.....-

Figure 1-6

SL =

Fax/ Am

Where: SL =

longitudinal stress, psi

Fax =

internaI axial force acting on cross-section, lb

Am =

metal cross-sectional area of pipe, in2

=

1t(do2 - di 2 ) / 4

=

1t dm t

do

=

outer diameter, in

di

=

inner diameter, in

dm =

mean diameter, = (do + di) / 2

1-8

FAX


A specifie instance of longitudinal stress is that due to internaI pressure:

Figure 1-7

=

PAil Am

P

=

design pressure, psig

Ai

=

internaI area of pipe, in2

=

1t

SL Where:

di 2 1 4

Replacing the terms for the internaI and metal areas of the pipe, the previous equation may be written as:

For convenience, the longitudinal pressure stress is often conservatively approximated as: SL

=

P do 1 4 t

Another component of axial normal stress is bending stress. Bending stress is zero at the neutral axis of the pipe and varies linearly across the cross-section from the maximum compressive outer fiberto the maximum tensile outer fiber. Calculatingthe stress as linearly proportion al to the distance from the neutral axis: Variation in Bending Stress Thru Cross Section Max compressive stress

M

Neutral Axis

1/2 max compressive stress Zero bending stress 1/2 max tension stress Max tension stress

Figure 1-8

1-9


Where: Mb = bending moment acting on cross-section, in-lb c

= distance ofpoint ofinterest from neutral axis of cross-section, in

l

=

moment ofinertia of cross-section, in 4

Maximum bending stress occurs where c is greatest - where it is equal to the outer radius: Smax

=

Where: Ro

=

Z

= section modulus of pipe, in3

outer radius of pipe, in

= 1/Ro Summing aH components oflongitudinal normal stress: SL

=

Fax / Am + P do / 4 t + Mb / Z

Hoop stress: There are other normal stresses present in the pipe, applied in directions orthogonal to the axial direction. One ofthese stresses, caused by internaI pressure, is called hoop stress. This stress acts in a direction parallel to the pipe circumference.

Figure 1- 9 The magnitude of the hoop stress varies through the pipe wall and can be calculated by Lame's equation as: SR = P (ri 2 + ri 2 ro2 / r 2) / (ro2 - q2)

1-10

COADE Pipe Stress Analysis Seminar Notes Where: SR

=

hoop stress due to pressure, psi

ri

=

inner radius of pipe, in

ra

=


r

=

radial position where stress is being considere d, in

The hoop stress can he conservatively approximated for thin-wall cylinders, by assuming that the pressure force, applied over an arbitrary length of pipe, l CF = P di 1), is resisted uniformly by the pipe wall over that same arbitrary length (Am = 2 t 1), or: SH =

SR

=

P di 1/ 2 t 1, or:

P di / 2 t, or conservatively:

Radial stress: Radial stress is the third normal stress present in the pipe wall. It acts in the third orthogonal direction, parallel to the pipe radius. Radial stress, which is caused by internal pressure, varies between a stress equal to the internal pressure at the pipe's inner surface and a stress equal to atmospheric pressure at the pipe's external surface. Assuming that there is no external pressure, radial stress may be calculated as:

= -p

Figure 1-10

Where: SR = radial stress due to pressure, psi Note that radial stress is zero at the outer radius of the pipe, where the bending stresses are maximized. For this reason, this stress componenthas traditionally been ignored during the stress calculations.

Shear stresses: Shear stresses are applied in a direction parallel to the face of the plane of the crystal structure of the material, and tend to cause adjacent planes of the crystal to

1-11


slip against each other. Shear stresses may be caused by more than one type of applied load. For example, shear stress may be caused by shear forces acting on the cross-section: Shear Distribution Profile

-----~)

----------~= j

---------- ~

V

~IN=O

/MAX

Figure 1-11

=

VQ/Am

'tmax

=

maximum shear stress, psi

v

=

shear force, lb

Q

=

shear form factor, dimensionless (1.333 for solid circular section)

'tmax

Where:

These shear stresses are distributed such that they are maximum at the neutral axis ofthe pipe and zero at the maximum distance from the neutral axis. Since this is the opposite of the case with bending stresses, and since these stresses are usually small, shear stresses due to forces are traditionally neglected during pipe stress analysis. Shear stresses may also be caused by torsionalloads:

T Figure 1·12 'tmax

=

1-12

COADE Pipe Stress Analysis Seminar Notes Where: MT

=

internaI torsion al moment acting on cross-section, in-lb

c

=

distance of point ofinterest from torsional center (intersection ofneutral axes) of cross-section, in

R

=

torsional resistance of cross-section, in4

=

2I

Maximum torsional stress occurs where c is maximized - at the outer radius:

=

'tmax

Summing the individual components of the shear stress, the maximum shear stress acting on the pipe cross-section is:

=

'tmax

v Q / Am + MT / 2 Z

Example Stress Calculations: As noted above, a number of the stress components described above have been neglected for convenience during calculation ofpipe stresses. Most V.S. piping codes require stresses to be calculated using some form of the following equations:

Longitudinal stress: SL

=

Shear stress:

=

Hoop stress:

=

Mb / Z + Fax / Am + P do / 4 t

Calculations are illustrated for a 6-inch nominal diameter, standard wall pipe (assuming the piping loads are known):

Cross sectional properties:

Piping loads:

da

=

6.625 in

Bending moment (Mb)

=

4247 ft-lb

di

=

6.065 in

Axial force (Fax)

=

33488 lb

t

=

0.280 in

Pressure (P)

=

600 psi

Z

=

8.496 in3

Torsional Moment (MT)

=

8495 ft-lb

Am

=

5.5813 in2

1-13


Longitudinal stress: SL

=

4247 x 12/8.496 + 33488/5.5813 + 600 x 6.625/4 (0.280)

=

15547 psi

Shear stress:

'[

=

8495 x 12/2 (8.496)

= 5999 psi

Hoop stress: SR =

600 x 6.625/2 (0.280) = 7098 psi

1.1.2 3-D State of Stress in the Pipe Wall During operation, pipes are subject to aIl ofthese types of stresses. Examining a small cube ofmetal from the most highly stressed point of the pipe wall, the stresses are distributed as so:

S4 SR

1

: '

SH

....

S

{SH SL

R

Figure 1-13 There are an infinite number oforientations in which this cube could have been selected, each with a different combination of normal and shear stresses on the faces. For example, there is one orientation of the orthogonal stress axes for which one normal stress is maximized, and another for which one normal stress is minimized - in both cases all shear stress components are zero. In orientations in which the shear stress is zero, the resulting normal components of the stress are termed the principal stresses. For 3-dimensional analyses, there are three of them, and they are designated as SI (the maximum), S2, and S3 (the minimum). Note that regardless of the orientation of the stress axes, the sum of the orthogonal stress components is always equal, i.e: SL + SR + SR = SI + S2 + S3 The converse ofthese orientations is that in which the shear stress component is maximized (there is also an orientation in which the shear stress is minimized, but this is ignored since the magnitudes of the minimum and maximum shear stresses are the same); this is appropriately called the orientation of maximum shear stress. The maximum shear stress

1-14


in a three dimensional state of stress is equal to one-halfofthe difference between the largest and smallest of the principle stresses (SI and S3). The values of the principal and maximum shear stress can be determined through the use of a Mohr's circle. The Mohr's circle analysis can be simplified by neglecting the radial stress component, therefore considering a less complex (i.e., 2-dimensional) state of stress. A Mohr's circle can be developed by plotting the normal vs. shear stresses for the two known orientations (i.e., the longitudinal stress vs. the shear and the hoop stress vs. the shear), and constructing a circle through the two points. The infinite combinations of normal and shear stresses around the circle represent the stress combinations present in the infinite number of possible orientations of the local stress axes. A differential element at the outer radius of the pipe (where the bending and torsional stresses are maximized and the radial normal and force-induced shear stresses are usually zero) is subject to 2-dimensional plane stress, and thus the principal stress terms can be computed from the following Mohr's circle: TMAX T

S2

'"

/

S,

S

-T TMAX

T Figure 1-14

The center ofthe circle is at (SL + SR) / 2 and the radius is equal to [[(SL - SR) / 2]2 + 't2 ]1/2. Therefore, the principal stresses, SI and S2, are equal to the centerofthe circle, plus or minus the radius, respectively. The principal stresses are calculated as: SI

=

(SL + SR) /2 + [ [(SL - SR) / 2]2 + 't2 ]1/2 and

S2

=

(SL + SR) / 2 - [ [(SL - SR) / 2]2 + 't2 ]1/2

As noted above, the maximum shear stress present in any orientation is equal to (SI - S2) / 2, or: 'tmax

=

[(SL - SR)2 + 4 't2 ]1/2

2

1-15


1.1.3 Failure Theories To be useful, calculated stresses must he compared to material allowables. Material allowable stresses are related to strengths as determined by material uniaxial tensile tests, therefore calculated stresses must also be related to the uniaxial tensile test. This relationship can he developed by looking at available failure theories.

crYield

Strain Tensile Test Results Unixial Tensile Test Machine

Tensile Test Specimen

Figure 1-15 There are three generally accepted failure theories which may he used to predict the onset of yielding in a material: 1 - OCTAHEDRAL SHEAR, or VON MISES THEORY 2 - MAXIMUM SHEAR, or TRESCA THEORY 3 - MAXIMUM STRESS or RANKINE THEORY These theories relate failure in an arbitrary three dimensional stress state in a material to failure in a the stress state found in a uniaxial tensile test specimen, since it is that test that is most commonly used to determine the allowable strength of commonly used materials. Failure of a uniaxial tensile test specimen is deemed to occur when plastic deformation occurs; i.e., when the specimen yields.

1-16

COADE Pipe 8tress Analysis 8eminar Notes

The three failure theories state: Octahedral 8hear - Von Mises Theory: "Failure occurs when the octahedral shear stress in a body is equal to the octahedral shear stress at yield in a uniaxial tension test." The octahedral shear stress is calculated as:

= 1/3 [ (SI - 82)2 + (82 - 83)2 + (83 - 8 1)2 ]112

'tact

In a uniaxial tensile test specimen at the point ofyield: 81

=

8Yield; 82 = 83 = 0

Therefore the octahedral shear stress in a uniaxial tensile test specimen at failure is calculated as: 'tact

=

1/3 [ (8Yield - 0)2 + (0 - 0)2 + (0 - 8Yield)2 ]1/2

=

2 112

X

8Yield / 3

Therefore, under the Von Mises theory:

Plastic deformation occurs in a 3-dimensional stress state whenever the octahedral shear exceeds 2 1/2 x 8Yield / 3. Maximum 8hear 8tress - Tresca Theory: "Failure occurs when the maximum shear stress in a body is equal to the maximum shear stress at yield in a uniaxial tension test." The maximum shear stress is calculated as: 'tmax

=

In a uniaxial tensile test specimen at the point ofyield: 81 =

8Yield; 82 = 83 = 0

'tmax

=

80:

(SYield - 0) / 2

= 8Yield / 2

Therefore, under the Tresca theory:

Plastic deformation occurs in a 3-dimensional stress state whenever the maximum shear stress exceeds 8Yield / 2.

1-17


Maximum Stress - Rankine Theory "Failure occurs when the maximum tensile stress in a body is equal to the maximum tensile stress at yield in a uniaxial tension test." The maximum tensile stress is the largest, positive principal stress, SI. (By definition, SI is always the largest of the principal stresses.) In a uniaxial tensile test specimen at the point of yield: SI

=

SYield; S2 = S3 = 0

Therefore, under the Rankine theory:

Plastic deformation occurs in a 3-dimensional stress state whenever the maximum principal stress exceeds SYield.

1.1.4 Maximum Stress Intensity Criterion Mostofthe CUITent piping codes use a slight modification ofthe maximum shear stress theory for flexibility related failures. Repeating, the maximum shear stress theory predicts that failure occurs when the maximum shear stress in a body equals SYield/2, the maxim um shear stress existing at failure during the uni axial tensile test. Recapping, the maximum shear stress in a body is given by: '(max = (81 - S3) / 2 For the differential element at the outer surface of the pipe, the principal stresses were computed earlier as: SI

=

(SL + SR) / 2 + [ [(SL - SR) / 2]2 + '(2 ]1/2 =

As seen previously, the maximum shear stress theory states that during the uniaxial tensile test the maximum shear stress at failure is equal to one-half of the yield stress, so the following requirement is necessary: tmax = [(SL - SR)2 + 4 1 2 ]112

<

2

2

Multiplying both sides arbitrarily by two saves the time required to do two mathematical operations, without changing this relationship. Multiplying by two creates the stress intensity, which is an artificial parameter defined sim ply as twice the maximum shear stress. Therefore the Maximum Stress Intensity criterion, as adopted by most piping codes, dictates the following requirement: [(SL - SR)2 + 4 '(2 ]1/2 < SYield

1-18


Note that when calculating only the varying stresses for fatigue evaluation purposes (as discussed in the following section), the pressure components drop out of the equation. If an allowable stress based u pon a suitable factor ofsafety is used, the Maximum Stress In tensity criterion yields an expression very similar to that specified by the B31.3 code: [ Sb2 + 4

S~

] 1/2 < SA

Where: Sb

=

longitudinal normal stress due to bending, psi

St

=

shear stress due to torsion, psi

SA

=

allowable stress for loading case, psi

Example Stress Intensity Calculations: Calculation of stress intensity may be illustrated by returning to our 6-inch nominal diameter, standard wall pipe for which longitudinal, shear, and hoop stresses were calculated. Reviewing the results ofthose calculations: Longitudinal stress: SL

=

15547 psi

Shear stress:

=

5999 psi

Hoop stress:

=

7098 psi

Assuming that the yield stress of the pipe material is 30,000 psi at temperature, and a factor of safety of 2/3 is to be used, the following calculations must he made: [(SL - SH)2 + 41:2 ]112 < 2/3 x SYield, or: [(15547 - 7098)2 + 4 x 5999 2 ]1/2 < 2/3 x 30000, or: 14674 < 20000 The 14674 psi is the calculated stress intensity in the pipe wall, while the 20000 is the allowable stress intensity for the material at the specified temperature. In this case, the pipe would appear to be safely loaded under these conditions.

1-19


1.2 Fatigue Failure The fail ure modes discussed above were sufficient to de scribe catastrophic failure based upon one time loadings. However, piping and vessels were also found to suffer from sudden failure following years of successful service. The proposed explanation for this phenomenon was fatigue failure ofthe material, resulting from propagation of cracks on the material crystal structure level due to repeated cyclic loading.

1.2.1 Fatigue Basics Steels and other metals are made up of organized patterns ofmolecules, known as crystal structures. However, these patterns are not maintained throughout the steel producing an ideal homogenous material, but are found in microscopic isolated island-like are as called a grains. Inside each grain the pattern ofmolecules is preserved. From one grain boundary to the next the molecular pattern is the same, but the orientation differs. As a result, grain boundaries are high energy borders. Plastic deformation begins within a grain that is both subject to a high stress and oriented such that the stress causes a slippage between adjacent layers in the same pattern. The incremental slippages (called dislocations) cause local cold-working. On the first application of the stress, dislocations will move through many of the grains that are in the local area ofhigh stress. As the stress is repeated, more dislocations will move through their respective grains. Dislocation movement is impeded by the grain boundaries, so after multiple stress applications, the dislocations tend to accumulate at grain boundaries, and eventually becoming so dense that the grains "lock up", causing a loss of ductility and thus preventing further dislocation movement. Subsequent applications of the stress cause the grain to tear, forming cracks. Repeated stress applications cause the cracks to grow. U nless abated, the cracks propagate with additional stress applications until sufficient cross sectional strength is lost to cause catastrophic failure ofthe material. Figure 1-16 illustrates this process.

1-20


••

Molecular pattern in unstressed grain

.....Slipping of one molecular surface over another after first application of stress

~_ '\

§§§§§§§§ ~ocati'" -+ §§§§§§§§

Slipping of a second molecular surface after a second application of

Slip'

stress

Dislocations beginning to interact and tangle

.~ ~

After many repeated applications of stress the dislocations are completelytangled and the grain is 'Iocked".

With another application of the stress, the grain "tears' and a fatigue crack is initiated.

Figure 1-16

One Cycl e TEST LOADING CURVE Tensile Test Specimen

Figure 1-17

1-21

COADE Pipe Stress Analysis Seminar Notes One important consideration is the fact that fatigue cracks usually are initiated at a free surface. Corrosive attack on a material often produces pitting ofmetal surfaces. The pits act as notches and produce a reduction in fatigue strength. In those specifie cases when corrosive attack occurs simultaneously with fatigue loading, a pronounced reduction in fatigue properties results which is greater than that produced by prior corrosion of the surface. When corrosion and fatigue occur simultaneously, the chemical attack greatly accelerates the rate at which fatigue cracks propagate. U nfortunately, fatigue failures can occur even when the stress in a material is below the yield stress. This is because localized stress concentrations can cause plastic deformation in a relatively few grains des pite the fact that the stress over a gross area ofthe section may be far below the material yield stress. If the section is subjected to a sufficient number of stress cycles, cracks can initiate in highly stressed grains and then propagate throughout the material, ultimately resulting in a fatigue failure of the section as a whole. The fatigue capacity of a material can be estimated through the application of cyclic extensive/compressive displacement loads with a uni axial test machine, as shown in Figure 1-17. SampIe results for typical ferrous material (with a yield stress of5 7,000 psi) are shown below:

Applied Cyclic Stress (psi)

Cycles ta Fa il ure

300,000

23

200,000

90

100,000

550

50,000

6,700

30,000

38,000

20,000

100,000

1.2.2 Fatigue Curves A plot of the cyclic stress capacity of a material is called a fatigue (or endurance) curve. These curves are generated through multiple cyclic tests at different stress levels. The number of cycles to failure usually increases as the applied cyclic stress decreases, often until a threshold stress (known as the endurance limit) is reached below which no fatigue failure occurs, regardless ofthe number of applied cycles. The endurance limit (for those metals that possess one) is usually quantified as the value orthe cyclic stress level which may be applied for at least 108 cycles without failure. Typical ratios of the endurance limit to the ultimate tensile strength of various materials are 0.5 for cast and wrought steels; about 0.35 for several nonferrous metals such as nickel, copper and magnesium; and 0.2 to 0.3 for rough or corroded steel surfaces (depending on the degree of stress intensification). An endurance curve for carbon and low alloy steels, taken from the ASME Section VIII Division 2 Pressure Vessel Code is shown in Figure 1-18. 1-22


tO'I:"""------r-----T""""-----r-----T""""-----., NOTH:

1" E- . . . . . . . 131 _ _ _ _ . _ ..... _ _ IZI T_5-11O.1 _ ""_ U1S 1II-1151to1.

-..--of __

w Cl :::>

1-

:::i

c...

~

(f) (f)

w

cr:

1(f)

U

:::i U >U

FIG. 5-110.1 DESIGN FAnGUE CURVES FOR CARIION, Law ALLOY, SERIES ~IOC, HM ALLOY STEELS AllO HIGH TENSILE S1ULS FDII TEMPERATURES NOT EXCEEDING 7UO'F

Figure 1-18 Note that according to the fatigue curve, the material doesn't fail upon initialloading, despite enormously high stresses that appear to be weIl above the ultimate tensile stress oftypical carbon and low alloy steels. The reasons for this are: 1

The highly stressed areas under fatigue loading are normally very localized. Catastrophic failure under one-time loading will normally occur only when the gross cross-section is overloaded.

2

Fatigue curves are usually generated through cyclic application of displacement, rather than force, loading. Displacement loads are "self-limiting". If a pipe is overloaded with an imposed displacement, plastic stresses will develop, deforming the pipe to its displaced position. At that point there will be no further tendency for displacements to occur, and therefore no continuation ofthe load, or further deformation leading to catastrophic failure. In the case of an applied force (which is not a self-limiting load), deformation of the pipe does not cause the force to subside, so deformation continues until failure.

3

The stress shown in a fatigue curve is a calculated stress, based upon the assumption that Hooke's law is applicable throughout the range of applied loading; i.e., S = E E, where:

E

=

modulus of elasticity ofmaterial, psi

E

=

strain in material, in/in

1-23

COADE Pipe Stress Analysis Seminar Notes In reality, once the material begins to yield, stress is no longer proportional to the induced strain, and actually is much lower than that calculated.

1.2.3 Effect of Fatigue on Piping A. R. C. Markl investigated the phenomenon offatigue failure ofpiping during the 1940's and 1950's, and published his results in papers such as "Piping Flexibility Analysis", published in 1955. He tested a number of configurations (straight pipe, and various fittings, such as pipe elbow, miter bend, unreinforced fabricated tee, welding tee, etc.) by using cyclic displacements to apply alternating bending stresses. Plotting the cycles to failure for each applied displacement, he found that the results of his experiments followed the form of fatigue curves.

16"

~ 1

(TYP,

41" (TYP)

~IL-,......_ _ _ _ _.....

~

~ Girth butt weld

-a-

f

1

Range of imposed displacements to impose complete stress reversaI.

~t--...,IL.Ô.-J-I____- -.....I -a- displacements _....:.....-----.l~R 1 • angeo f ou tpane

RangeOfinPlaneL~

~

...&...~ _ _ _ _ _ _...J

ml!-

-a -a. . .

/

displacements

~~ Range of outplane

....... Range of inplane / displacements ~_

{(-'------....

•

Range of inPlane/

displacement~

displacements

7'

·

......... Range of outplane

"'Placements

Figure 1-19 If an initially applied displacement load causes the pipe to yield, it results in plastic

deformation, producing a pre-stress in the system, which must be overcome by subsequent stress applications, resulting in lower absolute stresses during later load cycles. Because of the system "relaxation", the initial values of the thermal stress are allowed to exceed the material yield stress, with the aim being that the system "self-spring" during the first few cycles and then settle into purely elastic cycling. This "self-springing" is also called Elastic Shakedown. As shown in Figure 1-20, the maximum stress range may be set to 2SYieid (or more accurately, the sum of the hot and the cold yield stresses) in order to ensure eventual elastic cycling.

1-24


2Sy~~~"".r-----------------

2Sy -Sy~--------~--~~~~----~~--------~--

-2Sy~-----------------------------

Time~

Stress

Figure 1-20 Based upon this consideration, the initial limitation for expansion stress design was set to the sum ofthe hot and the cold yield stresses - the maximum stress range which ensured that the piping system eventually cycled fully within the elastic stress range. Incorporating a factor of safety, this resulted in the following criterion: SE <= F (SYe + Syh) Where: SE

=

expansion stress range, psi

F

=

factor of safety, dimensionless

SYe

=

material yield stress at cold (installed) temperature, psi

Syh

=

material yield stress at hot (operating) temperature, psi

1.2.4 Cyclic Reduction Factor At sorne point, in the vicini ty of 7,000 cycles, the (SYe + SYh)limita tion intersects the fatigue curve for carbon and low alloy steel. The allowable stress range must therefore be reduced to fit the fatigue curve for cyclic applications with 7,000 cycles or more: SE <= F f(SYe + Syh)

1-25


Where: f = cyclic reduction factor, as shown in the accompanying table

CYCLIC REDUCTION FACTOR TABLE Cycles N

Factor f

1

7,000

1.0

7,001

14,000

0.9

14,001

22,000

0.8

22,001

45,000

0.7

45,001

100,000

0.6

100,001

200,000

0.5

200,001

700,000

0.4

700,001

2,000,000

0.3

1.2.5 Effect of Sustained Loads on Fatigue Strength In almost an cases the material fatigue curves are generated using a completely alternating stress; i.e., the average stress component is zero. Research has shown that the magnitude of the mean stress can have an effect on the endurance strength of a material, the trend of which is shown below:

cr",

.. ::..•

~

~ ,~

;~

CIl

,5

d

..

·c

!

Ci

'j ~

~ ~

~

C~cles

alternating stress

For Design

crllll < 17'IIIIZ < crllt, <0"4 4

t0 5

Sa from endurance

=< ûS./ curve for completely

Tensile

1

1

10'

107

cr Yield

10 foilure (b)

Figure 1-21

1-26

Mean Stress Axis

cr Yield


Note that as the mean stress increases the maximum permissible absolute stress (Sa + Sm) increases, while the permissible alternating stress decreases. The relationship between the allowable alternating stress and the average stress is described by the Soderberg line, which correlates fairly weIl wi th test data for ductile materials. The equation for the Soderberg line is: SaCAllowed) = SaCfor R=-1) xCI - Sm/SYield) Where: R

=

Smin / Smax

Sa

=

(Smax - Smin) / 2

Sm

=

(Smax + Smin) / 2

Note that during the development of the ASME Boiler and Pressure Vessel Code Section III rules and procedures for analysis ofnuclear piping, the Special Committee to Review Code Stress Basis concluded that the required adjustments to a strain-controIled fatigue data curve based on zero mean stress, occur only for a large number of cycles Ci.e. N > 50,000 100,000) cycles for carbon and low-alloy steels, and are insignificant for 18-8 stainless steels and nickel-chrome-iron aIloys. Since these materials constitute the majority of the piping materials in use, and since most cyclic loading events comprise much fewer than 50,000 cycles, the effects of mean stress on fatigue life are negligible for piping materials with ultimate strengths below 100,000 psi. For materials with an ultimate strength equal to or greater than 100,000 psi, such as high strength bolting, mean stress can have a considerable effect on fatigue strength and should he considered when performing a fatigue analysis. For a piping application, the implication of the Soderberg line on the fatigue allowable is implemented in a conservative manner. The sustained stress Ci.e., weigh t, pressure, etc.) can be considered to be the mean component of the stress range after system relaxation, and as such is used to reduce the allowable stress range: SE <= F f(SYe + Syh - Ssus)

1-27


1.3 Stress Intensification Factors As noted previously, Markl's fatigue tests generated endurance curves for various fitting configurations, such as straight pipe, butt welded pipe, elbows, miters, welding tees, unreinforced and reinforced fabricated tees, mostly using 4" nominal diameter, size-on-size fittings. Markl noticed that the fatigue failures occurred not in the middle ofhis test spans, but primarily in the vicinity ofthe fittings, and in those cases, they also occurred at lower stress/cycle combinations than for the straight pipe alone. Earlier theoretical work pointed to a possible explanation. It had been shown that elbows tend to ovalize du ring bending, bringingthe outerfibers closerto the neutral axis ofthe pipe, thus reducing the moment of inertia (increasing flexibility) and the section modulus (increasing developed stress). Ovalization of Bend x

Section

Figure 1·22 The stress intensifica tion factors (the ratio of actual ben ding stress to the calculated ben ding stress for a moment applied to the nominal section) for elbows was known to be: =

0.75/ h2/3

=

0.9/ h 2/3

10

=

out-of-plane intensification factor

li

=

in-plane intensification factor

h

=

flexibility characteristic

=

t R/r2

t

=

pipe wall thickness, in

R

=

bend radius of elbow, in

r

=

mean radius of pipe, in

10

li

Where:

1-28


Markl found this to correlate fairly weIl wi th his test data and so adopted it. Tests on mitered bends correlated weIl with those for smooth bends, providing an equivalent bend radius R was used in the above equation for h. Markl's estimates of equivalent bend radius are shown below: Re

=

r(l + 0.5 sIr cot D) (for closely spaced miters)

Re

=

r(l + cot D) 1 2 (for widely spaced miters)

Re

=

equivalent bend radius, in

s

=

miter spacing at the centerline, in

D

=

one-half of angle between cuts

Where:

Markl found that the unreinforced fabricated tees could be modeled using the same formula as that for single (widely spaced) miter bends could be use d, if a half angle of 45 degrees was used. This produces a flexibility characteristic of: h

=

tir

For butt welded tees (such as ANSI B16.9 welding tees) Markl again adapted the bend equations, this time computing an equivalent radius (Re) and an equivalent thickness (te). Markl's equation for weI ding tees was: h

=

c ( te Re 1 r 2 )

=

ratio of tee-to-pipe section modulii, dimensionless

=

(tJt)3/2 (Markl's recommendation)

=

equivalent pipe wall thickness, in

=

1.60t (Markl's recommendation)

=

equivalent bend radius, in

=

1.35r (Markl's recommendation)

Where: c

te

Re

Inserting these values into the expression for h yields: h

=

4.4 tIr

This is precisely the expression used today for ANSI B16.9 welding tees.

1-29


For reinforced fabricated tees, Markl used the expression he had previously used for welding tees, with different equivalent wall thickness and bend radius: h Where: c te tp Re

=

c ( te Re / r 2 )

= = = =

(teft)3/2 (Markl's recommendation)

t + tp thickness of reinforcing pad or saddle, in r

The following tables compare the stress intensification factors suggested by Markl's test results versus the values calculated with his equations (results are for 4" nominal diameter, standard schedule pipe):

Bend in-plane

(in

tR/r 2

Test

Calculated

0.062

4.49

5.7428

0.210

2.17

2.5476

0.129

4.38

3.5238

0.320

2.02

1.9238

0.319

2.10

1. 9286

0.316

1.90

1.9381

0.328

1. 70

1.8904

0.331

1.53

1.8809

0.324

1.36

1.9095

0.332

1.28

1.8762

0.328

1.46

1.8904

1-30


Unreinforced tee (io): tir

Test

Calculated

0.0390

Il.04

10.84

0.0455

6.12

7.06

0.0947

2.95

4.33

0.1111

2.34

2.89

Reinforced tee:

in-plane (i;) Calculated Test

out-plane (io) Test Calculated

0.12

2.21

2.63

2.43

3.17

0.237

1. 78

1. 74

1.83

1.98

0.5

1.10

1.14

1.08

1.18

tpad

These fonnulas for intensification factors were adopted (and expanded) by the piping codes. Specifie fonnulas and/or fittings recognized by the individual ASME/ANSI B31 codes are usually shown in Appendix D ofthose codes (see Figure 1-23).

1-31


APPENDIX D FLEXIBILITY AND STRESS INTENSIFICATION FACfORS TABLE 0-1' FLEXIBIUTY FACTOR t AND STRESS INTENSIFICATION FACTOR 1 _ _ Ion

la)

fini"'" fortor

0Ut~_

k

i.

DoKrillllon

w~."'G

.._

or plpt _

CHotts 121. (C)-(711

Ill)

,-_Id __ J

< '2 Cl +

--Ji

bond ..

~ '2 (1

+

..w..,.

FledIllillJ thoozurlstk k

O.,

~r2

TR,

0.75

"

Jill'

-;m

-;;;-

fUlfS:,,7- --..

1.52

O.'

O.,

coti sf

.~_.!J".J, --y Z

.!,g

,,'/1,.

\In"

1....... ;,

1.6~

~

bit "

CII_ (2). 1.). (5), m)

Si_ ""'' '

r..... l _ 121. D)I

"li'

,,:1./1

M

~

Ir'''

Jil"

O., Ir'''

:w.. i. +

OOod

.J!

r,'

2

.--

-

'--2-

~.f 2

li

CHotts 121. 1.'. (7))

lb)

W.ldlng lot .... ASIII E 81U_

'. ~ ~0t.. Tc. :i!: 1.5 T

4.~ r,

lit

~~.

'.

~:r'2 .

IN.... 121. 141. (6). Ill). (Ul)

I.~

tbt

~

Remforttd fabnc.t~ tn

~f"

+

~-;.

JiJi~

wrth pad .or !.adtllt

[Note' 121. 1.1. IBI. (12). Ill):

If + ",i)'" f 1) ~

~~2 T' ., Tt' Pad

te)

lb)

lb)

o.•

Urninforced f.ltricated Ue IN.m 1Zl. I~). IIZl. mll

F

~l()+ '"

r,

Ii'''

o.,

Exuvdod _1... too "Ith ~ 0.0506 Tt < l..5T

'It

~I~

+ \.Ii

s.ddIe'

(1,+-r.) -i

hW

12

o.•

•.• 1. r,

'2

{Notn (2), loC}, Cl})]

(b)

Wtlded~ln

contour InSfrt

wllII r. '2:: 'tWL

"''

T(~l.sr

(ffote1

(0)

en

(4), i1l). (131]

IIrIr1ch _Id·•• IiItlno IIn........ ...,torc;ocll

INo... (2). «l.

I~).

0.9

Ir'''

1121)

Flaibllty

Factor k Bult -.eIdi!d joint. Nducer, or _t'Id neck fI"\1e

lb)

1.0 1.2

D " " b _ ...on lWtgo

Cel

StmI IlIlIn,ifiutioo F_; INoto Illl

FI'" wtIdH jOint, or socttt M'!d ftarlge or fittlng

Note (14)

Up )oint """"" (wlth ASME B16.9 Iap j,nt .tub)

Lb

Tht'tAdfod pipe joint, Or U.rudfd flallQf

Z.l

C_ _ _

S""I~ ~I ...

or ComIPtld ......... d bond [Not. no»

Figure 1-23 1-32

B


Subsequent research has demonstrated that Markl's formulas, having been based on a limited numher configurations (significantly having omitted reduced outlet tees) and disregarding any need to intensify torsional stress, are inaccurate in some respects. The major problem with reduced intersections tees lies in the out-of-plane bending moment on the header. Stresses due to these moments can never he predicted from the extrapolation of size-on-size tests. Figure 1-24 below illustrates the origin of this problem.

Area of high bending ............. stresses

Mob

Size-on-size

Mob

Reduced Intersection

Figure 1-24 Errors due to these moments can be non-conservative by as much as a factor oftwo or three. Furthermore, when the rlR ratio is very small, the branch connection has little impact on the header, so use oflarge stress intensification factors for the header can produce unreasonably large calculated stresses. R.W. Schneider ofBonney Forge pointed out this inconsistency for reduced branch connections. His paper on the subject states that the highest stress intensification factors occur when the ratio ofthe branch to headerradiiis about 0.7, at which point the nonconservativism (versus Markl's formulas) is on the order oftwo.

i from Markl

-

0.7

1.0

1.0

r/R Ratio of Actual i to Markl's i vs Ratio of 8ranch to Header Radius

Figure 1-25

1-33


1.4 Welding Research Council Bulletin 330 The Wei ding Research Council's Bulletin 330, "Accuracy of Code Stress Intensification Factors for Branch Connections" documented a major attempt to re-assess the existing code requirements for the intensification of stresses at tees and other branch connections. The difficulty ofthis task was summed up in the bulletin by author E. C. Rodabaugh, who stated: ''We would rate the relative complexity ofi-factors for pipe, elbows and branch connections by the ratios 1 :5:500. These comments on relative complexity, we think, are relevant at this point because at least sorne readers will be looking for simple answers to what they perceive to be a simple subject. They will not find any simple answers in this report." Summarizing the findings ofWRC 330 in order ofincreasing importance: 1)

The following note should be added wi th regard to branch connection flexibilities:

"In piping system analyses, it may be assumed that the flexibilityis represented byarigidjointatthebranch-to-runcenterlinesjuncture. However, the Code user should be aware that this assumption can be inaccurate and should consider the use of a more appropriate flexibility representation." 2)

ASME 2/3 and B31.1 users can use the ''Branch Connection" expressions for unreinforcedfabricated tees wheneverrlR< 0.5. (Markl's formulas specified that the same stress intensification factor be used on both the branch and header legs of a tee, regardless of relative sizes. The codes noted above permit the reduction ofthe stress intensification factor at the branch for relative diameters. CAESAR fi automatically considers the effects ofreduced intersections on the stress intensification factors for these codes unless directed otherwise by the user through the setup file.)

3)

B31.1 erred when including the calculations for branch connection stress intensification factors; instead they should have included the calculations as they appeared in ASME III. (Further clarification of this note is given in note 10 herein.)

4)

B31.3 should include the stress intensification factors for branch connections as per ASME III. (B31.3 uses Markl's original formulas, thus specifying the same stress intensification factor for both the branch and header of a tee, regardless of relative sizes.)

5)

B31.3 should intensify the torsional moment at branch connections, with the torsional intensification factor estimated as: it = (rlR)i o.

6)

B31.3 should eliminate the use of ii = 0.75io + 0.25 for branch connections and tees. It can give the wrong relative magnitude for header moments, and may underestimate the difference between Mo and Mi for rlR ratios between 0.3 and 0.95, and perhaps over-estimates the difference for rlR ratios below 0.2 and for rlR = 1.0.

1-34


7)

B31.3 and B31.1 should add restrictions to the stress intensification factor tables indicating that they are valid for RIT < 50.

8)

The codes should add notes that indicate that the stress intensification factors are developed from tests and/or theories based on headers being straight pipe with about two or more diameters length of pipe on either side of the branch.

9)

The codes should also add notes to indicate that for branch connections/tees the stress intensification factors are only applicable where the axis ofthe branch pipe is within 5 degrees ofnormal to the surface of the header pipe.

10)

The stress intensification factors for unreinforced fabricated tees, weldolets, and sweepolets should be changed to:

For (rlR) < 0.9: lb

=

1.5(RIT)2/3 (rlR)1/2 (r/rp ), with ib(tIT) > 1.5

For (rlR) = 1.0: 0.9 (RIT)2/3 (r/rp ), with ib(tIT) > 1.0

And:

=

0.8 (RIT)2/3 (rlR), with ir > 2.1

lb

=

intensification factor for branch (to be linearly interpolated for rlR ratios hetween 0.9 and 1.0)

R

=

mean radius ofheader pipe, in

T

=

thickness ofheader pipe, in

r

=

mean radius ofbranch pipe, in

rp

=

outer radius ofbranch pipe, in

t

=

thickness ofbranch pipe, in

Ir

=

intensification factor for run (header) pipe

Ir

Where:

Additionally, if a radius of curvature r2 is provided at the connection, which is not less than the larger of t/2, (Tb'+Y)/2, or T/2, then the calculated values of ib and ir may be divided by 2.0, but with the restriction that ib>1.5 and ir >1.5. Also, where reduced outlets are discussed, branch ends should he checked using Z = p (r2)t and i(tIT) in place ofi, with i(tIT) > 1.0.

1-35


Il)

There was not sufficient data available onreinforcedfabricated tees for Rodabaugh to make any definitive recommendations regarding them. Rodabaugh does however suggest that the normal usage whereby the width of the pad is taken to be at least equal to the radius ofthe nozzle should be observed even though not explicitly directed by the code.

12)

For t/T ratios of about one or more, stresses tend to be higher in the header, and are fairly independent ofthe wall thickness ofthe nozzle. As the tlI' ratio gets much smaller than one, the largest stresses shift to the branch. (This finding originally came out of the research for WRC 297.)

Comparisons ofWRC 330's proposaIs for stress intensification factors for various types of tees, versus B31.3 calculated values are shown on the following pages.

1-36


NO INTERSECTION RADIUS "831.3" VS. 'WRC 330' UNREINFORCED, FA8RICATED TEE STRESS INTENSIFICATION FACTORCOMPARISON

HEADER NOM

BRANCH SCH

WRC

--B31.3---

330 b

i ib

WRC

iob iOb

--B31.3---

~

i oh

1 330 h

330 h

330 b

ioh

1 40.

40.

2.433

2.874

2.433

.853

1.081

2.433

.959

1.125

1 48. 2 48.

1 48. 2 40.

4.184 3.359

2.769 2.769

3.359 3.359

..m .B24

.822 1.010

2. tHe 2.986

2.769 2.769

3.359 3.359

1.319 .927

1.6"" 1.125

J 40. l 40. J 43.

1 414. 2 414.

3.479 4.769 3.4811

2.860 2.860 2.868

3.488 3.488 3.488

.822 .U8

1.011 .738

3.488 3.481 3.4811

l.b57 1.657

1.881

2.86B 2.868 2.86"

1.362 1.362

.822

2.111 2.111 3.893

.925

1.125

4 40. 4 40. 4 40.

1 48.

3.416 4.682 5.694 3.B92

3.169 3.169

3.892 3.892

.928

1.139

2.11111

.677

.831

3.169 3.169

3.892 3.892

.557

.684 l.0U

3.892 3.a92

1.5119 1.5B9 1.189 .916

1.953 1.853 1.468

.814

2.lem 2.665 3.46m

3.169 3.169 3.169 3.169

3.891

Z 40. 3 4B, 4 414. 1 48. 2 48.

4.589

3.441 3.Hl

1.1128 .758 .617 .54J .889

1.271 .927 .763 .669

3.441 3.441 3.441 3.441 3.441

".255 4.255 4.255 4.255 4.255

1.639 1.639

2.826 2.826 1.817 1.488 1.125

.816

!.lm

2.m

.671 .589 .528 .885

.834 .732 .656 1.81l@

2.11111

.764 .671 .681 .549

.954 .837 .751 .686

.BU

1.811!

.747 .669 .612 .535

.936 .839 .671

.797

1.l!ea

.915 .837 .732 .654

4 49. 5 40. S 48. ~ 411!. 5 40. 5 411!.

3 40.

3 411. 4 48.

5 48.

6 49.

2 48.

b 411!. 6 411!. 6 40. 6 4\J.

:) 48.

S 48.

:) 4f.1. .. 48.

a 48. B 411!. 49. B 48.

e

4 U. 5 411. 6 48.

5 411. b 4@. B 411.

19 48.

.. 411.

III! 4B.

S 48.

18 48. lB 48. 19 48.

8 411.

12 48.

12 48. 12 41.

12 48. 12 48.

14 14 14 14 14

48. 4B. 4B. 49. 4B.

b 48. 18

n.

5 48. h 4B. B 48.

18 48. 12 48. b 48. 8 48. 10 48.

12 48. 14 48.

5.5BII 6.359 4.255

3.441 3.44J

4.255 4.255 4.255 4.255 4.255

4.477 5.444 6.282 6.919 4.548

3.655 3.655 3.655 3.b55 3.655

4.540 4.548 4.541 4.541 4.541

5.187 5.918 6.592 7.218 4.94'1

3.961 3.961 3.961 3.9bl 3.961

4.949 4.949 4.949 4.949 4.949

5.642 6.294 6.884 7.875 5.284

4.213 4.213 4.213

5.284 5.284 5.284

4.213 4.213

5.284 5.284

6.834

4.392 4.392

.728 .666 .592 .52@ .795

5.599

.697 .689 .545

3.348

l

3.441

b.bBB 7.549 8.443 5.523

·40592 4.392

5.523 5.523 5.523 5.513 5.523

6.383 7.382 B.IM 8.569 5.599

4.458 4.45B 4.458 4.4Se 4.4511

5.599 5.599 5.599 5.599

4.392

.5J9 .795

Lltllll

.768

1.all'!

.877

.767 .686 .653

2.nll

2.1111 2.342 3.14B 3,783

2.711

3.374 4.836

3.655 3.655 3.655 3.655 3.655

3.an

1.478 I.H2

.91B

1.125

4.548

1.741

4.548

1. 741

4.548

1.348 1.883 .986

2.162 2.162 1.674 1.346 1.125

1.886 1.754 1.410 I.PS .911

2.356 2.191 J.76J 1.472 1.125

1.122

2.516 2.282 l.a41 1.407

4.540

4.541

3.961 3.961 3.961 3.961 3.961

4.949 4.949

2.111 2.399 2.871 3.755 4.697

4.zn

2.@86 1.756 1.468

~.213

5.284 5.284 5.28" 5.284 5.284

.997

1.125

2.118 2.523 3.312 4.138 4.919

4.392 4.392 4.392 4.392 4.392

5.523 5.523 5.523 5.523 5.523

2.882 1.741 1.338 1.1163 .895

2.b18 2.189 1.673 1.337 1.125

2.328 3.847

4.458 4.45B 4.458 4.4511 4.458

5.599 5.599 5.599 5.5'19 5.599

1.911 1.468

2.405 1.839

1. Ib8

1.469

.982 .894

1.125

2.111 2.258 2.811

3.361 4.399

3.811

4.538 U188 . 4.977

1-37

4.213 4.213

4.213

4.949

4.949 4.949

1.236


NO INTERSECTION RADIUS "B31.3" VS. 'WRC 330" UNREINFORCED, FABRICATED TEE STRESS INTENSIFICATION FACTOR COMPARISON

HEADER NOM

16 40.

16 48.

lb 4". 16 4@, lb 48.

8RANCH SCH

WRC

-·831.3···

330 b

i ib

6.825 7.633 8.322 8.723 5.595

4.449

18 40.

7.281 7.B50 8.229 8.797 5.598

8 40. I@ 48.

12 48. 14 40. lb 48.

18 18 18 lB 18

48. 48. 40. 48.

20 20 20 2" 20

40. 48. 4". 48. 48.

12 40. 14 40. 16 4~. 18 4~. 20 40.

7.711 8.882 8.640 9.165 5.801

24 48. 24 40.

16 48. 18 40.

4~.

18 40. 12 40. 14 48. lb

40.

i ib

iob

WRC

···831 .3···

iOb

330 b

330 b

330 h

i ih

4.446

5.595

.928

2.664

4.446

5.595

4.446

5.595 5.595

.651 .583 .534

.733 .672

.510 .795

.641 1.098

.618 .567

4.449 4.449 4.449

5.598 5.598 5.598 5.598 5.598

4.681 4.681 4.681 4.681 4.601

5.BBI 5.a61 5.B01 5.881 5.8@1

.597 .56Q .532 .592 .793

4.707

.583 .549 .521

4.707

5.943 5.943 5.943 5.943

4.446 4.446

4.4~9

5.595

.541

.586 .795

3.332

I.6b9 1.334

2.1011 1.b79

3.961 4.352

4.446 4.446

5.595 5.595

1.123 1.822

1. 413 1.286

4.973

4.446

5.595

.894

1.125

.777 : 2.964 .713 : 3.523 .6811 3.871 .636 '4.423 1. 80~ l 4.976

4.449

5.598

5.598 5.598 5.598

1.5&1 1. 263 1. 1-49 1. "lib

1.889

4.449

.752 .718 .671

5.156

4.601

.736

3.512

.694 .658

3.951

6.52(1 6.520

i

1.113 .887

1.411 1.125

4.783 6.801

5.6lB 5.670

1.186 .945 .883

1.511

6.424

7.227· 7.227 7.227

.783

.640 .733 1.89B

5.879 6.293 6.695

5.899 5.899 5.899

7.532 7.532 7.532

1.083 .937 .B81

.51l!

.m

5.446

• 49'1 .585 .784

.637 .746

5.830 6.283 6.563

5.788 S.7B8 S.788

7.384;1 U63 7.3B4· .993 7.384 .933

5.788

7.384

6.480 6.4110 6.40& 6.489 6.480

B.280 8.280

.788

1.800

4~.

10.394 10.134 7.227

5.670 5.670

5.m

7.227 7.227 7. '227,

.546 .560 .785

11.763 !lU17 7.532

5.899 5.899 5.899

7.5321

.501

7.532. 7.532!

.572

7.384 j 7. 384: 7.384, 7.384

.56 48. 36 41l. 36 48.

30 -4B. 32 49. 34 48.

36 48.

36 48.

42 42 42 42

30

48. 48,

40. 4@.

42 40.

40.

32 U. 34 48.

36 40. 42 40.

11.210

5.788

11.599

5.788

9.902 7.384

5.788 5.788

11.5(18 11.9(17

b.400

12.231 12.633 8.209

b.UB 6.488

b.m b.480

.79'l.

~

8.208 ; 8.208 ; 8.20~ : 8.2@0 ; 8.209

1.252 1.125

5.140 5.140

.672

34 40.

.'193

.B92

4.619 5.796

.530

30 4~. 32 411.

!.b10

1.4&9

4.707

6.520 6.520

34 411. 34 48.

1.m

1.483 1. 277 1.117

5.BiH 5.a81 5.BIH 5.801

4.391

5.140 5.140

14 40.

5.m

1.009 ; 5.282

9.782 6.528

32 40.

5.598 1 .894

1.446 1.2116 1.125

5.943 5.943 5.'143

24 4~. 3" 40, 24

4.601

1.589

4.797 4.787 4.7@7

30 48. 39 48.

36 40. 32 40.

4.601 4.601 4.681

,633

20

32 41Ll.

4.449 4.449

Lel8

24 40.

32 49.

3.281

4.449

3.604 4.1lB 4.633

24 40,

4.m

330 h

5.595 5.595

24 4@.

40.

330 h

ioh

4.446 4.446

8.076 8.566 9.@37 5.943

4.707

i ih ioh

.,m ' .713 I.m

1.888 .713

.537 .521 .507

.780

1-38

.089 .6&8 .649 1.BS8

5.lbB 5.533 5.a86 6.228 7.289

S.670

5.943:

\,:)48

l.h92

1.191

1.504

!.lm

Lm

.B'11

I.! 25

l i

i i

1.204 1.125 1.281

1.197 1.125

1.35a 1.266

.8B2

1.198 1.125

1.238 ! .157 8.m 1.087 S.2U 1 1.828 B.2BILl 1 .878

1.587 1.482 1.393 1.316 1.125


NO INTERSECTION RADIUS

"831.3" VS 'WRC 330" WELDOLET STRESS INTENSIFICATION FACTOR COMPARISON HEADER NOM

BRAN CH SCH

3 48. 3 48. 3 48.

..·931.3..i ib

i ib iOb

WRC

iOb

---B31 ,3---

i ih

i Oh

330 b

330 h

i ih

i oh

330 h

330 h

40.

1

2.433

1.097

1.897

.~51

.451

2.162

1.Il97

l.097

.588

.518

1.516 1.516

.371 .451

2.1110

1.516

1.516

1 ;::::

1.516 1.516

.371

:::

.451

2.986

1.516

1.516

.722 .588

.722 • SilS

411. 2 4B. 3 411.

3.363 4.769 3.483

1. 570 1. 578 1.5711

1.570 1.570 1.570

.467 .329 .451

1 2.1110

.748

.748

1 2.11l0

1.571 1.571

.748

.748

i 3.093

1.57@ 1.570 1. SHI

1.578

.329 .451

.508

.588

1 3.366 1 1 4.682

1.756 1. 756

1.756

.522

.522

2.110

1,756

1.75b

.83b

.B3b

1.756

.375

1. m

5.694

1.756

1.756 1.756

.308

.375 i 2.18e .308: 2.665

1. 756 1. 756

.836 .659

.836 .659 .588

A0. '1 40. 2 48.

WRC

, 330 b

,

330 b

.467

1

4 48. 4 411. 4 48.

1 40. '1 411. 3 40.

4 41l.

4 40.

5 S 5 5 5

48. 4@, 48. 4@. 411.

il

40.

6 ~8. 6 4@. 6 4@.

1 411. '1 411. 3 411. -4

411.

5 411. 2 40. 3 40. 4 40.

1

I

!.?Sb

l :. :9: 1

.::r • .) __,

1.920

1 4.589

1. 920 1.928 1. 920 L92@

1.920 1.928 1.920 1.920 1.920

2.848 2.048 2.B48

2.048 2.848 2.048

2.848 2.048

2.848 2.1148

i 5.580 [ 6.358 : 4.255 4.47i ,5.444 • 6.2112 6.919 4.540

6 40.

5 40. 6 411.

8 40.

3 48.

,5,187

2.233

2.233

8 40.

4 5 b S

! 5.910

2.233

6.592 7.210 4.'149

2.233

2.233 2.233 2.233 2.233

8 40. Il 40. B 40.

40. 4iJ.

48. 40.

10 40.

4 U.

10 40. 10 4~. 10 40. !~ 40.

S 48.

5.642 6.294

12 12 12 12 12

40. 40. 40. 41l. 40.

14 14 14 14 14

40. 40.

n.

40. 40.

2,233 2.233

il 40.

6.884

2.384 2.384 2.384

B 411.

7.S75

2.384

III U.

5.284

2.384

S 40. (, 4@. 8 40.

6.034

III 41l.

8.443

2.492 2.492 2.492 2.492

6.6@0

7. 54'?

2.492

12 411.

6 40.

UB3

8 40. 10 40. 12 40.

7.3112 8.1bb S.569

2.526 2.526 2.526 2.526

14 40.

5.599

2.526

.451 ~5t

2.492 2.492 2.492 2.492 2.492 2.526

I.ni 1. 928 I.cm 1. 928

.579

2.188

.418 .344

2.11'" 2.342

.382 .451

3.11411 3.783

Lm

t.m

1.920

2.1148 2.1148 2.848 2.1148

'4.@3b

2.848 2.848 2.148 2.848 2.148

2.1811

2.233

.451

.458

.458

2.108

.376 .3311 .296

.376 .331 .296 .451

2.101l 2.711 ! 3.374

i

.451

.430 .378 .339 .318 .451

.422

.422

.379

.379

.451 .4311 .378

.339 .318

.346

.346

.303

.303

.m

.451

.378

.413 .378

.3311

.338

.~ .J

.295

1 .451

.451

\ .396

.346

.346

2.526

l

.396

i

.319

2.526

1

.309 .295

2.526

1 .451

2.526

l.·m 1.92i! 1. Ç7~

.579

.302

.~13

;!

3.m_~ ._1!756_~_~~m8

.418 .344

2.384 2.384 2.384 2.384 2.384

1

1.756

'9~

.295 .451

' 2.258 \ 2.811

3.361 ~.399

2.233 2.233 2.233

.975 .755

2,848

.5118

.5118

2.233

!.IIb3

t.m

2.233 2.233

.989 .794 .664 .598

,989 .794

1.135 .994 .831 .635 .588

L 135 .994 .831 .635 .588

1.181 .987 .755

1.181

2.233

.., ,.,.,..,. L.J..0·~1

2.492

2.492

2.492

2.492

2.492

2.492

2.492

2.492 2.492

2.328 3.847 3.811

2.526 2.526

4.5311

2.526 2.526

4.917

1-39

2.526

.588

.755 .687

\2.lH!

2.492

.B20

.975

i 2.8711 ! 3.755 ! 4.697 2.523 '3.m 4.138 4.9119

.914

.'m

2.384 2.384 2.384 2.384

i

.914

.975

2.384 2.384 2.394 2.394 2.384

2.lBB 2.399

.914 .914 .820 .632 .528

2.384

2.526

2.526 2.526 2.526

2.526

.b~3

•sile 1.885 .829 .663 .558 .588

.607

.664 .58B

.987 .152 .603 ,588 1.885

.829 .663 .558 .508

COADE Pipe Stress Analysis Seminar Notes t:l.Q INTERSECTION RADIUS "831.3" VS 'WRC 330" WELDOLET STRESS INTENSIFICATION FACTOR COMPARISON

HEADER NOM

1& 48. 16 48. 16 48.

16 4@. 16 48.

8RANCH SCH

WRC

·-831.3..·

330 b

i ib

330 b

330 h

8 11 12 14 lb

6.825 7.633 8.322 9.723 5.595

2.524 2.524 2.524 2.524 2.524

2.524 2.524 2.524 2.524 2.524

.3711 .331 .383

.371 .331

.289 .451

.289 .451

2.664 3.332 3.961 4.352

1 l 7.2111 ; 7.858

2.526 2.526 2.526

2.52l, 2.526 2.526 2.526 2.526

.351 .322 .387 .287 .451

.351 .322 .387

2.9/,4 3.523 3.871

.287

4.423

.451

.339 .324 .383 .286

.339 .324 .31113 .286 .451

.332 .313 .297

.332 .313 .297 .451

48. 4a. 48. 48.

48.

1

i ib iob

330 b

WRC

iob

.313

-831.3..·

i oh

i oh

330 h

330 h

2.524 2.524 2.524 2.524 2.524

2.524 2.524 2.524 2.524 2.524

.947 .758 .637 .5B8 .509

.947

2.526 2.526 2.526 2.521: 2.526

.952

4.976

2.526 2.526 2.526 2.526 2.526

3.28B

2.617

3.6114

2.617

4.118 4.633 5.156

2.617

4.973

i ih

~

.759 .637 .588 .519

i

19 48.

18 40. 18 4B.

18 41. lB 40.

28 48. 20 41. 20 4!1.

21 40. 28 48.

24 48.

24 48. 24 48. 24 40.

10 12 14 16

• ,II. 4". 48. 46.

i1 8.229 ! 8.797 18 411.! 5.598

2.~m

2.526

7.71l 8.182 8.648 18 411. 1 9.165 28 4&. i 5.811

2.617 2.617 2.617 2.617

2.617 2.617 2.617 2.617 2.617

8.176 8.566 2" 40 •. 9.137 5.943 24 48.

2.681 2.681 2.691 2.6Bl

2.681 2.681 2.691 2.681

12 48. 14 n. 16 48.

16 40.'

18 48.

2.617

.451

.451

33 U. 38 48.

24 48.; 38 40.,

.6S3 .571 .588

.653

.798 .726 .635 .565 .518

.;98 .726 .635

2.617

2.617 2.617 2.617 2.617 2.617

3.512 3.951 4.397 5.282

2.681 2.681 2.681 2.681

2.6Bl 2.681 2.681 2.681

.763 .679

.763 .679 .619 .5@8

2.617

.618

.568

.383 .451

.383 .451

4.619 5.796

2.942 2.942

2.942 2.942

.51B

10.3114 [email protected]

3.261 3.261 3.261

.314

.m

4.783 6.881 6.424

3.261 3.261 3.261

3.261 3.261 3.261

.692 .543 .518

.682

.322 .451

.314 .322

7.227

3.261 3.261 3.261

.578 .549 .508

.6l]

24 40.

34 40.

3" 40.

11. 763

3.:m

3.398

.289

34 48. 34 40.

32 40. 34 40.

18.317 7.532

3.39!l 3.398

3.:m

.329 .451

.289 .329 .451

5.879 6.293 6.695

3.398 3.398 3.398

3.3118 3.399 3.398

.578 .540

36 40. 36 48.

3.331 3.:)31

3.331 3.331 3.331

.297 .287 .336 .451

.297 .287 .336 .451 ·

5.446

5.8311 6.283 6.563

3.331 3.331 3.331 3.331

3.331 3.331

.612 .571

56 40. 36 40.

30 40. ; Il.218 32 40. 1 II. 599 34 41!1. 1 9.962 7.384 36 40,

3.331

.:)37

3.331

.sltS

42 48. 42 40.

38 48. 1 11.588 32 40. 1 11. 9117

.321 .311 .381

.321

5.168

.311 .3U

i 5.533 i 5.886

.293 .451

.293 .451

6.228 1.289

3.699 3.699 3.699 3.699 3.699

3.699 3.699 3.&99 3.b99 3.&99

.716 .669 .628 .594 .5"8

4~.

42 48. 42 48.

i

34 4@.! 12.28!

12.633 42 48. 1 8.2111

36 40.

.637 .508

2.942 2.942

31 40. 32 4111.

42

.5&8

2.942 2.942

32 48. 32 43.

,

•SilS

9.782 6.528

32 40.

!

.571 .588

1

l

i

.717

.852 .717

3.331 3.:331

3.699 3.699 3.699 3.6119 3.699

3.398

3.331

3.6119 3.699 3.699 3.699 3.699

1 1

"

1-40

.sIB

.543

.58a

.612 .571 .537 .508 .716 .669 .628 .594 .516


NO INTERSECTION

RADIUS "831.3" VS 'WRC 330" SWEPOLET STRESS INTENSIFICATION FACTOR COMPARISON HEADER

8RANCH

NOM

SCH

WRC

-·831.3···

330 b

i ib

iob

330 b

i ib

WRC

-.. 831.3..·

330 b

330 h

i ih

ioh

iob

i ih 330 h

ioh 330 h

1 48.

1

48.

2.43;>

.929

.986

.382

.372

2.1b2

.929

.9ib

.43i

.419

2 48. 2 48.

48. 2 48.

4.884 3.359

1.188 1.188

1.251 1.251

.291

.354

.316 .372

2.188 2.986

1.188 1.188

1.251 1.251

.566 .398

.596 .419

3 48. 5 48. 3 48.

48. 2 48.

2.618 4.155

1.222 1.222

1.296 1.296

.467 .294 .551

1.296 1.296 1.296

.617

1. 296

1.222 1.222 1.222

.582

1. 222

2.188 2.188 3.893

.582

3 40. 1 3. 480

.495 .312 .372

.395

.617 .419

4 48. " 48. 4 48. -4 48.

1 48. 2 48. 3 41.

1.337 1.337

1.450 1.450

.372

3.461

1.337 1.337 1.337 1.337

1.458 1.4SB 1.45" 1.450

.637 .582 .386

.544

1.458

.;m

2.188 2.188 2.665

.698

1.458

.566 .356 .255

.637

1.337 I •.n?

.522 .329 .235

5 4t1.

1 48.

.685

1.585

1.585

1.439

1.585

1.439

1.585 1.585

.4112 .284 .249 .372

1.439 1.439

1.439

.579 .365 .258 .226 .338

1.585

2 3 4 5

1.439 1.439

1.585

5 48.

1.439

1.585 1.585

.685 .614 .473 .380

.416 .279 .245

.311 .273

.219

.244

2.111 2.108 2.711 3.374

.334

.372

4.036

1.518 1.518

.355 .312 .290 .256 .372

2.1'8 2.258 2.B18 3.361 4.399

1.632 1.632 1. 632 1.632 1.632

.349 .313 .286 .258 .372

2.188 2.399 2.87i 3.7:15 4.697

1.726 1.726 1.726 1.726 1.726

.341

2.m

.312

2.523

.272

3.382

1.793 1.793 1.793

.244 .372

4.138

5 48. 5 48. 5 40. il 4@. il 48.

b 40.

,

4

2.563 4.868 5••m

48'1 3.sn

! 2.483 41. i 3.940 411. , 5.588 41. ~ 6.358 48. : 4.255

3.738 2 40. 3 40. 5.444 ., 41. ; 6.282 •

1.439

1.518 1.518 1.518 1.518

1.585

1.691 1.691

Lm

:; 48. i 6 48. !

6.919 4.540

l.m

1.691 1.691

3 48. ,

5.187

1.632

1.843

4 5 6 8

.,8. 48. 48.!

5.910

1.632 1.632

1.843

48.

4.949

10 48.

4 48.

10 48. 18 48.

5 48. b 411.! 1 8 48.! UI 411'1

ô 48.

ô

4~.

a 48. 8 4m. B 48.

a 4m. 3 40.

10 48. 10 48. 12 48.

12 48. 12 48. 12 48.

5 b 8 111

1.632 1.632

1.843 1.843 1.843

.315 .276 .248 .226 .330

5.M2 6.294 6.884 7.875 5.284

1.726 1.726

1.968 1.968

.3@b .274

1.726

1. 96B

1.726 1.726

1. 968 1.968

.251 .219

6.834 48. 1 6.688 41. 7.549

1.793 1.793

.297 2.857 2.857 .272 2.857 .237 2.857 1 .212 2.857 .325

!

6.592 7.218

48.\

48. Ii

8.443

12 48.

12 48.

14 4I!. 14 40.

48. i 6.3S3 7.382 8.166 18 48. 12 48. 8.569 14 4B. 5.599

14

40.

14 40. 14 41.

5.523

1.193

1.793 1.793

6

1.814

Il 48.

1.814 1.814

1.814 1.S14

2.885 2.885 2.885 2.885 2.085

.~27

.2B4 .248 .222 .212 .314

.638

.452

! i

2. IIi

2.1811 2.342 3.848 3.783

4.989

.327 2.328 .286 3.847 .255 3.811 .243 . 4.530 .372 1 4.977

1-41

1.439

1.518 1.518 1.518

1.793 1.793

1.814 1.814 1.814 1.814 1.B14

Lm

.6911 .419

.755 .755 .677

.521 .419

.723 .723

.805 .885

.568 .458

.624

1.691

1.691

.376

.419

1.843

.i77 .723

1.691 1.691

.581

1.843

.581

.878 .816 .656

1.843

.486

.548

1.843

.371

.419

1.968

.828

1.843

.937

L968

.822 .719 .681

1.9b8 1. 968

.468 .367

2.1157

.850 .719

.975 .815

.543 .434

.623

1. 968

2.i57 2.1157 2.857 2.857 2.085 2.885

2."85 2.885

2.885

.365

.6B6 .524 .419

.498 .419

.779

.896

.595

.684 .547 .46"

.476 .480 .364

.419


NO INTERSECTION

RADIUS "831.3" VS 'WRC 330" SWEEPOLET STRESS INTENSIFICATION FACTOR COMPARISON HEADER

9RANCH

WRC

-931.3---

NOM

SCH

330 b

i ib

!6 40.

8 48,

6.825

1.813

16 48. 16 48. 16 411.

18 411. 12 48. 1-4 48. 16 48.!

7.633

l.813

8.322 8.723 5.595

1.813 1.813

2.884 2.IS4 2.884 2.184

1.813

2. B84

18 48.

18 48.:

;.2111

18 4f1.

18 48. lB 4f1.

7.858 J.C 1 8.229 16 41. i 8.797

1.814 1.814 1.814

2.885

1.814

18 40.

18 48.,

5.:5'18

1.814

1.@8S 2.885

1.878 1.878 1.878 1.878 1.B78

2.168 2.168 2.160 2.160 2.160

1. 9t8 1.918 1. 918

2.213 2.213

16 48.

12 41.

4".

i

28 48.

12 48.

7.711

28 48.

14 48.

8.882

2S 48. 28 48.

16 4e.

a.bU

28 48,

28 49.!

24 24 24 24

lb 48. i 8.176 18 4111.! 8.566 2e 48.: 9.1137 24 U.! 5.943

48. 48. 48. 40.

lB

48. 1

9.165 5.881

330 b

.2116

.231

--931.3---

330 b

330 h

i ih

ioh

330 h

330 h

.385 .273 .258

3.:m

2.664

1.813

2.884

1.813

.680 .544 .458 .417 .364

.782 .625 .479 .419

.612

.704

.515 .469 .418 .364

.592

3.961

l.813 1.813

.372

4.352 4.973

1.813

2.884

2.964

1.814

.zn

3.:i23 3.871

1. 81~

.2211 .286

.291 .266 .253

1.814

.324

.372

4.423 4.976

2.885 2.885 2.1185 2.085

1.814

2.885

.243 .231

.28S .267 .251 .236 .372

3.288 3.614 4.118 4.633 5.156

1.878 1.878 1.878

2.168 2.163 2.163

.214 .372

3.512 3.951 4.:591 5.282

.25111 .372

.218

.288 .324

.239

2.885

.252

2.085

.231

2.213 2.213

.216

.284 .322

.236 .223 .211 .321

.258

.245

2.1b'

2.168

t.918 1. 918 1.910

2.213 2.213 2.213 2.213

.544 .483 .434 .362

.503' .419

4.619 5.796

2.971 2.171

2.428 2.428

.448 .357

.419

4.783 6.811 6.424

2.269 2.269 2.269

2.692

2.692 2.692

.474 .378 .353

.563 .449 .419

2.354 2.354 2.354

2.B!5 2.885

.4r18 .374

.477 .446

2.B85

.352

.419

2.750 2.756 2.750 2.751

.425 .397

.585 .472 .443 .419

3.854 3.1154 3.854 3.854

.492 .459 .432 .418

.519 .4911

3.854

.349

.419

.213

UI.394

2.269 2.269 2.269

2.692 2.692 2.692

.218 .224 .314

.259 .266 .:512

2.354 2.354

2.885 2.885 2.885

.28M .228 .315

.238 .272 .572

5.879

18.134

32 48.

32 48.

t 7.227

34 48.

48.! 11.763 32 48.1 10.317 34 48.! 7.532

1

2.354

.539 .471 .419

1.878

2.428 2.428

24 41l. 3111 48.

.526

1.870

2.m

.318

1.814

ioh

.570 .519 .454 .404 .363

2.1m

32 4i1. 32 48.

i ih

2.884 2.884 2.884

9.71l2

24 48,' 30 41l.

34 48. 34 48.

iob

WRC

iob

6.528

38 4!11. 38 48.

30

1.918

i ib

6.293 6.695

1.911

.659

.633 .560

.599

.525 .466 .419

.526

1

36 48.

J

3b 4f1.

30 48. f 11. 218 32 48. 1 Il. 599 34 48. i 9.982 7.384 36

2.312

42 48. 42 40.

32 48.

3B 48., 11. 588 11. 987 34 48. 1 12.281 3& 48. i 12.633 42 4@. f 8.28@

2.548 2.548 2.540 2.541 2.540

3b 48.

36 40.

42 48.

42 48. 42 48.

48.1

2.312

2.312 2.312

2.751 2.758 2.751 2.750 3.054 3.854

3.854 3.854 3.854

.2116

.245

15.446

.199

.237

5.838

.234.

.279

8~;) 3·

.372

6.213 6.563

2.312 2.312 2.312 2.312

.221 .213 .287 .2S1 .310

.265 1 5.168 .256 i 5.533 .249 5.886 .242 1 6.228 .:m l 7.289

2.548 2.548 2.548 2.548 2.548

1-42

l

.373

.352

.591 .552


1.5 Code Compliance 1.5.1 Primary vs. Secondary Loads

Markl's investigation of the fatigue problem, following the earlier recognition of the maximum stress theory offailure, led to identification of the basic problem in the design of piping systems. Not one, but two different criteria must be satisfied, one for primary loads, which may lead to single application catastrophic failure, and one for cyclic, dis placementdriven loads that may lead to fatigue failure (especially in the vicinity offittings and other discontinuities) after repeated applications. The main characteristics ofthese two different types of loads are described below:

Primary Load Characteristics: 1 -

Primary loads are usually force driven (gravity, pressure, spring forces, relief valve, fluid hammer, etc.).

2

-

Primary loads are not self-limiting. Once plastic deformation begins it continues unabated until force equilibrium is achieved (through change of the external boundary conditions or through material strain hardening), or until failure of the cross section results.

3 -

Primary loads are typically not cyclic in nature (and those that are, such as pulsation or pressure variation, show characteristics of both primary and secondary loads).

4

-

Allowable limits for primary stresses are related, through failure modes such as those advanced by the Von Mises, Tresca, or Rankine theories, to the material yield stress (i.e. the point where plastic deformation begins), the ultimate strength, or, for sustained loads only, to time-dependent stress rupture properties (such as creep characteristics).

5 -

Excessive primary load causes gross plastic deformation and rupture. Failure may occur with a single application ofthe load. Note that failures that occur due to single load applications usually involve pressure (hoop stress) design failures and are not directly addressed by CAESAR n or by the flexibility stress requirements ofthe codes. Such pressure design requirements are encompassed in the minimum wall thickness requirements discussed in detail in separate sections of the codes.

Secondary Load Characteristics: 1 -

Secondary loads are usually displacement driven (thermal expansion, imposed anchor movements, settlement, vibration, etc.).

2

-

Secondary loads are aImost always self-limiting, i.e. the loads tend to dissipate as the system deforms through yielding or deflection.

3

-

Secondary loads are typically cyclic in nature (except settlement).

1-43


4

Allowable limits for secondary stresses are based upon cyclic and fatigue failure modes, and are therefore limited based upon requirements for elastic cycling after shakedown and the material fatigue curve.

5

A single application of the load never produces failure. Rather catastrophic failure can occur after some (usually high) number of applications of the load. Therefore, even if a system has been running successfully for many years, it is no evidence that the system has been properly designed for secondary loads.)

Several examples should help illustrate:

Primary Stress Failure: Springs were improperly sized to support the weight of the valve operator on a system. When the line was fùled for hydrotest, everything (stresses and displacements) appeared fme, since the pipe could support the moment imbalance at ambient temperature. However, heating up the fluid (and pipe) during startup, the valve sagged and the guardrail was crushed in less than 30 minutes due to the decrease in strength at the operating temperature. Steps ta failure: 1

Weight loads were improperly accounted for. (The primary stresses were tao high.)

2

At operating tempe rature there was a resulting drop in material strength.

3

Gross deformation began almost immediately and continued until force equilibrium was achieved (the spring bottoming out).

Secondary Stress Failure: After 12 years of successful operation, inspection of the inside surface of a vessel revealed fatigue cracks in the vicinity of a piping nozzle connection. A subsequent analysis showed that a temperature increase in the adjacent vessel and piping system (alongwith a relocation of pipe restraints for the new operating conditions) made several years ago caused the stresses to exceed the expansion allowables. Even though the calculated stress range at the

1-44

COADE Pipe Stress Analysis Seminar Notes junction was weil over 470,000 psi, thejunction survived several years hecause of the selfrelieving nature of the thermalload, and the fact that the system cycled fewer than a dozen times over the two year period. Steps to failure: 1

Thermal allowables were exceeded by mistake.

2

After about a dozen applications of the excessive load, cracks formed on the highly stressed inside surface of the vessel at the junction with the nozzle.

Therefore, code compliance requires that the piping system be checked for both types of loading - primary and secondary. The basic steps involved in doing code compliance are outlined below: 1

Compute the primary stresses, i.e. the stresses due to the sustained primary loads, usually weight and pressure, and those due to the occasional primary loads, such as earthquake, wind, fluid hammer, etc.

2

Compute the range of the varying stress, i.e. the expansion stress range.

3

Compare the primary stresses to their allowables, which is based on a factor of safety times the hot yield stress.

4

-

Compare the expansion stress range to its allowable, which is a factor of safety times a value varying with the number of cycles such that it fits the material fatigue curve (adjusted for mean stress), but never exceeds the sum of the hot and cold yield stresses.

Note that due to the shakedown effect, and the fact that the primary and secondary stresses have different failure criteria, these two load types are reviewed in isolation. Therefore, it should he stressed that, as far as most codes are concerned, there is no such thing as "operating stress".

1.5.2 Code Stress Equations The piping code stress equations are a direct outgrowth of the theoretical and investigative work discussed above, with specific limitations established by Markl in his 1955 paper. The stress equations were quite similar throughout the piping codes (i.e., between B31.1 and B31.3) until the winter of 1974, when the power codes, having observed that Markl was incorrect in neglecting intensification of the torsional moment in a manner analogous to the bending component, combined the bending and torsional stress terms, thus intensifying torsion. It should be noted that the piping codes exactly calculate the stress intensity (twice the maximum shear stress) only for the expansion stress, since this load case contains no hoop or radial components, and thus becomes an easy calculation. Including hoop and radial stresses (present in sustained loadings only) in the stress intensity calculation makes the

1-45


calculation much more difficult. When considering the hoop and radial stresses, it is no longer clear which of the principal stresses is the largest and which is the smallest. Additionally, the subtraction of Sl-S3 does not produce a simple expression for the stress intensity. As it turns out, the inclusion of the pressure term can be simplified by adding only the longitudinal component of the pressure stress directly to the stress intensity produced by moment loadings only. This provides an equally easy-to-use equation and sacrifices little as far as accuracy is concerned. The explicit stress requirements for the piping codes addressed by CAESAR II follow below. Note that most codes allow Pdi2 / (d0 2 - di 2 ) to be used in place ofPdo / 4t.

1.5.3 831.1 Power Piping The B31.1 code requires that the engineer calculate sustained, expansion, and occasional stresses, exactly as defmed below:

Sustained: 0.75i MA

P do +

z

4t

Where: Ssus, SI

=

sustained stress, psi

1

=

intensification factor (single factor for aIl types of moments), as per Appendix D ofB31.1 Code (note that 0.75i may not be less than 1.0)

MA

=

resultant moment due to sustained (primary) loads, in-lb

=

[Mx2 + My2 + M z 2 ]1/2

=

basic allowable material stress at the hot (operating) temperature, as per Appendix A ofB31.1 Code. Sh is roughly defined as the minimum of:

Sh

1)

1/4 of the ultimate tensile strength of the material at operating temperature;

2)

1/4 of the ultimate tensile strength of the material at room temperature;

3)

5/8 ofthe yield strength ofthe material at operating temperature (90% of the yield stress for austenitic stainless steels);

4)

5/8 ofthe yield strength of the material at room temperature (90% of the yield stress for austenitic stainless steels); and

5)

100% of the average stress for a 0.01 % creep rate per 1000 hours.

1-46


Expansion: iMc

z Where: SE

=

expansion stress range, psi

Mc

=

resultant range ofmoments due to expansion (secondary) loads, in-lb

SA

= =

Allowable expansion stress, psi

Sc

basic allowable material stress at the cold (installation) temperature, as per Appendix A ofB3!.1 Code.

Occasional: Soce

=

z

Pdo

0.75iMB

0.75i MA +

z

+

4t

Where: Soce = occasional stresses, psi MB

=

resultant moment due to occasionalloads, in-lb

k

=

occasionalload factor

=

1.2 for loads occurring less than 1% of the time

=

1.15 for loads occurring less than 10% of the time

1.5.4 831.3 Chemical Plant and Petroleum Refinery Piping

Sustained: B31.3 does not provide an explicit equation for sustained stress calculations, but only requires that the engineer compute the longitudinal stresses in the pipe due to weight and pressure, and then ensure that these do not exceed Sh. This is most commonly interpreted to mean:

Fax +

+

z

4t

1-47


Where: Fax

=

axial force due to sustained (primary) loads, lb

Mi

=

in-plane bending moment due to sustained (primary) loads, in-lb

Mo

=

out-plane bending moment due to sustained (primary) loads, in-lb

li> 10

=

in-plane, out-plane intensification factors, as per Appendix D ofB31.3 Code

Sh =

basic allowable material stress at the hot (operating) temperature, as per Appendix A of B31.3 Code. Sh is defined as the minimum of: 1)

1/3 of the ultimate tensile strength of the material at operating temperature;

2)

1/3 ofthe ultimate tensile strength of the material at room temperature;

3)

2/3 of the yield strength of the material at operating temperature (90% of the yield stress for austenitic stainless steels);

4)

2/3 ofthe yield strength ofthe material at room temperature (90% of the yield stress for austenitic stainless steels);

5)

100% of the average stress for a 0.01% creep rate per 1000 hours;

6)

67% of the average stress for rupture after 100,000 hours; and

7)

80% of the minimum stress for rupture after 100,000 hours.

Expansion: [(ii Mi)2 + Cio Mo)2 + 4MT2]1/2

z Where: Mi

=

range of in-plane bending moments due to expansion (secondary) loads, in-lb

Mo

=

range of out-of-plane ben ding moment due to expansion (secondary) loads, inlb

MT

=

range oftorsional moment due to expansion (secondary) loads, in-lb

Sc

=

basic allowable material stress at the cold (installation) temperature, as per Appendix A ofB31.3 Code.

1-48


Occasionsl: The equation for calculating occasional stresses is undefined by B31.3, which simply states that the sum of the longitudinal stresses due to sustained and occasionalloads shall not exceed 1.33Sh. The default interpretation ofthis requirement is to calculate the sustained and occasional stresses independently (as per the equation given for sustained stresses above) and then to add them absolutely. Note the differences between these two codes: 1

-

B31.I intensifies torsion, while B31.3 doesn't.

2

-

B31.3 calculation methods are undefined for sustained and occasionalload cases, while they are explicit for B31.1.

3

-

In its most common interpretation, B31.3 neglects torsion in the sustained case, while B31.I includes it.

4

-

B31.I neglects all forces, while in the default interpretation, B31.3 includes Fax in the sustained case.

5

-

Allowable stresses are different for each code.

6

-

Stress increase for occasionalloads are different for each code.

Note that both codes additionally cite a conservative value of SA, f(1.25S c + O.25Sh), which may be used instead ofthe more liberal allowable off(1.25Sc + 1.25Sh - SI). This is a carry over from pre-computer days, when sustained stress calculations were rarely done, so SI was not known explicitly, and conservatively estimated to be at its maximum allowable level of Sh. Specific requirements of other common codes are shown below as weIl.

1.5.5 ASME Section III, Subsections NC & ND (Nuclear Class 2 & 3)

Sustained:

=

BI Slp + B2 Mal Z < 1.5 Sh

Bl,B2

=

primary stress indices for the particular product under investigation

Slp

=

longitudinal pressure stress

Ma

= =

resultant moment on the cross-section due to sustained (primary) loads rMx2 + M~ + Mz2]112, in-lb

Sh

=

basic aIlowable material stress at the hot (operating) temperature, as per ASME III Code. Sh is roughly defined as the minimum of:

Ssus Where:

= P di2 / (d0 2 - di2 ), psi

1-49


1)

1/3 of the ultimate tensile strength ofthe material at operating temperature;

2)

1/3 of the ultimate tensile strengthofthe material at room temperature;

3)

2/3 of the yield strength of the material at operating temperature (90% of the yield stress for austenitic stainless steels);

4)

2/3 of the yield strength of the material at room temperature (90% of the yield stress for austenitic stainless steels);

5)

100% of the average stress for a 0.01% creep rate per 100 hours;

6)

60% of the average stress for rupture after 100,000 hours; and

7)

80% of the minimum stress for rupture after 100,000 hours.

Expansion: SE

=

i Mc / Z < f( 1.25 Sc + 0.25 Sh ) + Sh - SL

=

resultant range of moments on the cross-section due to variations in loading (usually due to thermal effects)

=

[M~

Where: Mc

SL =

+ M; + Mz2]1I2, in-lb

Slp + 0.75 i Ma / Z (where 0.75 i >= 1.0)

Occasional: The occasional stress equations are: For Service Level C (Emergency): Socc = BI x Slpmax + B2 (Ma + Mb) / Z < 1.8 Sh <= 1.5 Sy For Service Level D (Upset): Socc = BI x Slpmax + B2 (Ma + Mb) / Z < 2.4 Sh Where: Slpmax

= pressure stress due to the peak pressure, psi

Mb

=

=

resultant moment on the cross-section due to occasional (primary) loads [Mx2 + M; + M z2]1/2, in-lb

Sy

=

yield stress ofmaterial at temperature, psi

1-50


1.5.6 831.4 Fuel Gas Piping The B31.4 piping code requires that the engineer calculate and check the sustained, expansion, and operating stress.

Sustained: SL

=

Slp + Sb < 0.75 x 0.72 x SYield

Where: Slp =

Sb

=

the longitudinal pressure stress, psi

bending stress due to sustained loads, psi

Expansion: Se =

(Sb2 + 4 St2 )1/2 < 0.72 SYield

Where: Sb

=

range ofbending stress due to varying loads, psi

St

=

range of torsional stress due to varying loads, psi

=

Mt /2Z

SYield = specified minimum yield stress ofmaterial, psi.

Operating: Sope = FIE a dT - v SH 1 + Se + SL( 1-F ) < 0.9 SYield Where: F

=

% of pipe axial restraint (long buried pipelines are considered to be fully axially restrained, i.e. F = 1; while pipelines above ground on slide plates are not axially restrained, i.e. F = 0

E

=

modulus of elasticity ofpipe material, psi

a

=

thermal expansion coefficient of pipe material, in/in/oF

dT = v

=

temperature change of pipe from ambient, oF Poisson's ratio

1-51


SR

=

hoop stress, psi

=

Pdo / 2t.

Occasional: Soee

* 0.72 * SYield * k

=

Slp + Sb < 0.75

=

longitudinal pressure stress

=

resultant moment due to occasionalloads

=


Where:

1.5.7 831.8 Gas Transmission and Distribution Piping Code Like the B31A code, the B31.8 piping code requires that the engineer calculate and check the sustained, expansion, and operating stress.

Sustained:

=

Slp + Sb < .75 S x F x T

Slp

=

longitudinal stress due to pressure, psi

Sb

=

bending stress due to sustained loads, psi

St

=

torsional stress due to sustained loads, psi

S

=

specified minimum yield strength of pipe material, psi

F

=

Construction Type

SL Where:

1-52


Factor F Wasteland, Deserts, Mountains, Grazing Land, Farmland, Sparsely populated areas and Offshore. Fringe areas around cities, Industrial areas, Ranch or Country Estates. Suburban Housing Developments,

A B C

Shopping Centers, Residential Areas. Multi-Story Buildings are prevalent, Traffic is heavy and where there may be numerous other utilities underground.

D

Temperature Derating Factor T

T

250 or l ess 300 350 400 450

Expansion:

=

(Sb 2 + 4 St2 )1/2 < 0.72 S

Where: Sb =

bending stress due to varying loads, psi

St

range of torsional stress due to varying loads, psi

=

Operating: Sope = Se + SL < S Where: terms are as defined previously.

Occasional: Socc

* SYield * F * T * K

=

SI + Se

=


< 0.75

Where: K

0.60 0.50 0.40

Pipe Temperature deg. F.

1.0 0.967 0.933 0.9 0.867

Se

0.72

all others as defined previously

1-53


1.5.8 Canadian Z1831Z184 Oil/Gas Pipeline Systems

Sustained: SL

=

0.5

* Shoop + SB

~ S

*F * L *J *T

Where: Shoop = hoop stress

=

Pd/2tn

=

resultant bending stress

SB

=

specified minimum yield strength

F

=

design factor

L

=

location factor

J

=

joint factor

T

=

temperature derating factor

Expansion: SE = (Sb2 + 4St2 )112

~

0.72

*S *T

Where: Sb

St

=

resultant bending stress

=

iMb/ Z

=

torsional stress

=

Mt/ 2Z

Occasional:

* Shoop + SB

*F *L *J *T *K

=

Fax / A + 0.5

Fax

=

axial force due to sustained and occasionalloads

A

=

cross sectional area of the pipe

SB

=

resultant bending stress due to sustained and occasionalloads

K

=


Socc

~ S

Where:

1-54


1.5.9 RCC-M C Sustained: SL =

Pdo /4tn+O.75*i*MA/Z$;Sh

Where: p

=

design pressure

do

=

outside diameter of pipe

tn

=

nominal wall thickness

1

=

stress intensification factor

MA = resultant moment

Z

=

(Mx2 + M; + Mz2

=

section modulus

)112

Sh = material allowable at design temperature

Expansion: SE = i Mc / Z $; f(1.25S e + .25Sh ) + Sh - SL Where: Mc =

range of resultant moments due to expansion loads

Sc

material allowable at room temperature

=

Occasional: Sace = P max do / 4tn + 0.75 * i

* (MA + MB ) / Z $; 1.2 * Sh

Where: P max = maximum pressure occurring MB

=

resultant moment due ta occasionalloads

1-55


1.5.10 Stoomwezen Sustained: SL =

P (De - d ) / 4d + 0.75

* i * MA

/ Z
Where: p

=

design pressure

De =

outside diameter

d

=

formula wall thickness

1

=


MA =

resultant moment

Z

=

section modulus

f

=

sustained allowahle, the minimum offive equations (see code for details)

Expansion: SE

=

i

* MB / Z < fe

Where: MB = fe

=

resultant moment due to expansion loads expansion allowahle, the minimum oftwo equations (see code for details)

Occasional: Socc =

SL + 0.75

* i * (MA +

MB) / Z < 1.2f

Where: MB

=

resultant moment due to occasionalloads

1.5.11 Special Considerations of Code Compliance 1

Many of the non-power codes separate the in-plane and the out-of-plane stress intensification factors (and do not intensify torsion). For the power codes the SIF's can he computed for in-plane, out-of-plane, and torsional moments using SIF = 0.9/ h 2/ 3 . For the petrochemical and other non-power codes:

1-56


2

-

li

0.9/h2/3

1·1

0.9/h2/3

li

0.9/h2/3

la

=

0.9/h2/3

la

Miters:

la

Welding tees:

la

Reinforced tees: Unreinforced tees:

0.9/h2/3

li

= = = =

li

=

0.75i o + 0.25

0.9/h2/3 0.75i o + 0.25 0.75i o + 0.25

The power codes do not recognize the extruded weI ding tee, the sweepolet, or the weldolet. The non-power codes do, and if any ofthese fittings are used in a power application, the engineer must consider the validity ofusing the stress intensification factors from the chemical codes. The power codes explicitly define the equation to use for the sustained stresses. The non-power codes do not. The non-power codes do however tell the user to compute the longitudinal stresses due to sustained loads, and B31.3 Interpretation 4-10 issued May 8, 1985 instructed the user to include the axial force term in this longitudinal stress. The power codes explicitly omit this axial force term from the definition of the sustained stress calculation.

3

4

0.75/h2/3

la

= = = =

Bends:

-

5

Power codes do not include pressure stiffening effects on bends, while the petrochemical and related codes do. Note that the power codes use the term 0.75i in the sustained stress equation, while the non-power codes historically have not. In Interpretation 1-34 issued February 23,1981 the B31.3 code permitted its us ers to employ the 0.75 i stress intensification term for sustained and occasionalloads. (CAESAR II provides this as an option.) In Interpretation 6-03 issued December 14, 1987, the B31.3 code permitted its users to ignore the stress intensification term for sustained loads. It is recommended that this latest interpretation be ignored and that i or 0.75i be used as the stress intensification factor for sustained and occasional loads.

6

-

Power codes provide special formulas for reduced branch connections. Nuclear and fossil codes have not come together otl. their interpretations as ofthis time, however. These rules come into effect whenever the branch diameter is less than 0.5 times the header diameter.

7

-

Class 1 piping rules also allow flexibility coefficients to be computed and inserted into reduced branch intersections in order to consider the flexibili ty ofthe branch relative to the header. No other piping code at this time includes this as an option.

8

B31. 3 was the first piping code to instruct the user to rem ove corrosion allowance from the section modulus before making sustained and occasional stress calculations. Other piping codes simply warned of the deleterious effect of corrosion when joined with cyclic loadings. Arguments are that B31.3 is

1-57


instilling a false sense of security when dealing with corrosion in this manner, im plying that B31.3 is leading the user to believe that he is properly considering the effects of corrosion. Other arguments state that B31.3 considers corrosion in the wrong stress calculation, and that if anything it should have been incorporated into the expansion stress calculation. 9

Some ofthe pi ping codes incl ude longitudinal weldjoint efficiency factors in their allowable stress tables. In the majority of the codes where these factors are included, the codes also instruct the user to divide the allowable stress by the joint efficiency before using the table value for flexibility calculations, thus increasing the allowable stress. Note that post-1980 B31.3 codes DO NOT include the longitudinal weld joint efficiencies in the stress tables.

10 -

The piping codes are unanimously silent on the point ofBourdon pressure effects. If included, the pressure will cause some distortion of the piping system. If excluded there will be no displacements due to pressure.

Il -

European piping codes for the most part are formulated differently than the D.S. codes. The Europeans use an effective stress calculation and compare the results directly to an allowable without emphasizing the concern for fatigue. The Swedish piping code does have a provision that allows its users to employ the ASME B31.1-1977 code providing the Swedish allowables are used.

12 -

Almost all piping codes allow the exact expression for pressure stress to be used in place ofPdJ4t in the sustained stress calculations. The exact pressure stress value is:

13 -

Most piping codes also allow the use of a increased section modulus for the stress calculations at the branch end of a reduced intersection. The reduced section modulus is calculated as:

z =Cp) Cr2) te Where:

=

r

mean radius of the branch, in lesser of tnh or i tnb, in

14 -

tnh

=

nominal wall thickness of header, in

tnb

=

nominal wall thickness ofbranch, in

Note that the B31.3 and related piping codes do not intensify the torsional stress term in the expansion or sustained stress equations, while the power codes do. This is considered an oversight on the part of the code as Markl's tests clearly indicate that the torsional moment should he intensified. This was con:firmed by the research documented in WRC 330.

1-58


15 -

Stress indices are used in ASME Section III, Class 1, 2 and 3 piping codes. There are three different indices: B

Related to gross plastic deformation (sustained stress intensification factor).

C

Gives the magnitude of the primary plus secondary stress range (elastic shakedown stress intensification factor).

K

When used with C, gives the magnitude of the primary plus secondary plus peak stress range (C*K is the fatigue stress intensification factor). There are three subscripts used with the stress indices: 1

Used for pressure loadings

2

Used for moment loadings

3

Thermal gradient loadings

As a rule ofthumb, 2i =C2 * K2, where i is the stress intensification factor for the

B31 codes (as discussed above). 16 -

In almost aIl cases, the cold modulus of elasticity and nominal dimensions are to be used in the flexibility analysis of piping systems. Using the cold modulus produces larger, and therefore more conservative stresses. BS 806 and ASME Class 1 codes have provisions for using material properties in the hot condi tion. NEMA SM23 also provides for using the hot modulus of elastici ty for evaluating loads on turbine nozzles.

1.5.12 Evaluation of Multiple Expansion Range Cases It is often the cases that the tempe rature ofthe piping system is not constant throughout the operating cycle, or there is potentially more than one operating cycle (i.e., pump A on, pump Bon, both pumps on, both pumps oro.

In these cases it is common to find that the tempe rature rises on sorne occasion to a maximum value, say Te; then, as events occur during the normal course of operation the temperature varies between Te and other lower temperature states, say Tl, T2, ... , Tn. In these cases the piping codes have provided a simplified method by which the cumulative damage due to the various thermal cycles may be evaluated by converting reduced thermal expansion cycles into equivalent full temperature cycles. The user will find that cumulative damage rules usually become important only either the number ofthermal variations is large, or when the magnitude of the temperature variation is a large percentage of the maximum design temperature expected. The following rules should be followed when evaluating systems with multiple temperature variation cycles:

1-59


•

Te should be selected as the highes t operating tem perature of the pi ping system, even if the startup cycle does not go directly to this temperature.

•

The expansion allowable stress should be based on Te, i.e. SA should be calculated from Sh for temperature Te.

•

The range dTe is determined as the difference between Te and the ambient temperature. Ne should be estimated as the total number oftimes during the life of the unit that the temperature will be expected to vary from ambient to Te.

•

The temperature ranges between Te and each of the other reduced temperature states should be calculated, i.e.:

dTn •

=

The number of cycles associated with each operating mode should be estimated as: Temperature change dTl occur Nl times, Temperature change dT2 occur N2 times, ... , Temperature change dTn occurs Nn times

•

The total number of equivalent full temperature cycles that these partial cycles represent can be calculated as:

•

The cyclic reduction factor f should be selected based upon the number of equivalent cycles, N, while other components of SA and SE should be based upon tempe rature Te. Example: A particular process line varies in temperature as the quality of the catalyst varies. The particulars of the operation are outlined below:

Ambient = 700 F Startup goes to 5600 F It is estimated that the maximum temperature ever required will be 650 0 F and the minimum temperature required during operation will be 430 oF. The temperature will fluctuate between 560 0 F and 650 0 F perhaps 10 times per day, and between 560 0 F and 450 0 F perhaps 5 times per day. The design life ofthe unit

1-60


is 12 years, and it is estimated that the unit will be shut down at least once each month for maintenance. •

Te should be selected as the highest operating temperature ofthe piping system. In this case, it is equal to 650 0 F.

•

The range dTe is determined as the difference between 650 0 and the ambient temperature of70 0 F, so dTe = 580 0 F. The estimate of Ne, the total number of times that the temperature will be expected to vary through this range is: Ne

= 1 shutdown/month X

12 months/yr X 12 yr

= 144

•

The temperature ranges between Te and each ofthe other reduced temperature states are:

•

The number of cycles associated with each operating mode are: NI = 10 times/day X 365 days/yr x 12 yr = 43800 N2 = 5 times/day x 365 days/yr x 12 yr = 21900

•

The total number of equivalent full temperature cycles is: N = 144 + (90/580)5 x 43800 + (200/580)5 x 21900 = 255

•

The cyclic reduction factor fis selected based upon 255 cycles, so f =1. 0 (for fewer than 7000 cycles). As noted, the material allowable stresses SA and Sh are based upon 650 0 F, and the expansion stress, SE, is calculated for the system operating at 650 0 F. Warning: These cumulative damage rules don't fully address those cases where one part of the piping system stays at Te while another part ofthe piping system undergoes a temperature fluctuation. In these cases it is common to simply analyze each case separately. The ASME Section III, Subsection NB (Nuclear Class 1 Piping) Code provides rules which may he followed should the user be concerned about the cumulative damage where different parts of the piping system cycle through different temperature states. The requirements are described below:

Cumulative Damage: If there are two or more types of stress cycles which produce significant stresses, their cumulative effect shan be evaluated as stipulated in Steps 1 through 6 below: 1

Designate the specified number oftimes each type of stress cycle oftypes 1,2,3, ... , n, will be repeated during the life of the component as nI, n2, ng, ... , nn, respectively. In determining nI> n2, ng ... , nn consideration shall be given to the

1-61


superposition of cycles of various origins which produce the greatest total alternating stress range. For example, if one type of stress cycle produces 1000 cycles of a stress variation from zero to +60,000 psi and another type of stress cycle produces 10,000 cycles ofa stress variation from zero to -50,000 psi, the two cycles to be considered are shown below: (a) Cycle type 1: nl=1000; and Sa1tl=(60000+50000)/2 (b) Cycle type 2, n2=9000; and Salt2=(50000+0)/2 2

-

For each type of stress cycle, determine the alternating stress intensity Salt, which for our application is one half of the range between the expansion stress cycles (as shown above). These alternating stress intensities are designated as Saltl , Sa1t2, ... , Saltn.

3

-

On the applicable design fatigue curve fmd the permissible number of cycles for each Salt computed. These are designated as NI, N2, ... , N n.

4

-

For each stress cycle calculate the usage factors VI, V2, ... , Vn, where VI = nl/ NI, V2 = n21N2, ... , V n = nnlNn· Calculate the cumulative usage factor V as V = VI + V2 + ... + Vn.

5 6

-

The cumulative usage factor shall not exceed 1.0.

1-62

2


2.0

Piping Design For Loading Types ............................................................................. 1

2.1

Designing For Sustained Loads - Pressure ............................................................ 2 2.1.1 Minimum Wall Thickness Requirements ...................................................... 2 2.1.2 Pressure Design ofElbows and Miters .......................................................... 4 2.1.3 Pressure Design of Flanges and Blanks ........................................................ 5 2.1.4 Pressure Design of Branch Connections ....................................................... 6 2.1.5 Restraint ofUnbalanced Expansion Joint Pressure Loads ....................... 8-9

2.2

Designing For Sustained Loads - Weight ............................................................. 10 2.2.1

Calculation ofWeight Stresses .................................................................... 10

2.2.2 Use of Standard Weight Spans .................................................................... 13 2.2.3 Consideration of Nozzle Loads .................................................................... 19 2.3

Designing For Expansion Loads ............................................................................. 22 2.3.1 Magnitude of Thermal Load ........................................................................ 22 2.3.2 Guided Cantilever Method .......................................................................... 24 2.3.3 Refining the Model Through the Use ofRestraint Stiffnesses ................... 26 2.3.4 Use of Expansion Loops ............................................................................... 27 2.3.5 Simplified Expansion Stress Check ........................................................ 29-30 2.3.6 Stress Reduction through Use of Expansion Joints .................................... 30 2.3.7 Expansion Stress - Other Solutions ..................................................... 33-33

2.4

Ranger Design ....................................................................................................... 34 2.4.1 Variable Spring Ranger Design Basics ....................................................... 35 2.4.2 Load Variation ............................................................................................. 37 2.4.3 Ranger Selection Table ............................................................................... 37 2.4.4 Ranger Design Process - Restrained Weight, Free Thermal, and More .. 39 2.4.5 Restraint Placement Using Distance to First Rigid Criteria ..................... 40 2.4.6 Notes on Ranger Design .............................................................................. 43 2.4.7 CAESAR II Ranger Design Control and Options .................................. 45-49

2.5

Designing For Occasion al Loads (Static Equivalent of Dynamic Loads) ............... 50 2.5.1 Wind Loading ............................................................................................... 50 2.5.2 Earthquake Loading ............................................................................... 54-56 2.5.3

Quickly Applied Loads ................................................................................. 56


2.0 Piping Design For Loading Types As described in Section 1.0, the pipe stress analyst is concerned with two types ofloadsprimary and secondary. Not only are the causes and the failure modes ofthese two loading types quite different, but not surprisingly, the solutions to these two types ofloading are usually different as weIl. In fact, the solution to a problem caused by one of the loading types often causes a problem with the other loading type. Therefore, a compromise must often be reached in order to find the solution to these two types ofloading.

Note that primary loads are usually classified further, according to their duration ofloading. Those primary loads which are nearly always present throughout operation are called sustained loads, while those which occur less frequently are called occasionalloads. The methods ofresisting these two types ofloads are similar, with the main difference beingfound in the use of a higher allowable stress for occasionalloads (as seen in Section 1).

2-1


2.1 Designing For Sustained Loads -

Pressure

AlI piping systems must be designed to withstand sustained loadings. Sustained loads are classified as those caused by mechanical forces which are present throughout the normal operation of the piping system. Therefore the se loads: •

Are force driven, as opposed to displacement driven, and

•

Are present for relatively extended periods of time, as opposed to those which change dynamically.

Typical sustained loads consist of: •

Pressure -loads due to operating (or design) pressure,

•

Weight - uniform loads due to the weight ofthe pipe, fluid, and insulation, and concentrated loads due to the weight of in-line components (such as valves, flanges, etc.), and

•

Spring hanger pre-Ioads and other applied forces.

2.1.1 Minimum Wall Thickness Requirements Since hoop pressure stresses are approximately twice as large as longitudinal pressure stresses, pipe wall thicknesses are initially sized for hoop stresses. Because ofthis, pressure design of components is usually done far before, and therefore in isolation, from the pipe stress analysis phase of piping design. Because of this, pipe stress software such as CAESAR II does not normally handle this part ofthe design effort. A discussion of pressure design of components is included here for the sake of completeness, and is based upon an amalgam of the requirements of various codes. Note that pressure design of piping components must be done according to the requirements of the user's specifie code, not to the rules described here! Because the pipe wall is sized for the large hoop stress, this usually provides sufficient margin between the allowable stress and the longitudinal pressure stress to accommodate the weight stresses. The requirement for the minimum pipe component wall thickness is:

=

t +c

tm

=

minimum wall thickness, in

t

=

minimum wall thickness required for pressure, in

c

=

sum of allowances for thread or groove depth, corrosion, erosion, and manufacturer's tolerance, in

tm Where:

2-2


For thin wall (t < D/6), straight pipe under internaI pressure, t may normally he calculated, through various approximations of Lame's equation, as: t

=

PD 1 2(SE + PY), or:

t

PD 1 2SE, or:

t

= =

(D/2) x (1 - [(SE - P) 1 (SE + P)]1/2), or:

t

=

P (Di + 2c) 1 [2(SE - P(l-Y)]

P

=

design pressure, psig

D

=

outside diameter, in

Di

=

inside diameter, in

S

=

basic allowable stress at design temperature, psi

E

=

casting or longitudinal weld quality factor (typically ranges from 0.8 to 1.0)

y

=

material coefficient, with a value (depending upon the specific code requirements) to he interpolated from:

Where:

Temperature. oF Material

<=900

950

1000

1050

1100

1150

1200

>1250

Ferriti c

0.4

0.5

0.7

0.7

0.7

0.7

0.7

0.7

Austenitic

0.4

0.4

0.4

0.4

0.5

0.7

0.7

0.7

Nickel All oys

0.4

0.4

0.4

0.4

0.4

0.4

0.5

0.7

Other ductil e

0.4

0.4

0.4

0.4

0.4

0.4

0.4

0.4

Cast iron

O.

Requirements for pressure design of other piping components are described in the following sections. (For B31.3 y = 0.0, for B3l.l y =.4. The CAESAR II check uses 0.4 for ail codes except B3l.3, where y = 0.0.)

2-3


2.1.2 Pressure Design of Elbows and Miters When using elbows, the minimum wall thickness after bending shall not fall below that calculated for straight pipe. For mitered elbows, the maximum allowable pressure is calculated differently depending on whether the angle of the miter cut is less than or greater than 22.5°. For 0 < 22.5 0 , the allowable maximum pressure is the lesser of: Pm

=

[SE(T - C)/r2]

X

[(T - c) 1 (T - c + 0.643 tan 0 (r2(T-c»1/2)]

or:

For 0 >= 22.5 0 , the allowable maximum pressure is:

=

[SE(T - C)/r2] x [(T - c) 1 (T - c + 1.25 tan 0 (r2(T-c»I/2)]

Pm

=

maximum allowable internaI pressure for miter, psig

T

=

minimum pipe wall thickness, in

r2

=

mean pipe radius, in

RI

=

effective radius of miter bend (defined as the shortest distance from pipe center line to the intersection of the planes of adjacent miter joints - see Figure 2-1), in

o

=

angle of miter cut (see Figure 2-1), degrees

Pm Where:

I - - H - - - R,,------i

Figure 2-1

2-4


2.1.3 Pressure Design of Flanges and Blanks Pressure design offlanges is a complex task, requiring consideration ofthe configuration and materials of the flange, bolts, and gasket. Potential causes of failure are bending stresses in the flange, localized stress concentrations in the hub, yielding of the bolts, or unloading of the gasket, causing leakage. Design offlanges is covered in detail in Section VIn of the ASME Boiler an Pressure Vessel Code; however, due to the complexity, it is rarely done by the pipe stress engineer. Instead, the most common piping codes endorse the use offlanges conforming to recognized standards such as ANSI B16.5 "Pipe Flanges and Flanged Fittings". This standard designates standard pressure classes of flanges, which are recognized by the codes to be acceptable for the following combinations of pressure and temperature:

Design Pressures (psig) for Flange Pressure Classes Pressure Class Temperature. oF 150

300

400

600

900

1500

100

275

720

960

1440

2160

3600

150

255

710

945

1420

2130

3550

200

240

700

930

1400

2100

3500

250

225

690

920

1380

2070

3450

300

210

680

910

1365

2050

3415

350

195

675

900

1350

2025

3375

400

180

665

890

1330

2000

3330

450

165

650

870

1305

1955

3255

500

150

625

835

1250

1875

3125

550

140

590

790

1180

1775

2955

600

130

555

740

1110

1660

2770

650

120

515

690

1030

1550

2580

700

110

470

635

940

1410

2350

750

100

425

575

850

1275

2125

800

92

365

490

730

1100

1830

850

82

300

400

600

900

1500

900

70

225

280

445

670

1115

950

55

155

220

310

465

770

1000

40

85

160

170

255

430

2-5


A more detailed discussion of flange analysis, with specific regard to determination of leakage under load, is provided in Section 3 of these seminar notes.

Blanks are designed based upon formulas for the calculation ofbending stresses for plates under pressure loading. The minimum thickness for a blank is calculated as: tm

=

dg [3P / 16SE]1/2 + C

=

inside diameter of gasket for raised or flat face flanges, or gasket pitch diameter for ring joint and fully retained gasketed flanges, in

Where: dg

2.1.4 Pressure Design of Branch Connections A pipe having a branch connection is weakened by the opening that is cut in it, so it may be necessary to provide reinforcement to replace the metal removed from the wall thickness at the opening. A typical fabricated tee is shown in Figure 2-2.

Limitlof

1

Nominollhickn...

M-_MiI__I I=__ -i-______ --:J~~t=-"-----..... pipe

--r--

1

ornozzr. Thicknes:ll. measured specification

.1 i

- - - - - - - - - t Pipe

------

Figure 2-2 For fabricated tees, with the angle between branch and header of at least 450, the area required to replace the area of the opening is calculated as:

2-6


Where: Al

=

area required to be replaced, in2

th

=

pressure design thickness ofheader pipe, in

dl

=

effective length of pipe wall removed from header at intersection, in

~

=

sm aller angle between axes ofbranch and run, degrees

This required area must he exceeded by the total available reinforcement area, or:

Where: A2

=

area resulting from excess thickness ofheader pipe, in2

d2

=

half-width of reinforcement zone, in

=

(Tb - c) + (Th - c) + dl/2, but not less than dl

Th

=

minimum wall thickness ofheader, in

Tb

=

minimum wall thickness of branch, in

Ag

=

area resulting from excess thickness ofbranch pipe, in2

L4

=

height ofreinforcement zone outside ofheader, in

=

lesser of 2.5(Th - c) or 2.5(Tb - c) + Tr

tb

=

pressure design thickness ofbranch pipe, in

Tr

=

minimum thickness ofreinforcing ring or saddle, if any, in

~

=

area ofwelds and reinforcement provided for the intersection within the area of reinforcement as defined as a parallelogram extending a distance of d2 on either side of the centerline of the branch, and from the inner wall of the header pipe to a distance ofL4 along the axis of the branch, measured from the outside of the header pipe.

2-7


2.1.5 Restraint of Unbalanced Expansion Joint Pressure Loads Pressure usually only creates stress in the pipe, rather than loadings on supports/restraints, because pressure loads are neutralized at the cross-section by the tension in the pipe wall. One exception to this is when the pipe is not continuous from anchor to anchor, such that tension is not present in the pipe wall at aIl locations of the system. (Note that a second exception occurs when the Bourdon effects of pressure are considered. The Bourdon effect is due to the axial extension of pipes either under high pressure or in long runs, causing displacements which must be absorbed by the piping system. Since this is a displacement load, it is a secondary load, and therefore is not considered here.) Tension in the pipe wall is not continuous when there are expansion joints or slip joints present in the system. These types of components are too flexible in the axial direction to transmit the pressure force, therefore the unbalanced pressure load must be handled by either tie rods or external pipe restraints. The unbalanced pressure load is calculated as: Fp

Ae

=

P

Fp

=

Pressure force, lb

P

=

Design pressure, psig

Ae

=

Effective area of expansion joint

De

=

Effective diameter of expansion joint, in

=

internaI diameter of pipe + depth of one corrugation, in

Where:

When using restraints to absorb the unbalanced pressure load, it is recommended that guides be located on the adjacent pipe runs in order to reduce the danger ofbuckling. The Expansion Joint Manufacturers Association recommends that the first guide be placed a distance no further than 4 pipe diameters from the expansion joint, with the second guide placed no further than 14 pipe diameters from the first. Figure 2-3 shows some typical piping layouts using expansion joints.

2-8

COf\DE Pipe Stress Analysis Seminar Notes

* O=Pipe ~-+--Vertical

....

0.0.

Support

/,f-----~-

/Pipe

Anchor

Expans~_11

Joint

--; C

~ Anchor

"" 1st Guide

'-----~E

Ail Other Guides

Figure 2-3 More information on the use ofexpansion joints is foundin Section 2.3.6 and Section 3 ofthese seminar notes.

2-9


2.2 Designing For Sustained Loads - Weight 2.2.1 Calculation of Weight Stresses Stresses due to weight loads acting on a supported pipe can be estimated through the use ofbeam theory. The simplest method of estimating pipe stresses due to weight is to first consider the pipe as being a continuous run, with supports located at constant intervals (this is a somewhat accurate model ofpiping traveling horizontally, mounted on racks, and with a minimum ofin-line components):

5

l l ~e

l L

·1· e--+-e~ Figure 2-4

Elementary beam theory can be used to determine stresses in a member due to loading on that member. Normally the member considered is one-dimensional, homogenous with respect to cross-sectional and material parameters, and restrained in a number of degreesof-freedom atone or bothends. This model can only be usedifthe effects examined are limited totwo adjacent support points and the straightrunofpipebetweenthose support points. The question is what beam stress equation should be used? Bearn theory states that ifboth support points are pinned (free to rotate):

w JJHHHHHBB BBUBBB!!. Figure 2-5 The maximum moment in the beam is in the center of the span, and has a value of:

=

w1 2/8

Mmax

=

maximum moment in the beam, in-lb

w

=

uniform weight of pipe, fluid, insulation, etc., lb/in

l

=

length of beam, in

M max

where:

2-10


Ifboth ends are fixed, or rigid (restrained against rotation):

~ 11/1II1Il!!!!Il III I!! 11111 Il ~ Figure 2-6 The maximum moment is at the ends ofthe span, and has a value of:

Mma

= w1 2/12

Which formula is more appropriate? Examining a typical pipe support detail:

@

@

Figure 2-7 The clamp/pin/rod hardware allows rotation of the pipe, therefore simulating a pinned connection. However, if an spans are of identicallength and loading, the reaction of the adjacent pipe span prevents rotation at the support, therefore simulating a fixed connection. The true condition is somewhere in between, so a compromise approximation is reached:

Mmax = w1 2/10 with the location of the maxim um moment being somewhere between the ends and the center (i.e., anywhere) on the span.

2-11


Ofcourse, there sometimes are concentrated loads (valves, flanges, etc.)in the pi ping system. The effect ofthese on the pipe stresses can he estimated as weil. For pinned connections: p lOI

"!"

a

C*J

LS.

e

1 ..

b----j h

.1

Figure 2-8 The maximum moment is located at the point ofloading, and has a value of:

=

Pab!l

a

=

longer portion of span, in

b

=

shorter portion of span, in

Mmax Where:

For fixed connections:

..-t-I-----

a -----t---

~I--------e------~

Figure 2-9 The maximum moment is located at the end nearer to the load, and has a value of: Mmax

=

pa2b!l2

In either case (or actually some case in between), the additional stress (MIZ) due to concentrated loads should be added to the stress from the uniform load in order to determine the total stress in the pipe. Or, examining the formulas above, it is evident that, as the shorter portion of the span (b) approaches zero length, the moment, and therefore the stress, approach zero as weIl. This points to an important rule of design - if supports are placed as near as possible to concentrated loads, the effect ofthese loads from a stress point ofview may be neglected. (They must still be considered for the support loads, of course.)

2-12


2.2.2 Use of Standard Weight Spans Implementation ofthe preceding analysis provides a simple way to design for weight loading. The engineer may first support all concentrated loads in the system as closely as possible, reducing the stresses due to those loads to near zero. Next, converting the formula Mmax = w1 2/10 into its corollary:

=

(10 Z SalI / w)1/2

LalI

=

allowable pipe span for weight loading, in

Z

=

section modulus of pipe, in3

SalI

=

approximate allowable stress ofpiping material for weight stresses (Sh, less pressure stresses, divided by intensification factor, for example), psi

LalI Where:

If the piping system is then supported, such that no straight span exceeds Lall, the engineer can be sure that allowable weight stresses are not exceeded in the system, and no analysis per se need be done. In order to save even the brief time required to calculate LaU, the Manufacturer Standardization Society of the Valve and Fitting Industry has calculated allowable piping spans for various piping configurations, and published them in their standard MSS SP-69 (Figure 2-10). They have calculated the maximum allowable piping weight spans based upon the following criteria: 1

the pipe is assumed to have standard wall, with insulation,

2

the maximum moment is calculated as Mmax =wI 2/10,

3

-

no concentrated loads are present,

4

-

there are no changes of direction in the spans, which are assumed to run in the horizontal plane,

5

-

the maximum allowable stress is assumed to be 1500 psi, combined bending and shear,

6

-

maximum deflection of the span under load is limited to 0.1", and

7

-

stress intensification factors of components are not considered.

Due to the low allowable stress value used, there is sufficient factor of safety that this standard span may he applied to a wide range of piping configurations. If the engineer supports a piping system such that no span in the system exceeds the standard spans listed in the table, it is virtually certain that the system is adequately supported for weight loading. However, it is rare that a piping system has no concentrated

2-13


loads, consists of only horizontal runs with minimal changes in direction, etc. Therefore, standard practice dictates that standard spans be applied subject to the following four caveats: 1

Supports should be located as close as possible to concentrated weights. The theoretically best location for a support is directly on the concentrated load; however, this is usually impractical.

2

A developed length of 3/4 of the standard span or less should be used when the piping run changes direction in the horizontal direction, in order to minimize eccentric moments. The theoretically best location for a support is on an elbow; however, this is not recommended due to the bend stiffening and increased local stresses associated with attachments on a bend.

2-14

TABLE 3.

o

MAXIMUM HORIZONTAL PIPE HANGER AND SUPPORT SPACING 3:

~

~....

'"d

1 NOMINAL PIPE OR TUBESIZE

t\:)

..... 1

01

l' ~

.....•

Q

4

J

5

COPPERTUBE

WATER SERVICE

VAI'OR SERVICE

ft

m

ft

m

ft

ft

m

1/4

7

2.1

8

2.4

5

I.S

5

I.S

J/8

7

2.1

8

2.4

5

I.S

6

1.8

1/2

7

2.1

8

2.4

5

I.S

6

1.8

WATER SERVICE

m

VAPOR SERVICE

7

2.1

9

2.7

5

I.S

7

2.1

1

7

2.1

9

2.7

6

1.8

2.4

1 1/4

7

2.1

9

2.7

7

2.1

8 9

2.7

1 1/2 2

9

2.7

12

J.7

8

2.4

.10

3.0

10

3.0

\J

4.0

8

2.4

21/2

Il

J.4

14

4.3

9

2.7

J/4

~

2

STD WT STEEL PIPE

FIRE OUCTILE PRO· ..ON TECTION PAESSlIRE

::lé! or-

:~ tri

1":11:1

g;.a as

i~

-tri

~~ ut

3

12

3.7

15

4.6

10

3.0

II 13 14

J 1/2

13

4.0

16

4.9

Il

3.4

15

4.6

4

14

4.3

17

5.2

12

3.7

16

4.9

5

16

4.9

19

5.8

13

4.0

18

5.5

6

~ r-

17

5.2

21

6.4

14

4.3

20

6.1

:!l

8

19

24

7.3

16

4.9

23

7.0

10

22

26

7.9

18

5.5

25

7.6

19

5.8

28

8.5

12

23

5.8 6.1 7.0

30

9.1

14

25

7.6

32

9.8

16

27

8.2

35

10.7

18

28

8.5

37

20

30

9.1

39

11.3 Il.9

24

32

9.8

42

12.8

30

33 10.1

44

13.4

3.4

6 psrlRoru

°"t'I

~ooj.~t.I zO~

1'1--

...

. ::1:

",

~ X

1ft III",

>:: "'n

~z

~e ~S; ~~

'"III

i ~

~

~

zZO ...... ...

. ::1:>

l'IX

III'"

>:: "'n ~z

>!i

~

Si

800j0°

~I 00

"'0

j!g

ASBESTOS CEMENT

ra

1ft

2:

CAST IRON SOIL

ra

4.J

:z:

8

Q~~ l''~;:' goi 5°9 ozz:z:- 300jiC

4.0

ooj

7

>"t'I n2: ::1: tri

Si!!3 r::~

~:s 2:;g

>g; ~Q

~~

>"t'I

... 0

~g

a~ trl

z OC') "1113 g

a~

ai! 2: tri

>'"

ilia

~a 1110

~n

~n

"'2

QG ~

"'2:

Qa

r-

~

'"III

=;; a: >

1ft

Z

!ii >

~ )III

tri

)III

rA )III

; !C

~

~

i

'"C CJ

10

Il

GLASS

PLASTIC

FIBERGLASS RJ:IN· FORCED

~OD

en ..,

l'I~

:IIIr-

§~ 0-

za: -> !",X

'"

j ~

1=

°1:lE > Z c ~

Q

"'~ triO

~i

~6 ."lE

8~ Zl"

"'> "'z >c ooj.., c> 'i:!:l c )III

...

tri

)III

rA :11:1

$q

i

~

rA )III

{3 CI

~

!C

!'>

c)III

~

i~

i ~

Q

~

:1 '"rA )III

~~

-z Oc z_

1

0

)III

~

~ ~

~

00 00

j....

:IIIr-

!è!;:: '"III!C

r:n

~

1'10

ooj:!

(1)

~

~

00

r:n

(1)

~

> zg >

'"

10

."

~

i

'""II i ~ 0

C'2:)

> ~

. NOTE: (1) FOR SPACING SUPPORTS INCORl'ORATING TYPE 40 SHIELDS, SEE TABLE S. (2) OOB NOT APPLY WHERE SPAN CALCULATIONS ARE MADE OR WHERE THERE ARE CONCENTRATED LOADS BE1WEEN SUPfORTS SUCH AS FLANGB, VALVES, SPECIALTIES, ETC., OR CHANGES IN DIRECTION REQUiRING ADDmONAL SUPI'ORTS.

~

s.I:S

~

Z

~

00


3

-

The standard span doesn't applyon risers, since no moment (and thus no stress) develops regardless of the riser length. The number and location of supports should be determined by the location and strength of building steel. However, it is preferable to locate supports above the center ofgravity oflong risers in order to prevent toppling.

4

-

Support locations should be selected as close to building steel as possible in order to simplify support configuration.

The steps involved in supporting a piping system for sustained loads can he illustrated with an exam pIe. In Figure 2-11, the system consists of a 12" diameter, standard schedule steel pipe filled wi th water, wi th a design pressure of 150 psi, and a design tem perature of 3500 F, which runs hetween two equipment nozzles. The engineer first must determine the standard span for the system. For 12" diameter, water filled pipe, the standard span is shown in MSS SP-69 to he 23 feet. For changes of direction, 3/4 of this span is 17 feet-4 inches. Next, the engineer locates supports. The first concern is to locate them near concentrated loads - supports should he located as close as possible to the two valves (for example, near node points 20 and 70). The first ofthese is optional, depending on whether the nozzle at node point 10 is assumed to act as an anchor, and whether it is desirable to minimize the nozzle loads on the equipment. The next concern is the placement of supports on the riser. Assume that the capacity of the building steel dictates that the weight of the riser be split hetween two supports. It is recommended that one ofthese be placed above the center ofgravity ofthe riser (for example, 15 feet below the top of the riser). Now supports can be located elsewhere in the system, starting at the nozzle at node point 10. A support was located near node point 20 earlier; we now want to locate the next one downstream within the standard span. It is evident that pipe changes direction within 23 feet, so the developed length to the next support should be maintained as less than 17 feet4 inches. The next run ofpipe accommodates a full 23 foot run, so two supports can be located between node points 30 and 40. The line of action ofthe supports on the riser provide support to the end of the horizontal 30-40 run, so no additional support is required at node point 40. Support locations can he continued to he selected in this manner until alilocations meet the selection criteria; one solution is shown in the Figure 2-12. Once completed, what does this accomplish? By using the standard span criteria, the engineer can assume that the maximum stress in the piping system due to weight loading does not exceed 1500 psi. Therefore, substituting this value for the weight component ofthe stress equation: Ssus

=

PA/Am + 1500 = 150(113.1)/14.58 + 1500 = 2664 psi < 20,000 psi

2-16


12" DIA - STD SCH PIPE MAT'L - A106 GR B FLUID - WATER PRESSURE - 150 P~ TEMP - 350 DEGREES F ELBOWS - LONG RADIUS INSULATION - 2" CS VALVES - 150# GATE VALVES (WT =826#) NOZZLES (ANCHOR POINTS) @1 0 & 90 Sc = 20,000 PSI Sh = 20,000 PSI THERMAL CYCLES <7000

Figure 2-11 12" DIA - STD SCH PIPE MAT'L - A106 GR B FLUID - WATER PRESSURE - 150 PSI TEMP - 350 DEGREES F ELBOWS - LONG RADIUS INSULATION - 2" CS VALVES - 150# GATE VALVES (WT=826#) NOZZLES (ANCHOR POINTS) @10 & 90 Sc = 20,000 PSI Sh = 20,000 PSI THERMAL CYCLES <7000

PROPOSED HANGER LOCATION

Figure 2-12

2-17


Piping sag is not a problem, since dis placement is limited to 0.1 inches.

Therefore the engineer has demonstrated that this piping system meets the sustained stress criteria, without having to do any actual "work". This can be confirmed by actually doing an analysis of the supported system. The results in Figure 2-13 show that the maximum sustained stress actually calculated for the configuration shown in Figure 2-12 by CAESAR n is 2418 psi, showing that the shortcut analysis is reasonably accurate, yet conservative. The CAESAR n analysis also shows a maximum vertical dis placement under weight of 0.0046", which is also conservative.

LICENSED TO: TETRACOM SERVICES CAESAR II DISPLACEMENT REPORT CASE 2 (SUS) W+P1 NODE 10 20 22 28 29 30 34 36 38 39 40 44 46 48 49 50 55 60 70 72 78 79 80 85 90

10/1 13269 Ver 3.18

PAGE:1

FILE:SUPT01 DATE:NOV 4.1992

- - - - -Transl ati ons (in. ) - -------Rotations(deg. ) - DX DY OZ RX RY RZ .0000 .0000 .0000 .0000 .0000 .0001 .0005 .0024 .0034 .0034 .0032 .0009 -.0014 -.0024 -.0024 -.0023 -.0015 -.0006 -.0005 -.0004 -.0001 .0000 .0000 .0000 .0000

.0000 .0000 .0000 -.0040 -.0046 -.0039 .0000 .0000 -.0012 -.0008 -.0005 .0000 .0000 -.0005 -.0009 -.0016 .0000 -.0024 -.0010 .0000 -.0001 .0000 .0001 .0000 .0000

.0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 -.0002 -.0009 -.0067 -.0073 -.0012 -.0004 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000

2-18

.0000 .0000 .0010 .0024 .0034 .0052 -.0065 .0075 .0004 .0030 .0027 .0013 -.0012 -.0032 -.0041 -.0017 -.0010 .0042 .0042 .0029 -.0003 .0007 .0004 .0004 .0000

.0000 .0000 .0000 .0000 -.0001 -.0002 -.0003 -.0004 -.0004 -.0005 -.0005 -.0005 -.0005 -.0004 -.0004 -.0004 -.0004 -.0003 -.0003 -.0003 -.0002 -.0001 .0000 .0000 .0000

.0000 .0000 -.0017 -.0044 -.0031 -.0024 -.0022 -.0015 -.0012 -.0011 -.0010 -.0007 -.0004 -.0003 -.0002 -.0002 -.0001 .0000 .0000 .0000 .0000 .0003 -.0010 -.0012 .0000


LICENSED TO: TETRACOM SERVICES CAESAR II STRESS SUMMARY CASE 2 (SUS) W+P1

10# 13269 Ver 3.18 FILE:SUPT01 DATE:NOV 4,1992

PAGE: 10

*****CODE STRESS CHECK PASSED PIPING CODE: B31.3 (1990) HIGHEST STRESSES: (lb./sq.in.) CODE STRESS %:

2418.

@NODE

34

BENDING STRESS:

1255.

@NODE

34

54.

@NODE

22

AXIAL STRESS:

1300.

@NODE

46

3D MAX INTENSITY:

2631.

@NODE

22

TORSIONAL STRESS:

ALLOWABLE:

20000.

Figure 2-13 A further implication of this approach is that in order to eliminate a stress or deflection problem due to weight loadings, the best solution is usually to reduce the unsupported span of the piping- i.e., add more supports.

2.2.3 Consideration of Nozzle Loads The previous discussion has primarily concerned the effect of supports on system stresses. The engineer is also interested in determining loads on supports and nozzles, in order to select the appropriate support hardware, to check the overloading of equipment, or to calculate vessel stresses. Areview of the restraint loads shows that the hanger loads are on the order of2000 to 3000 pounds. These loads would be used as an upper limitfor the selection ofthe support hardware - for example, the rods, clamps, brackets, supporting steel, etc. must be capable ofresisting these loads at a minimum. A review of the weight load (Y-force) on the nozzle at node point 10 (see Figure 2-14) shows a relatively small load, of only 237 pounds, which should be acceptable for most types of equipment. However, closer inspection shows that the sign is positive, indicating that the piping system is pushing up on the support, rather than down. This seems unnatural for a gravity load, and in fact is due to the unbalanced elbow at node point 30 pivoting about the hanger at node point 22. Therefore, even though the nozzle load is low, this is not an optimally supported system.

2-19


LlCENSED TO: TETRACOM SERVICES CAESAR II RESTRAINT REPORT CASE 2 (SUS) W+P1

lOf! 13269 Ver 3.18 FILE:SUPT01 DATE:NOV 4,1992

NODE

FX

FY

Forces(lb. ) Fl

10 22 34 36 44 46 55 72 85 90

O. O. O. O. O. O. O. O. O. O.

237. -2024. -2300. -2190. -3014. -3054. -1606. -2044. -803. -604.

-l. O. O. O. O. O. O. O. O. l.

--- Moments(ft.lb.) MY Ml

MX

-3. O. O. O. O. O.

397. O. O. O. O. O. O. O. O. 3l.

o.

O. O. -3.

-100. O. O. O. O. O. O. O. O. ll10.

PAGE:

5

TYPE Rigid Rigid Rigid Rigid Rigid Rigid Rigid Rigid Rigid Rigid

ANC +Y +Y +Y +Y +Y +Y +Y +Y ANC

Figure 2-14 The system support can probably be improved by moving the hanger a bit closer to the elbow to reduce pivoting - but how close is enough? Figure 2-15 shows the restraint loads for a configuration with the restraint at node point 22 moved 2'·0 closer to the elbow (i.e., 3'·0 from the end of the valve). The sign is now correct (indicating a reasonably balanced system), but the load on the nozzle is now 495 pounds, larger than before. It is not certain that this is an improvement.

lOf! 13269 Ver 3.18 FILE:SUPT01 DATE:NOV 4,1992

LlCENSED TO: TETRACOM SERVICES CAESAR II RESTRAINT REPORT CASE 2 (SUS) W+P1

- - - - Forces(lb. ) NODE 10 22 34 36 44 46 55 72 85 90

FX O. O. O. O. O. O. O. O. O. O.

FY -495. -1348. -2224. -2219. -3005. -3055. -1606. -2045. -804. -603.

Fl -l. O. O. O. O. O. O. O. O. l.

----

MX 239. O. O. O. O. O. O. O. O. 30.

MY 4. O. O. O. O. O. O. O. O. 5.

Figure 2-15

2·20

PAGE:

1

- - Moments (ft. lb. ) - -

Ml 17l. O. O. O. O. O. O. O. O. ll09.


ANC +Y +Y +Y +Y +Y +Y +Y +Y ANC


However, this exercise demonstrates that support and nozzle loads may be tailored by adjusting the locations of the supports. The best location for the hanger may be estimated by interpolating between the two results, in order to minimize the load acting on the nozzle. This shows that the best (where ''best'' is defined as minimizing Ydirection weight force on the nozzle at node point 10) location for the hanger is: d

=

1.0 - (3.0 - 1.0) x 2 x (237) / (-495 - 237)

=

distance ofhanger from valve, ft

= 1.648 ft = 1'-7-3/4"

Where: d

Analyzed Case #1

237#

/

( 1" - 7 - 3/4" )

Location

Distance of Support from Valve, Feet

--+-----+--......::.;0:--+------1 3.0

/ Interpolation Une

-495#

'1 Analyzed Case #1

Figure 2-16 Tuning nozzle loads may also be done by varying the support loads, rather than the support locations. This is done by refusing to allow the system weigh t to settle on its own, but rather by forcing weight unbalance at certain support locations. In this way, ifthe support at node point 22 is underloaded, the system is less likely to push up on the support. For example, if the support at node point 22 only takes -1725 pounds, the shortfall will be split up between the nozzle at node point 10 and the support at node point 32, with the bulk ofthe shortfall going to the nozzle, which is doser. This shortfall, of approximately -300 pounds, will reduce the upward load at node point 10 by approximately 225 pounds (with the support at node point 34 being reduced by the remaining 75 pounds), down to approximately zero pounds. (Proof of this is left to the reader.) The load at selected supports can be forced to be unbalanced through the use ofpre-Ioaded springs (i.e., the loads are set to something other than the naturally distributed weight load), thus influencing the resulting loads on the nozzles. This is most easily done by releasing degrees-of-freedom at anchor points during the restrained weight phase ofhanger design, as discussed in Section 2.4 of these seminar notes.

2-21


2.3 Designing For Expansion Loads Note: It must always he remembered that the engineer must consider the RANGE of expansion stresses (between the cold and hot conditions of the pipe, for example) when considering expansion loadings. The absolute stress value is not a particularly meaningful parameter when discussing expansion stresses, due to the shakedown (self-springing) effect. 2.3.1 Magnitude of Thermal Load A piping system, when heating up, normally tries to expand against its restraints, resulting in internaI forces, moments and stresses: p

Figure 2-17 The axial force generated in the above configuration can be estimated to be the axial force required to compress the pipe back to its originallength after it has been allowed to grow freely. Its free growth is:

Figure 2-18 A

=

cd

Where: A

=

thermal axial extension ofunrestrained pipe, in

ex

=

linear thermal expansion of material from ambient to operating tempera ture, in/in

1

=

length of pipe, in

2-22


The axial force required to compress that growth is:

1~1 ..

e----.l-I

p

Figure 2-19 ~

=

Pli AE

~

=

axial compression of pipe under load, in

P

=

compressive load on pipe, lb

A

=

cross-sectional area of pipe, in 2

E

=


Where:

Equating the deltas, the axial force can be estimated as: 00

=

PlIAE, or: P = AEa.

Considering a rather benign operation - a 12-inch diameter, standard wall pipe (A = 14.58 square inches, E = 29E6 psi) operating at 3500 F (a. = 1.88E-3 in/in) - the axialload is calculated as: P

= 14.58 x 29E6 x 1.88E-3 = 800,000 pounds

From the point ofview ofmost piping codes, there is no stress, since no moment is produced in the axial run (although the codes do state that the possibility of buckling must be considered); however, this is not a good design. An alternate is no restraint at one end, allowing the pipe to grow unimpeded; therefore no load develops. However this is not good design either, since the pipe must normally attach to sorne relatively fixed piece of equipment, and cannot usually be floating in space. What is the solution to this problem? It is necessary to have sorne restraint on the system, but too much may cause excessive forces, moments, and stresses. Looking at the examples above, allowing no movement produces a force of about 800,000 pounds. Allowing 100% of the pipe's desired free movement causes no force. Interpolating, ifwe could devise a means by which the piping system remained intact, yet allowed 99.8% of the pipe's desired free movement, the developed force would be approximately: (1.0 - 0.998) x 800,000 = 1600 pounds This is a much more manageable situation.

2-23


2.3.2 Guided Cantilever Method One proposed means of allowing nearly all of the pipe's free movement (while still holding the system together) is to provide adjacent, perpendicular legs to absorb the thermal growth through bending, as shown in Figure 2-20.

1

10 '-0

1

\

12" cp STD Les @350·F

1

10'-0

~ Cold

Hot

Figure 2-20

Each leg can he modeled as a guided cantilever. According to beam theory: ,i

=

Pl3 / 12EI = cd

M

=

P1I2

--r"""=-

P ~

=

PI ...... 3 /

= PI

/

Figure 2-21

2-24

2

12EI

= ex

1


Solving for P: P

=12EI

d /13

= 6EI d /12 SE = 6EI d /12Z = 6ER d /12

Solving for M: M Solving for SE: Where: l

=

moment ofinertia of pipe cross-section, in'

1

=

length of leg absorbing thermal growth, in

Z

= section modulus of pipe cross-section, in3

I/R R

=


Note that the calculated expansion stress range SE is independent of the wall thickness of the pipe (on a system-wide basis). Therefore, increasing (or decreasing) the pipe wall is usually not an adequate solution to an expansion stress problem. This equation also points out that the stress range decreases with the square of the length of the absorbing leg, so the longer the leg absorbing the displacement, the lower the stress range. For the configuration shown in Figure 2-20: d

=

1.88E-3 x (10 x 12) = 0.23"

SE

=

6 x 29E6 x 6.375 x 0.23 / (10 x 12)2 = 17,700 psi

An expansion stress range of 17,700 psi is normally not a problem, however it must he

rememhered that this equation did not take into account the stress intensification factor (SIF) at the elbow at the top of the leg. Considering an in-plane stress intensification factor for a long radius hend oftypical SIF value of2.8, this would result in a stress range of about 49,600 psi, which is probably excessive for typicallow carbon steel applications. (Note that this value is actually conservative, since the guided cantilever model does not take into account the fact that thejunction ofthe two legs will rotate some under the load, and further neglects the additional flexibility of the elbow.) Against what do we compare the 49,600 psi stress range? We compare it against SA, which is: SA

=

f[1.25 (Sc + SH) - Sil, or, conservatively:

=

f(1.25 Sc + 0.25SH)

For a typicallow carbon steel (A106 Grade B, for instance) and a typical piping code (B31.3, for instance), Sc and SH are both 20,000 psi, giving a conservative value for SA of30,000 psi (the non-conservative value of SA cannot he calculated without knowing the sustained stress S} at the point of interest). The expansion stress range can he approximated for any run of pipe using the guided cantilever equation shown above, as long as the displacements to be absorbed are known. 2-25


2.3.3 Refining the Madel Through the Use of Restraint Stiffnesses What if the calculated expansion stress range is too high? How can we reduce it? Consider the initial example, with the pipe fully anchored at its ends. What would happen in reallife? The restraints would probably bend sorne under the enormous load, allowing sorne piping expansion, which would then reduce the internalload (since expansion loads are selflimiting). What happens if the bending of a support - i.e., its flexibility - is explicitly considered in the analysis? (Normally a pipe stress program by default considers a restraint to be "infinitely rigid". For exam pIe, CAESAR n's default restraint stiffness is in the range of1E12 pounds per inch.) If the restraint actually has a lateral stiffness of 10,000 pounds per inch (instead of1E12), the thermal growth is partially absorbed by the pipe and partially absorbed by the restraint:

- - - t._.

due to pipe leg

t::. t::.

= PL~3/12EI

due to support

+ P/10000

P = t::./(L~3/12EI + 1/10000) M t::. L/2(L~3/12EI + 1110000) SE = t::. L/2(L~3/12EI + 1/1 oooo)/Z

=

) Lateral Stiffness of Support = 10000 lb/in

Figure 2-22

For a 12" diameter standard wall pipe, the calculation is as follows: 1

=

279.3 in4

Z = 43.8 in3

SE

=

0.23 x 120/ [2(1203 / (12 x 29E6 x 279.3) + 1110000) x 43.8]

=

2675 psi

2-26


This significantly reduces the stress range (from the previous value of 17,700 psi) - not through any actual modification, but simply through a refinement of the model. From this we can gain two insights: 1

-

It is sometimes a good idea to provide actual restraint (and nozzle) stiffnesses in the model- the doser to reality the model becomes, the more accurate are the results. Refinement of the model may save the cost ofmodifying piping systems which initially appear to be over stressed.

2

-

If a system really is over stressed, a potential fix may be the introduction of flexibility at the restraints, either by removing restraint or by providing less than infinitely-rigid restraints (or gaps).

Restraint stiffnesses may he calculated through any means and then entered by hand, or simulated in the piping model through the use ofstructural or pipingelements. Vessel nozzle stiffnesses may be calculated manually using Welding Research Council Bulletin 297 or sorne equivalent. Modeling ofrestraints using CAESAR II's structural modeler and use of WRC Bulletin 297 are discussed in Section 3 of these course notes.

Note that it is best not to selectively enter flexibilities for some restraints and not for others. This willresult in the inaccurate distribution ofloads, resultingin nonconservative results.

2.3.4 Use of Expansion Loops In the event that model refinement is not sufficient to solve the problem (i.e., there is a real problem, and notjust one on paper), something must be done. Re-examining the equation for the guided cantilever model:

It is evident that the stress analyst cannot easily change the terms 6, E, R, or delta. This leaves only l, the length of the leg absorbing the thermal growth. This can be done through the addition of an expansion loop. In this case, the thermal growth is partially absorbed by each of the legs running orthogonally to the thermal growth:

=

P(120)3/ 12EI + P(240)3 / 12EI

=

P (1203 + 2403) / 12EI

P

=

12EI Ll / (1203 + 2403)

SE

=

6ER LlI / (1203 + 2403)

Ll

2-27


/'>, /'>,

= =

P(120)~3/12EI + P(240)~3/12EI P(120~3 + 240~3)/12EI

~10'-O

1

10'-0

i

10'-0

~

Figure 2-23 The expansion stress range in each ofthe legs is linearly proportional to the length ofthat leg, so: SEl

SE2

=

6 x 29E6 x 6.375 x 0.23 x 240/ (1203 + 2403 )

=

3937 psi

=

6 x 29E6 x 6.375 x 0.23 x 120/ (1203 + 2403 )

=

1918 psi

The stress range calculated in the longer leg is only 3937 psi (note that the maximum expansion stress is found in the longest leg resisting the displacement), compared to 17,700 psi without the loop. Generically, the stress range in a legj, due to thermal expansion in a direction perpendicular to that of leg j, is:

Where: SEj

=

stress range in a legj Oegj must be orthogonal to the direction of the thermal growth to be absorbed), psi

lj

=

length of leg j, in

li

=

length ofleg i (where leg i represents each leg helping to absorb the thermal growth; normally, aIl legs running orthogonally to the thermal growth), in

Therefore, the calculated stress range should always decrease if expansion loops are added in any direction perpendicular to a direction of thermal growth, since the denominator in the expression for the expansion stress will increase.

2-28


2.3.5 Simplified Expansion Stress Check The concept that addition ofexpansion loops reduces the expansion stress range in a system is recognized by the B31.3 code (and others). This is codified in the requirement that expansion analysis need not be explicitly done for a system meeting the following conditions: 1 -

the system is all of the same size, material, etc.,

2 -

the system has no branches, and consists of only a single run between two anchors,

3 -

there are no intermediate restraint points (note that hangers are traditionally excluded from consideration as restraints), and

4 -

D y / (L - U)2 < 0.03 Where: D

=

pipe outer diameter, inches

y =

resultant thermal growth to he absorhed, inches

L =

totallength of piping, feet

U

=

straight line distance between anchors, feet

The term (L - U) represents the amount of extra pipe (i.e.,loops) in the system. Examination of this equation reveals that, after factoring through constants, it is simply a form of the guided cantilever stress equation:

This simplified check can be illustrated by applying it to the system shown in Figure 2-12. It is clear that this system meets the first three criteria - the system is aIl of the same size, material, etc.; the system has no branches, and consists of only a single run between two anchors; and there are no intermediate restraint points except hangers. For the fourth requirement: D

=

12.75 in

y

=

[«(11+12)x12x1.88E-3)2+(50x12x1.88E-3)2+«45+33)x12x1.88E- 3)2]1/2 x-growth y-growth z-growth

=

2.154 in

=

11 + 45 + 50 + 33 + 12 = 151 ft

L

Dy / (L - U)2

= 12.75 x 2.154/ (151- 95.46)2 = 0.0089 <

2-29

0.03


Therefore the system illustrated in Figure 2-12 need not he explicitly analyzed for expansion stresses. How accurate is this simplified expansion stress check? Based upon the ratio of 0.0089 to 0.03 = .2968, one can infer that this system is stressed to approximately 29.68% of its allowable stress. Assuming a low carbon steel of type A106 Grade B or similar and fewer than 7,000 expected cycles, the allowable stress ofthe system at 3500 F can be conservatively estimated as: f(1.25 Sc + 0.25SH) = 1.0(1.25 x 20,000 + 0.25 x 20,000) = 30,000 psi This would imply that the maximum expansion stress in the system is somewhere in the range of: SEmax

=

0.2968 x 30,000 = 8903 psi

Figure 2-24 shows the actual maximum expansion stress found in this system through an actual CAESAR II stress analysis - 9051 psi (within 1 %), which demonstrates the accuracy (at least for this particular case) ofthis simplified method.

LICENSED TO: TETRACOM SERVICES CAESAR II STRESS SUMMARY CASE 3 (EXP) D3(EXP)=D1-D2

ID# 13269 Ver 3.18 FILE:SUPT01 DATE:NOV 4,1992

PAGE: 10

**** CODE STRESS CHECK PASSED PIPING CODE: B31.3 (1990) HIGHEST STRESSES:

(lb./sq.in.)

CODE STRESS %:

90S!.

@NODE

48

BENDING STRESS:

90S!.

@NODE

48

90.

@NODE

50

330.

@NODE

46

9252.

@NODE

48

TORSIONAL STRESS: AXIAL STRESS: 3D MAX INTENSITY:

ALLOWABLE:

48681.

Figure 2-24

2.3.6 Stress Reduction through Use of Expansion Joints In extreme cases, expansion joints may be added to a system in order to increase flexibility when there is insufficient room for a loop. Expansion joints resemble bellows and provide very flexible pipe legs within very short leg lengths.

2-30


Pipe ,-

\

1

\) Corrugations expand and contract

8ellows

Figure 2-25 Use of expansion joints is usually a last resort solution since expansion joints present maintenance problems due to their fatigue failure mode. (lt should he noted that expansion joints may on occasion actually offer an economical solution in extreme cases, such as when the alternative is expansion loops ofvery large diameter pipe ofexpensive material such as alloy or stainless steel.) Expansion joints also present problems due to the fact that they cannot transmit tension through their wall. This problem is discussed in Section 2.1.5. Since the failure mode of expansion joints is fatigue, the relative expansion displacements hetween the start and end ofthe expansion joint must be checked against the manufacturer's allowables. Note that the allowables provided will not he absolute values, but will be based upon a specifie number of cyclic applications. The manufacturer must always provide a fatigue curve or sorne other type of adjustment factor in order to determine the allowable displacementforadifferentnumberofcycles. Forexample,amanufacturermayrequirethat the allowables be divided by a factor based upon the numher of load cycles:

Factor

Cycles 1

0.49

200

0.56

1000

0.81

3000

1.00

15000

1.38

If the manufacturer provides allowable displacements in the axial, bending, and lateral directions, aIl three movements should be evaluated using a linear interaction formula: ~act

}Cact

+

1ract

+

<= 1.0

2-31


Where: Xact

=

actual axial dis placement of expansion joint, in

eact

=

actual bending angle of expansion joint, degrees

Yact

=

actuallateral displacement of expansion joint, in

Xall

=

allowable axial dis placement of expansion joint, in

Sall

=

allowable bending angle of expansion joint, degrees

YaIl

=

allowable lateral displacement of expansion joint, in

Ablorption 01 Axial M_enl

AbIorption 01 Anguler Ralalion

Absorption 01 Laierai Movemenl

Figure 2-26

AbIorption 01 Multl-DilKtiona1

(Rnu"ant) Lateral Deflection

In the event that the manufacturer only provides allowable axial movements, the other two displacements may he converted to equivalent axial displacements. In that case, the following requirement must be met: Xact + 0.00872665 De act + 3 DYact / 1 <= Xall

Where: D

1

=

equivalent internal diameter of expansion joint, in

=

distance from inner edge of convolution to outer edge of convolution on opposite side of expansion joint

=

flexible length of expansion joint, in

Expansion joints are especially weak in torsion, so this type ofloading should be kept to a minimum. Means of modeling various expansion joint configurations is discussed in Section 3 of this document.

2-32


2.3.7 Expansion Stress -

Other Solutions

Often expansion stress problems are due to the use offittings with large stress intensification factors (SIF). In these cases, it may be possible to upgrade fittings to those with lower SIF values. For example, miters can be upgraded to bends, which can be further upgraded to bends with longer radii of curvature. Unreinforced fabricated tees can be upgraded to reinforced fabricated tees, which can be further upgraded to welding tees. Threaded pipe or socket weld connections may be upgraded to butt welds. SIFs are discussed in more detail in Section 1.3 ofthese seminar notes. Modification ofthe restraint configuration may also solve expansion problems. For exam pIe, in most cases, removing restraints increases flexibility, decreasing expansion loads. Limit

stops may be used to allow some movement, thus reducing internaI generation of expansion loads. Strategically placed restraints can be used to force thermal growth from areas ofhigh stress to are as of lower stress.

2-33


2.4 Hanger Design As seen in Section 2.2, as more restraint is provided to a piping system, weight stresses decrease. Conversely, Section 2.3 demonstrated that as restraint is remoyed from a system, expansion stresses decrease. This contradiction must be resolved through some type of compromise.

Likewise, in systems supported only with rigid supports, it is possible that the pipe might lift off of some supports and lock up against others once it thermally expands. This is demonstrated through a review of the restraint loads during the operating load case for the system shown in Figure 2-12, which shows that the pipe lifted off of the supports at node points 36, 44, and 72 (rendering them inactive) and had a partiallock up at node point 55, overloading the support.

LICENSED TO: TETRACOM SERVICES CAESAR II RESTRAINT REPORT CASE 1 (OPE) W+T1+P1 NODE 10 22 34 36 44 46 55 72 85 90

FX -22. O. O. O. O. O. O. O. O. 22.

IDfl 13269 Ver 3.18 FI LE: SUPT01 DATE:NOV 4,1992

--- Forces(lb.) FY FZ 1074. -3023. -253I. O. O. -4940. -6238. O. -1143. -602.

505. O. O. O. O. O. O. O. O. -505.

MX 676. O. O.

o.

O. O. O. O. O. -18I.

-- Moments(ft.lb.) MY MZ -4700. O. O. O. O. O. O. O.

o.

-5200.

325. O. O. O. O. O. O. O. O. 1103.

PAGE:

7


ANC +Y +Y +Y

+Y +Y +Y +Y +Y ANC

Figure 2-27 It would be preferable if the pipe could move to its new position and, at the same time, onto its supports; or even, if the support could move with the pipe. One solution is a weight and pulley assembly:

Figure 2-28

2-34


The drawback to this is that the assembly is bulky, and requires restraint design for twice the pipe load. The mechanical implementation of this concept is the constant spring, or constant effort support (Figure 2-29). This support has an internal pivot arm attached to a spring; as the pipe moves up or down, the moment arm about the pivot and the spring force vary inversely, creating a constant moment about the pivot, and therefore a constant force acting against the pipe weight. The drawback of constant springs is that they are often too expensive for the application; therefore they are usually used only when pipe movements are very large.

Figure 2-29 2.4.1 Variable Spring Hanger Design Basics A less expensive alternative is a variable spring hanger, in which the spring load varies somewhat as the pipe moves. From the analysis of the system shown in Figure 2-12 it was evident that some change in the support loads as the pipe goes from the cold to the hot condition is tolerable from a stress point ofview. The trick is to design the system with an acceptable load range. A variable spring hanger (Figure 2-30), pre-set to some load, provides support throughout the range ofpipe movement; as noted above, there is some change in load as the pipe moves from cold to hot position. As the pipe moves up, the load plate on the spring moves up, allowing the spring to decompress, thus decreasing the load which the spring puts on the pipe. As the pipe moves down, the load plate on the spring also moves down, further compressing the spring, and thus increasing the load of the spring on the pipe.

2-35


Figure 2-30 The objectives of spring hanger design are to choose a spring which: 1

will provide the weight support load necessary to balance the pi ping system after the pipe has moved from its cold (installed to its hot (operating) position,

2

permits the total movement ofthe pipe from its cold to hot position, and

3

does not cause an excessive expansion stress range in the pipe as the spring load ranges from its cold to hot load.

Since the variable spring hanger load changes as the pipe moves from its cold to its hot position, and one objective of hanger design is usually to provide the weight support load necessary to balance the piping system in its hot position, it is necessary to install the spring with an unbalanced "cold load". This unbalanced load can be determined by:

CL

= HL + k L1 th

CL

=

cold load (unbalanced installation load of spring), lb

HL

=

hot load (desired target load to support balanced weight at spring location), lb

k

=

spring constant of variable spring use d, lb/in

il th

=

travel, or expected thermal movement of pipe at spring location, going from installation to operating, where up is positive, in

Where:

2-36


2.4.2 Load Variation Under certain circumstances, specifications may recommend that the relative load change be minimized, by limiting the Load Variation to aval ue such as 10% or 25%. Load Variation is defined as: Var

=

1 HL - CL 1 / HL = 1 k Ll th 1 / HL

=

Load Variation (ratio of 0.0 to 1.0)

Where: Var

Since the hot load and thermal movement are dictated by the piping system configuration, the variability of an individual spring can be controlled only by varying the spring rate. Most manufacturers provide springs with three (or more) different spring rates per load size, recommended for short-range (0 to 1/2 inch), mid-range (0.5 to 1 inch), long-range (1 to 2 inch) displacements. Since all springs in a given load size support the same range ofloads over their total travel, the spring rate (and therefore the variation) of a long range spring is typically one-half ofthat of a mid-range spring, which in turn is one-halfthat of a short range spring. The use of a spring load variation criteria is normally a holdover from an earlier era, when it was used as justification for not including spring stiffnesses in expansion loading cases. If the spring stiffnesses are included in the analysis (as is normally the case with CAESAR II), load variation criteria may probably be considered to be an unduly restrictive requirement.

2.4.3 Hanger Selection Table Springs are selected from a table such as that shown in Figure 2-31. This table shows the load capacities within the workingranges of each spring size, and the spring constants of the short-, mid-, and long-range spring series for each ofthe sizes. Knowing the hotload, thermal movement, and variability requirement, the process in selecting a spring from the table is: 1 -

Calculate the maximum permissible spring rate as: k max

=

Var HL / 1 Ll th 1

2 -

Determine the spring load size by finding the hot load in one ofthe columns of the hanger selection table.

3 -

For that size spring, select the spring series with a spring rate less than or equal to that calculated above.

4 -

Calculate the cold load (from CL = HL + k Ll th) and verify that the cold load also falls within the working range ofthe spring.

5 -

Ifthis is not the case, then try again with a different spring series of the same size, or a spring of an adjacent size.

2-37


This process may he illustrated by an example. Assume that the target hot load of a spring is computed as 613 pounds, the computed expected travel from cold to hot positions at that point is 1.5 inches up, the load variation limitation is 20%, and the spring manufacturer provides the selection table shown in Figure 2-31: 1 -

The maximum permissible spring rate is: k max = 0.20 x 613 /

1

1.5

1

= 82 lb/in

2 -

The hot load of 613 pounds is within the range of spring load size 7 (with a recommended range from 392 to 672 pounds), as shown in the selection table. Therefore, a spring of this size is a preliminary candidate for selection.

3 -

Looking at the bottom three lines of the column for size 7 shows that the only springwith a stiffness helow the permissible springrate is the long-range spring, with a stiffness of 56 lb/in.

4 -

For this spring, the Cold Load would be: CL = 613 + 56 x 1.5 = 697 lb This load is outside of the recommended range ofthe spring (but does fall within the total range of336 to 728 pounds, which should really only be used in extreme circumstances). Therefore, the size 7 spring should not be used.

5 -

It is now necessary to go through the calculation again, trying another spring. The hot load of 613 pounds also fans within the range of a size 8 spring (recommended range from 525 to 900 pounds), making this size also a candidate for selection. Looking at the spring rates for the size 8 spring, again only a longrange spring has a stiffness (75 lb/in) below the permissible rate.

The Cold Load calculation is: CL = 613 + 75 x 1.5

= 726 lb

This load is within the recommended range of the spring; therefore a long-range, size 8 spring pre-set to a Cold Load of 726 pounds must be specified.

2-38


.......... Ag.

•

0

.... fig.

1

fig.

12 0

1

2

3

4

83 81 lOS 141 118 841011 147 118 .114 153 71 91 118 1511 0 74 96 123 185 7888 127 110 54 7'11 101 131 178 sa 81 lOS 138 182 v. sa 84 108 140 188 50 87 111 144 1114 81 88 115 149 83 112 118 153 \1085 115 122 158 '2 87 .7 125 182 217 et 100 128 1811 71 102 132 171 '1(073 105 135 175 74 108 138 178 78 110 142 184 247 78 113 146 1811 1 80 Il. 1411 1SC1 82 118 152 197 j284 84 121 155 201 88 123 159 208 114 118128 lez 210 282 811 129 1116 214 -111 131 1118 2111 .. 113 134 172 223 _ 95 137 176 228 43 44 48 48 50 52

0

"..

1

3

,....

4

2

5

21011

Il

~

~

7

• " 450

10

"

12

13

14

15

1.

l i t 252 336 lIDO 780 1020 1350 1800 2400 3240 4IiOO 197 283 350 489 1!2!5 813 1083 14015 1875 2500 3375 4888 488 S!iO 845 110S 1483 l11!iO 2IlOO 31510 4875 501 875 878 1148 151. 2025 2700 31545 5083 525 700 1110 1190 1575 2100 2800 37110 5250 544 ~ 1M3 1233 1831 2175 2900 3915 S438 238 315 420 583 750 1/75 1275 1888 2250 3000 4050 5825 244 3211 434 581 77S 100811318 1744 2325 3100 4185 5813 252 338 448 IlOO 800 1040 1380 1800 2400 3200 4320 eooo 2eO 347 482 819 825 1073 1403 18511 2475 3300 4455 8188 288 357 478 838 850 1105 1445 1913 2550 3400 4580 lI37S 1_ 1 _ 2825 3500 4725 11583 858 675 1138 ' 1170 1530 2025 2700 2114 378 675 48eO 8750 278 3118 1203 1573 zoel 2775 291 138t 518 4885 l1li38 2991399 532 713 950 1235 11115 2138 2850 3800 5130 7125 3071410 548 731 975 1288 111S11 21114 2825 3900 5285 7313 315420 5110 750 1000 1300 1700 2250 3000 4000 5400 7500 323 431 574 719 1025 1333 1743 230S 3075 4100 55315 7888 331 441 5Il8 1B8 1050 13115 1785 2383 3150 4200 SII10 7875 339 452 802 aoe 1075 1311B 1828 2419 322S 4300 5805 11083 347482 B16 825 1100 1430 1670 2475 4400 5IMO B250 354 473 830 844 1125 1483 1913 2531 4500 11075 8438 3821.483 844 883 1150 1485 1956 2588 3450 4800 8210 81125 1988 2844 3525 4700 8345 8813 37014114 IISII 881 1175 3781 seM 872 900 1200 '5110 2040 2700 3IlOO 4900 84110 9000 388 515 688 1118 1225 15113 20113 2758 3875 4900 11815 3114 525 700 SCIa 1250 1825 2125 2813 3750 5000 8750 8375 402 536 714 966 1275 1868 21l1li 2l18li 3825 5100 688S 11S113 2210 21125 3900 5200 7020 9750 410548 728 875 1300

=

l30e

•

---..,In.

208 273 384 213 284 378 221 54 392 228 305 40S

= = ~ 12

2

...

_ _ 1iH

wMdng

;r:

-

::

~

,_ 1: ,_ 8pItng _ _ -

.,88

17

::: S!iOO 6750 1000 7250 7500 7750 l1OOO

1•

1.

20

1/750 10000 10250 10500 10750 11000 11250 11500 11750 12000 12250 12500 12750 13000

22

1OS10 14'00 18750 25005 11053 14888 19631 28017 11496 15275 203'3 27089 1111C1B 15883 21094 28131 8320 12380 111450 21675 211173 12823 17038 226S6 30215 9SS2 l1li85 13265 17825 23438 31258 10317 13108 18213 24219 322118 IOSS0 14150 18800 25000 33340 101182 18388 25781 343112 11315 lS035 1l1li75 2eII3 35424 11847 15477 20583 27344 118110 15e20 21150 28125 37508 12312 1113e2 21738 281108 38549 12845 18805 22325 211888 385111 12977 17247 22e13 304et 4CMI33 13310 171110 23IiOO 31250 41875 13842 1.,32 24088 32031 42717 13975 18575 24e75 32813 43759 14307 111017 33!iIN 44801 14840 1lI4II0 34375 45843 141172 111802 28438 3151sa 48885 15305 20345 Z7025 3I5IICI8 47112e 1S1137 20787 27513 387111 4l1li88 151170 21230 28200 37500 50010 111302 21872 28788 38281 51052 Il1t13S 22115 29C!75 3lIOII3 52094 ll111t17 22567 2l1li83 38844 53136 17300 23000 305SO 40825 54178

7990 8322 886S 8t67

:: ,_

:= ::

21

fig.

a

0

fig. 0

V.

-

==

.... ....• 0

1011

1

1011

1

2

'1(0

11011

3

1

2

4

114

21011

5

11o'l

3

Il

1'1(0

31011

7

lb. pel' ln.

30 42 54 70 114128 188 224 300 400 520 880 900 1200 1600 21110 3000 4000 5320 7080 Il4OO 12500 18670 15 21 7

10

27 35 47 63 84 112 150 200 260 340 450 800 13

17 23 31

800

1080 1500 2000 2880 3540 4100 tl250 6335

42 56 75 100 130 170 225 300 400 540 750 1000 1330 1710 2350 3125 4167

Figure 2-31

2.4.4 Hanger Design Process - Restrained Weight, Free Thermal, and More The procedures described above assume that the hot load and thermal movement required for spring selection are already known. How does the engineer calculate the hot loads and thermal movements? The procedure for the entire hanger design process is as follows: 1 -

Pick out support locations using standard span criteria, and do a weight analysis, assuming that there are rigid Y-restraints at each location. This analysis is called the "restrained-weight" analysis. The weight loads distributed to each of the restraint during this analysis are used as the hot loads during spring selection.

2 -

Next, remove the restraints from the support locations, and do a thermal expansion analysis. This analysis is called the "free-thermal" analysis. The thermal movements at each of the support locations are used as the thermal travels for selecting the springs. (Note that due to the technical effects of possible non-linear effects in the system, CAESAR II performs not a true "free-thermal" load case, but rather a load case called "operating for hanger travel", which includes the effects of thermalloads, weight loads, and the spring hot loads calculated in the restrained weight case. Since the piping weight loads and the spring hot loads essentially cancel each other out, this effectively results in a thermal only load case, but with non linear effects considered.)

2-39


3 -

Using the hot loads calculated from the restrained-weight case and the travels calculated from the free-thermal case, select a spring for each location from the spring selection table as described above. Use the spring constant to determine the required cold load (pre-set hanger load) for installation.

4 -

Alter the model to reflect the presence ofthe springs by adding ares traint at each support location with a stiffness equal to the spring constant ofthe spring, and by adding the pre-set spring load (cold load) as a force active during the sustained load case. Then re-analyze allload cases in order to get the effects ofhaving the actual springs present.

The four steps ofthis procedure (except the locating of the supports) are done automatically by CAESAR II whenever the user designates a hanger (i.e., a candidate for spring hanger design) in a piping system 2.4.5 Restraint Placement Using Distance ta First Rigid Criteria If the above procedure is followed exactly, it is likely that almost all support locations will show some vertical movement, and will therefore require springs. For economic reasons, it is best to try to limit the number of springs by using rigid supports at locations with small movements.

How can this be done? The analyst can potentially impose zero movement at points where it is safe from an expansion stress point ofview; ifthere is no vertical pipe run between points of zero growth, all supports along that run should have zero growth as well. For example:

-r---~i~n:1 ) Down Three Supports with No Uplift

) Position Four Supports with Uplift

Figure 2-32 The question is where can rigid restraints be placed without causing expansion problems? If the expansion displacement is known at a given point, the minimum distance to the first rigid restraint can be calculated using the guided cantilever stress formula: S

=

6ERA /1 2 , so:

Imin

=

(6ERd /

SaU)1/2

2-40


Where: lmin

= minimum distance to the first rigid restraint from a point ofknown vertical dis placement, in

SalI

= an approximate allowable expansion stress value, based upon SR, Sc, f, any intensification factors present, and the generallevel of expansion stresses present in the system prior to addition of the rigid restraint (often a nominal value such as 10,000 psi is used), psi

This can be illustrated by an example. In the system depicted in Figure 2-12, there are eight hanger locations. If the procedure described in Section 2.4.4 is followed to the letter, there will certainly be thermal growth at aIl ofthese locations, so eight springs will be selected. In order to reduce this numher, sorne engineers impose a rigid displacement criteria - for example, if the displacement at the hanger location is less than sorne value (such as 0.1 inches), they will select a rigid rod rather than a spring. (Note that this procedure is not foolproof, since using rigid rods at locations such as the tops of risers or near equipment nozzles may cause lock-up or lift-off despite having "free-thermal" displacements ofless than 0.1 inches; therefore application of a procedure such as this must he reviewed carefully.) The reader can confirm that re-running this problem with a rigid displacement criteria of 0.1 inches actually leads to a reduction in springs used, with only five selected. The engineer can do better than that by pre-selecting potential hanger candidates through the use of the distance-to-first-rigid criteria. AlI of the vertical thermal growth in this problem is generated by the riser between node points 40 and 50; we can calculate that growth as: il = 50 x 12 x 1.88E-3 = 1.13 in

The engineer can direct part of this growth upward, to be absorbed by the horizontal portion between node points 10 and 40, and the rest ofit downward to be absorbed by the horizontal portion between node points 50 and 90, by requiring a rigid restraint at one ofthe hanger locations on the riser. Which ofthese two hangers is a better candidate for a rigid support? Since the expansion stress is inversely proportion al to the square of the lengths of the resisting legs, it is logical to direct the greater part of the thermal growth upward, since a quick check reveals that the spans from node point lOto 40 are longer than the spans between node points 50 and 90. Therefore, the hanger at node point 46 should be selected as the rigid support, and the other one (at node point 44) must be a spring. This will force a thermal displacement of(35 /50) x 1.13 = 0.79 inches upward, and a displacement of(15 / 50) x 1.13 = 0.34 inches downward. Looking at the portion of the system from node point 10 to 40, one first considers the hanger located at node point 36. A rigid support may be placed here if the span from node point 36 to 40 is long enough to absorb the displacement of 0.79 inches, or, if the distance from node point 36 to 40 is greater than the calculated distance to the first rigid, which is (using SalI = 10,000 psi): lmin

= (6ER il / SaIl )1/2= (6 x 29E6 x 6.375 x 0.79 / 10,000)1/2 = 296 in = 24'-8

2-41


The distance between node points 36 and 40 is only 13 feet, which exceeds the calculated distance to the first rigid, so the support at node point 36 must he a spring. Looking at the next support point, at node point 34, the distance from the riser is 36 feet, exceeding the allowable distance to the first rigid. This indicates that a rigid support may be placed at node point 34 without causing undue expansion problems. Next, since the pipe is being restrained vertically at node point 34, there are no imposed displacements at the nozzle at node point 10, and there are no risers hetween node points 10 and 34, alilocations between the nozzle and the restraint will also have no vertical thermal growth. This me ans that a rigid support, rather than a spring, may be used at node point 22 as weIl. The lower part ofthe piping system must absorb only 0.34 inches. The minimum distance to the first rigid required to absorb this dis placement is: lmin

=

6ER il / Sall)1/2 = (6 x 29E6 x 6.375 x 0.34/10,000)1/2

=

194 in = 16'-2

The distance between the riser and the hanger location at node point 55 is 11'-4, which is wi thin the minimum distance to the first rigid. Therefore, the support at this location should he a spring. The distance from the riser to the support at node point 72 is 25 feet, which exceeds the minimum distance criteria, and may therefore he a rigid support. This rigid support holds the pipe to a zero vertical displacement, which when considered with the anchor at node point 90, and the lack ofrisers in-hetween, means that the hanger at node point 85 may also be a rigid support. By pre-declaring these supports to he rigid supports, the engineer limits the number of possible spring hanger candidates in this case to a maximum ofthree, a savings on hardware heyond that achieved though use of the rigid rod displacement criteria method. It should be noted that declaring that a support to he a rigid support does not mean that it must be constructed as a two-way restraint; ifthere is no net uplift force, it may be built as a simple rodhanger.

2-42


DISTANCE TO

20'-0

RIGID----r~

>16'-2~ \.\~Cl

\.\~l>.

DISTANCE TO FIRST RIGID

~

Figure 2-33

2.4.6 Notes on Hanger Design 1

-

In the event that a system which carries a fluid with a specific gravity less than 1.0 is to be hydro tested, the springs will generally have to remain pinned during the hydro test. The hanger hardware (clamps, rods, etc.) and supporting structure will have to he selected and/or designed to withstand the hydro test loads, which will normally he the controlling design loads for these supports.

2

-

When specifying the spring hanger's Hot and Cold Loads, the anticipated weight of additional hardware should be added to the loads calculated by CAESAR II, especially if it is expected to he significant (such as in the case of large stock clamps or a trapeze assembly made of structural steel). The spring must also support the hardware, and if this is not considered when specifying the spring parameters, the piping weight loading will he unbalanced by the weight of the hardware.

3

-

Horizontal movement at hanger locations must he considered when designing a support in order to assure that the pipe does not move 80 far that i t falls off of the support. Additionally, support manufacturers typically limit the range of a hanger rod's arc in to values such as 60, where the arc can be calculated as: Arc

= Tan-! (horizontal movement / rod length) 2-43


In cases where the horizontal movement is especially large, it may be advisable to install the support in an offset position to minimize the deviation of the line of support action from vertical in both the cold and hot positions. 4

-

In systems where installation is difficult due to flange fit-up problems caused by unbalanced cold loads, it may be preferable to adjust the springs in the field to carry the hot load once the system has been started up. In cases where nozzle operating loads are not critical, and fit-up problems are more of a concem, CAESAR II can provide Cold Load Design, where the weight loads are balanced in the cold, rather than the hot, condition. CAESAR II provides the option of calculating both the "theoretical" and the "actual" cold loads for springs. The theoretical cold load is the load to which the spring should be preset prior to installation (usually this is done at the factory, and the springis pinned tokeepitat this value). This is the load which the spring will exert on the piping system in the cold condition, as long as there is no vertical displacement ofthe system at this location. Since the cold load is almost always unbalanced vs. the piping weight load, there will be a net load on the system at this location in the cold condition. Ifthis net load is large, or the piping system is very flexible, the system may displace under the load, leading to extension or compression ofthe spring, and a corresponding change in the load plate reading. The new readingofthe spring load is what CAESAR II calculates as the "actual" cold load. Or more simply, the "theoretical" cold load is the cold load to be specified in the factory order of the spring, while the "actual" cold load is an approximation ofthe reading ofthe spring load after pulling the pins upon initial installation. The actual installed load case is important if the springs are to be adjusted or checked in the cold condition, or if the spring's cold load is being set in position, rather than at the factory.

5

6

-

Excessive use of spring hangers may create a dynamically unstable (low natural frequency) system due to lack of restraint stiffness. These systems have essentially no horizontal support, and typically small vertical stiffnesses resisting movement in the Y direction. Note that constant effort spring supports have no dynamic effect on a piping system.

7

-

Selected hanger locations may actually hold the pipe down during the restrained weight case due to unbalanced parts ofthe system pivoting about other supports. CAESAR II flags these with a warning during the analysis and reports them as zero load constant effort supports in the hanger table during output. When this occurs, the offending supports should be removed, or the support locations in the vicinity should be reconsidered.

8

-

There are special provisions to consider when cold spring and hanger design exist in the same job. Cold spring should be omitted from the restrained weight case, and included in the operating load case for hanger travel. The actual installed load case should be run with the cold spring in order to determine the installed hanger settings in the presence of cold spring. It is the user's responsibility to verify that the displacements during the actual installed case are still within the manufacturer's recommended load range. Problems usually only arise when

2-44


there is significant cold spring in a vertical run of pipe in the vicinity of one or more spring hangers. 9

-

In a liquid filled line, the springs may he installed when the system is empty. In this case it is necessary to ignore the "actual" cold load, and in some cases it may be preferable to adjust the springs in the field to carry the cold load once the system has been filled.

2.4.7 CAESAR Il Hanger Design Control and Options CAESAR II provides a number ofuser specified options for controlling its automatic hanger design. The control options may, for the most part, be applied to the system globally, or at specific locations. These options are fully descrihed in the CAESAR II U ser's Manual, but are discussed to some extent here: Actual cold load calculation - This is described in more detail above. The user should specify Yes, if: 1

-

The spring installation load is to he adjusted with the pipe resting on the spring and free to move vertically otherwise (i.e. there isn't a steel strap around the spring base and the load flange, preventing movement of the load flange when the spring is adjusted in the cold position).

2

-

The piping adjacent to the spring is very flexible and/or the stiffness ofthe spring is very high.

3

-

Fluid fùled systems are installed and set empty, and the user wishes to know the empty installation load.

Use short range springs - CAESAR II's hanger design algorithm first tries to select for an application a short range spring, followed by a mid-range, and then a long range, spring. On some construction sites short range springs are considered specialty items, and are only used where available spring installation clearance is small and where travel from cold to hot is small. In these cases, the user may instruct the design algorithm to bypass consideration of short range springs (and start with mid-range springs frrst) unless space limitations require it. Allowable Load Variation - As noted above, this is computed as: Var = 1 CL - HL 1 / HL = 1 k A th 1 / HL The maximum possible load variation inherent in recommended ranges of the spring tables approaches 100% when the Hot Load is less than the Cold Load, and is approximately 50% when the Hot Load is greater than the Cold Load. Typical values for the permissible load variation range from 10% to 25%. A constant support may he forced at a location by specifying a minuscule load variation requirement at that location.

Rigid Support Displacement Criteria - Where feasible, rigid supports are considered preferable to springs supports, for reasons of economy (purchase, installation, and mainte-

2-45


nance) and vibration prevention. Therefore, if a rigid support can be chosen instead of a spring at a location, the engineer will usually want this to occur. One definition of a spring support is: "a restraint that supports a given load through some thermal travel". If the thermal travel is zero, or very small, then it is hypothesized that a rigid support can he used in place of the spring. This is indeed true providing that the surrounding pipe is relatively flexible as compared to the rigid rod. The extent to which rigid supports are chosen can be controlled by this criteria. At any support location where the vertical displacement calculated during the operating load case for hanger travel is less than the specified Rigid Support Displacement Criteria, a rigid rod will be selected and used in subsequent load cases. Note that this may not be desired at spring locations in the vicinity ofpumps or other rotating equipment or on risers, since this may result in high nozzle loads or thermallockup/liftoff of the support. It is best ifthis criteria is used in conjunction with some pre-design of support locations, such as that discussed in Section 2.4.5 of these seminar notes.

Free AnchorslRestraints - Often a major objective ofhanger design is the minimization of equipment nozzle loads due to weight. This is done by forcing an unbalanced hot load (usually an overload) at the hanger location nearest to the equipment nozzle. This unbalanced force pulls on the nozzle, thus relieving it of some of the weight that would normally fall on it under a natural distribution - ideally, the hanger would be sufficiently unbalanced to make the load on the equipment nozzle as close to zero as possible. In an attempt to force this unbalance, anchors at equipment nozzles are often ''freed" during the restrained weight case, forcing all of its weight to the hot load of the nearest support. This technique should be used sparingly in those configurations where no hangers are located within three pipe diameters or so in a horizontal direction from the nozzle being released. It is also recommended that care be taken when releasing more than just the Ydirection force at a anchor/restraint, as release of additional degrees-of-freedom may cause gross angular and vertical displacements, resulting in unrealistic hanger design loads.

Manufacturer's Tables - This entry is used to designate the manufacturer of the springs (and thus the hanger table) to he used, as weIl as certain design criteria relating to selection of the hangers within this table. The selection criteria include: 1 -

use of maximum (vs. recommended) load range,

2 -

centering of the spring in the table, and

3 -

cold load (vs. hot load) design.

Most hanger vendors provide hanger tables with two ranges defined: 1) a restricted, or recommended load range, and 2) a maximum allowed load range. In order to provide margin against analytical uncertainties, it is best to use the recommended range. The maximum allowed load range may be used in certain situations, such as to permit the use of variable support hangers instead of the more expensive constant effort support, or when an alreadyowned spring is to be used over a new one.

2-46


In cases where the expected analytical uncertaintyis especially high, maximum margin may he provided by selecting the spring which most closely centers the loads in the hanger table. Cold load design balances the weight loads in the cold, rather than the hot, condition. This may he desired in those systems where installation is difficult due to flange fit-up problems caused by unbalanced cold loads, and where nozzle operating loads are not critical.

Available Space - In certain cases, the distance between the top ofthe pipe and the steel overhead; or between the bottom of the pipe and the foundation or platform below, govern the type (and number) of springs which may he used at a specific location. This value may be specified at individual hanger locations for use in spring selection. Figure 2-34 defines the available space as used in the CAESAR II spring design.

Available clearance for hanger. (Input positive number for hanger available space.)

Available clearance for cano (Input negative number for Cab available space.)

--1;.-

j Figure 2-34

The available space option together with the "number of springs allowed" option lets the user design multiple spring support systems.

2-47


Number of Allowed Springs - Ifthere is physicaUy more than one spring can at a given hanger location, that numher may be specified here. Likewise, the maximum number of springs that the user will permit may be specified (in the event that CAESAR II has to split the load in order to meet space criteria). In the case of multiple springs, CAESAR II will split the load evenly among aU springs. User Defined Operating Load - In some piping configurations the program selected operating (or hot) load on the spring doesn't unload the equipment nozzle sufficiently to satisfy manufacturers aUowables. In these situations the user can force a hot load (higher or lower), overriding the program calculated value in an attempt to tune weight distribution and bring the equipment loads within the allowables. The user's entry in this case should normaUy be a variation of the value initially proposed by the program spring selection algorithm. Before adjusting the operating load the user should determine if a preferable course of action is freeing the problem nozzle during the restrained weight case (as discussed above). Old Hanger Redesign - In cases where part of a piping system is redesigned, it is preferable that the hanger design algorithm re-select the existing springs in the system wherever possible. Where they can be re-used, new load ranges may he identified for them, and only a readjustment ofthe load flange in the field may be required. Where the existing springs can't be used, new ones will be recommended. The Old Ranger Redesign capability allows the user to do this. Multiple Load Case Spring Hanger Design - This option is useful whenever the piping system has multiple thermal states that are sufficiently different such that the results from each thermal state should he considered when doing the spring hanger design. Figure 2-35 illustrates this idea:

Figure 2-35

2-48


The hanger at "A" should he designed with the main pump running, and the hanger at "B" should be designed with the backup pump running. Once the springs are designed for their respective thermal cases they are inserted into the piping system and the various operating conditions run to check for stress or equipment overloads. The options available in CAESAR II for combining data from the various design load cases are shown below: 1

Design per thermalload case 1

2

-


3

-


4

-

Design for maximum operating load Design for maximum travel

5 6

-

Design for average load and average travel

7

-

Design for maximum load and maximum travel

2-49


2.5

Designing For Occasional Loads (Static Equivalent of Dynamic Loads)

As noted earlier, piping systems must he designed to withstand primary and secondary loadings. Sustained loads were discussed as beingthe most common types ofprimary loads. There are additional requirements for the evaluation of occasionalloads, or primary loads which are present for short time durations, typically 1% to 10% of the total operating time. Failure criteria are typically the same for occasionalloads as for sustained loads, except that creep failure is not a concern for occasionalloads. Because of this, the allowable levels for the absolute sum of sustained and occasional stresses are the same as those for sustained loads, but increased by a factor (typically 15% to 33%). For example, looking at the B31.1 equation for occasional stresses:

Slp + 0.75 i Ma/Z + 0.75 i MlIZ < k Sh Where:

=

longitudinal pressure stress, psi

l

=


Ma

=

resultant moment on cross-section due to sustained loads, in-lb

Z

=

section modulus of pipe cross-section, in3

Mb

=

resultant moment on cross-section due to occasionalloads, in-lb

k

=

occasional stress factor

=

1.2 for loads present less than 1% of time

=

1.15 for loads present less than 10% oftime

=

Basic allowable stress in hot condition

Slp

Sh

Typical of these types ofloads are wind loads, earthquake loads, and quickly applied loads (reliefvalve, fluidhammer, etc.). These are dynamic (meaningthat they change as afunction of time) loads, and are therefore discussed in greater detail in Sections 4 and 5 of these seminar notes. However, the easiest (but less accurate), and therefore most common means of analyzing dynamic loads is usually to model them as static (meaning that they are constant throughout time) loads, with the magnitude increased to reflect the dynamic load amplification.

2.5.1 Wind Loading Wind loading is caused by the loss ofmomentum of the wind striking the projected area of the pi ping system. The static linear force per foot generated by a steady-state, constant speed wind load can be calculated as: f

=

P eq * S

* D sine 2-50


Where: f

=

"pseudo static" wind force per length of pipe, lb/ft

P eq

=

equivalent wind pressure, psi

=

V2 /2g

=

densi ty of air, Ibm/ft3

=

0.0748Ibm/ft3 at 29.92 in Hg and 700 F

v

=

design velocity ofwind (usually the 100-year maximum wind speed), ft/sec

g

=

gravitational constant, 32.2 ft/sec 2

S

=

shape factor (or drag coefficient), based upon Reynolds number ofwind and shape of structure; this typically varies between 0.5 and 0.7, with a value of 0.65 characteristic ofpiping elements, dimensionless

D

=

pipe diameter (including insulation), ft

e

=

angle oforientation between pipe and wind, where 0 0 represents the pipe axis parallel to the wind direction

Since this represents the force associated with a steady-state flow of air, the calculated value is often increased by a gusting factor in the range of 1.0 to 1.3 to account for dynamic effects. The linear force per foot, f, is calculated for each end of the element and the average taken. The average is assumed to apply as a uniform staticload over the entire length ofthe element. ASCE #7 (formerly ANSI A58.1) modifies this concept slightly to consider facility importance, proximity of hurricanes, etc. Its formula for wind load is:

=

0.00256 Kz (1 V)2

Kz

=

Exposure coefficient, based upon height above ground level and congestion oflocal terrain (varies from 0.12 for 0-15 feet height in city environment to 2.41 for 500 feet height in wide open terrain), dimensionless

I

=

importance factor, based upon importance of structure and proximity to hurricane coast (varies from 0.95 for non-essential facility over 100 miles from a hurricane to 1.11 for essential facility on the hurricane coast), dimensionless

v

=

basicwindspeedCexcludingfrom theaverageabnormallyhigh windloading events such as hurricanes or tornadoes), from ANSI A58.1 map (rangingfrom 70 to 110), milhr

f

Gt Cd D sin

Where:

2-51


~

=

gusting factor, based upon height above ground level and congestion oflocal terrain (varies from 1.0 for 500 feet height in wide open terrain to 2.36 for 015 feet height in city environment), dimensionless

CAESAR II's ASCE #7 wind input screen requests a number ofparameters, from which the coefficients of the equation above are determined.

ASCE #7 provides a map of basic wind speeds in the Continental United States. The following is a crude summary of the map:

Region

Basic Wind Speed

California Other West Coast Areas Rocky Mountains Great Plains Non-Coastal Eastern U. S. Gulf Coast Florida - Carolinas Mi ami New England Coastal Areas

70 mph 80 mph 70 mph 80 mph 70 mph 100 mph 100 mph 110 mph 90 mph

ASCE #7 adjusts the importance factor according to the site's Distance from Hurricane Ocean line. This typically translates into the distance from the east coast or the Gulf of Mexico in the Continental U .S. Ifthe plant site is greater than 100 miles from either the east or the gulfcoasts, then a value of 100 miles should be used (no credit may he taken for any plant site greater than 100 miles from any ofthese hurricane prone areas). The importance factor is further influenced by the Structural Classification, where the options are:

CateQor.v

Description

1

Everything except the options below Primary occupancy (greater than 100 people> Essential facilities. i.e. hospitals Failure represents low hazard

II III IV

The exposure coefficient and gusting factor are influenced by the terrain's Wind Exposure type, where the options are: 1 -

Large city center

2 -

Urban and suburban

3 -

Open Terrain

4 -

Flat coastal areas 2-52


Wind is a static, horizontal uniform load. Itmay act in any direction, and as such the engineer has several items to consider: How many directions should be analyzed for sensitivity to wind?

1 2

-

Should both positive and negative directions be evaluated?

3

-

Should sorne skewed direction be evaluated?

4

-

Do nonlinear supports (i.e. horizontal guides with gaps) and/or friction affect the wind load?

5

-

Should the wind act on the piping system in the cold or hot condition?

The logic diagram shown in Figure 2-36 should serve as a guideline when setting up and analyzing wind load cases to satisfy piping code requirements. (Note: The load cases shown here only contain the basic analysis components. Other items such as imposed displacements, concentrated loads, etc. may need to he added to the load cases shown above for the user's particular job.) DOES THE PIPING SYSTEM CONTAIN FRICTION, 1-D RESTRAINTS, AND/OR GUIDES WITH GAPS?

l

YES

NO~ IS THE MOST SENSITIVE WIND DIRECTION OBVIOUS?

YES~ RUN: JOB1 1 (OPE) 2 (SUS) (OCC) 3 4 (EXP) 5 (OCC)

T+P+W P+W WIND Di - D2 S2+S3

l

RUN: JOB1 (OPE) T+P+W 2 (SUS) P+W 3 (OPE) T + P + W + WIND1 4 (OCC) D3 - Di 5 (EXP) Di - D2 6 (OCC) S4+S2

1

NO

RUN: JOB1 1 (OPE) (SUS) 2 3 (OCC) (EXP) 4 5 (OCC)

T+P+W P+W WINDX Di - D2 S2+S3

JOB2 (SUS) P+W 2 (OCC) WINDZ 3 (OCC) Si +S2

JOB2 * 1 (OPE) 2 (SUS) 3 (OPE) 4 (EXP) 5 (OCC)

T+P+W P+W T + P + W + WIND 2 D3 - Di S4+S2

*REPEAT THIS LOAD SET FOR ALL OTHER WIND DIRECTIONS (BOTH + AND -) OF CONCERN

Figure 2-36 For nonlinear systems an additional algebraic case may be required to extract the occasional bending moments from the operating hending moments. In perfectly linear systems an occasionalload case can he run alone, with this used for the stress component due to the

2-53


occasionalload. With nonlinear systems, the effect the occasionalload has on the system is linked to the effect of the operating loads on the system. The algebraic load cases shown in Figure 2-36 permits these two effects to be separated.

2.5.2 Earthquake Loading Earthquakes may be analyzed using either dynamic or static methods. Dynamicearthquake analyses, which will be covered in depth later, are not discussed here. Static earthquake loads are determined and applied in a manner very similar to static wind loads. The static loading magnitude is considered to be in direct proportion to the element's weight. Earthquake load magnitudes are given in terms of the gravitational acceleration constant, i.e. g's. If an earthquake is modeled as having a 0.5g load in the X direction, then a force equal to one-half of the system's weight is applied to the pipe uniformly in the X direction. Earthquake static load cases are set up and determined exactly as they are for wind occasional loads, i.e. by considering the same load case, non linearity, and directional sensitivity logic. In some cases the client specifies the magnitude of the earthquake loading in g's and the direction(s). In others, the analysis is left to the sole discretion of the engineer. It is not unusual to see only X or X-y components ofan earthquake. It is not uncommon to see Y only components, or X, Y and Z simultaneous components. When not provided by the client, there are a number of sources for obtaining the seismic gfactors:

Response spectrum: If seismic response spectra are available for the piping system, then, given the natural frequency of the lowest mode of vibration of a piping system, the analyst can find a corresponding acceleration from one of the curves. Ifthis acceleration lies on the right side of the peak, this acceleration may be conservatively used an overall g-factor. For more information on seismic response spectra, refer to Sections 4 and 5 of these seminar notes. Building code: Building codes provide ways to calculate seismic g-factors, based upon earthquake potential, structure type, and structure fundamental frequency. For example, the Uniform Building Code and the BOCA Basic/National Building Code calculates:

=

ZKCT

g

=

static equivalent g-factor to use for seismic design, multiples of gravity

Z

=

seismic coefficient based on earthquake zone, equal to 0.0 for Zone 0, 0.25 or Zone 1, 0.5 for Zone 2, and 1.0 for Zone 3

K

=

structure type constant, ranging from 0.67 to 3.0, dimensionless

C

=

0.051T 1/3 , but not greater than 0.1

T

=

fundamental period (inverse of frequency) of structure, sec

g Where:

2-54


ASCE #7: This standard calculates seismic g-factors in a manner similar to those of the building codes, based upon earthquake potential, structure importance, structure type, structure fundamental frequency, and soil parameters. The requirement is:

v

= ZIKCSW

Where:

v

= totallateral force or shear at the base, lb

Z

= seismic zone coefficient:

Seismic Zone

Coefficient, Z

4 3

1 3/4 3/8 3/16 1/8

2 1 0

I

=

occupancy importance factor:

Category

Description

1

1

Everything except the options below

1.0

II

Primary occupancy -

III

Essential facilities, i . e. hospitals

1.5

IV

Failure represents low hazard

NIA

> 100 people

1. 25

K

= structure type constant from Table 24 of ANSI A58.1, ranging from 0.67 to 2.5 (use K=2.0 for structures other than buildings)

C

= 1/(15 Tl/2), not greater than 0.12

T

= fundamental period (inverse of frequency) of structure, sec

S

= soil type coefficient from Table 25 ofANSI A58.1, ranging from 1. 0 to 1.5 (note that the product ofC and S neednot exceed the value 0.14, so this value should he used as a conservative maximum).

W

= total dead load

The "g'" factor can be found be dividing both sides ofthis equation by W, so: g

= V/W=ZIKCS

2-55


For piping, the generic equation for the maximum g-factor is: g

=

Z (1.0) (2.0) (0.14)

and, for the various values of Z:

Seismic Zone

Product

g-load

4

(1)(1)(2)(0.14)

0.28

3

(3/4)(1)(2)(0.14)

0.21

2

(3/8)(1)(2)(0.14)

0.105

1

(3/16)(1)(2)(0.14)

0.0525

o

(1/8)(1)(2)(0.14)

0.035

2.5.3 Quickly Applied Loads Loads that are applied near-instantaneously, and then remain constant for a reasonable duration oftime, such as fluid hammer and relief valve loads, effectively are applied with a DynamicLoadFactor(dynamicmultiplier)betweenO.Oand2.0. Thisisevidentbyassuming the worst case - no damping and instantaneous application of a constant force - and performing a time history analysis of the dynamic equation: M x(t) + K x(t)

= F(t)

Equating energies (where the kinetic energy added to the mass is Fx( t), while the crumpling energy of the spring is Kx(t)2/2): Fx(t) = Kx(t)2/2, or Kx(t) = 2 F(t) The term Kx(t) represents internally induced forces/moments within the system. The DLF is the ratio of the induced forces to the applied forces, or K x( t)max / F( t), which in this case has its maximum value of2.0. It is often highly conservative to apply twice the calculated force as a static load, but this is still often done. As the load ram p-u p time (such as the opening time of a relief valve) increases, or the load duration decreases (such as fluid hammer in a short piping leg), the DLF will decrease as well. In order to take advantage of the "true" (reduced) DLF, it is necessary to perform a dynamic analysis, such as a time history analysis or a response spectrum analysis. In lieu of a dynamic analysis, the user can only estimate a DLF, estimate the applied load, and apply a concentrated static force equal to the DLF times the applied load to the piping system.

Fluid Hammer: It is not always so easy to calculate the applied loads. One method of estimating fluid hammer loads is described in Crocker & King's Piping Handbook as: F

= P c dv A /144g

2-56


Where: F

=

fluid hammer force (exclusive ofDLF), lb

p

=

density offluid, Ibm/ft3

=

62.4 for water

=

0.0003 for saturated steam at atmospheric pressure

=

1.85 for superheated steam at 10000 F and 1500 psig

=

speed of sound in a fluid, ft/sec

=

for liquid: 12 [g Ef / (1 + D Ef / tEp)]

=

approximately 3000-4000 ft/sec for water in typical pipe sizes

=

for gas: (kgRT) 1/2

=

approximately 2000-2500 ft/sec for steam in typical pipe sizes

g

=

acceleration gravity, ft/sec 2

Ef

=

bulk modulus of fluid, psi

=

approximately 300,000 psi for water and other fluids

D

=

inside diameter of pipe, in

t

=

wall thickness ofpipe, in

Ep

=

modulus of elasticity of pipe material, psi

k

=

ratio of specifie heats for gas, dimensionless

=

1.3 for steam, 1.24 for ethylene, 1.27 for natural gas

=

gas constant, ft-Ib/lbm-oR

=

85 for steam, 55.1 for ethylene, 79.1 for natural gas

T

=

temperature of gas, oR

dv

=

change in fluid velocity causing fluid hammer, ft/sec

A

=


c

R

Relief valves: Relief valves are used in piping to provide an outlet on those occasions when pressure builds up beyond that desired for safe operation. When the pressure setting is

2-57


reached, the valve opens, allowing sufficient fluid to escape from the piping system to lower the pressure. This discharge initiates ajetforce, which must be resisted by the piping system. Valve opening time and duration of the jet load affect the dynamic response of the system, thus affecting the developed loads. Reliefvalve jet loads are normally provided by the valve manufacturer. Ifthis is not the case, the loads can he estimated by a thorough thermodynamic analysis. This methodis discussed in detail in Section 5 of these course notes. In lieu of thermodynamic and dynamic analyses, the B31.1 code provides a means of estimating the discharge force (as an equivalent static force) of a relief valve venting steam to atmosphere. The force is estimated as such:

=

DLF (M V 1 g + P A)

F

=

static equivalent discharge force, lb

DLF

=

dynamic load factor (as calculated helow), dimensionless

M

=

mass flow rate from valve x 1.11 (factor of safety), Ibm/sec

V

=

fluid exit velocity, ft/sec

=

[(2gJ)(ho - a) 1 (2b - 1)]1/2

J

=

conversion constant, 778.16 ft-Ib/Btu

ho

=

stagnation enthalpy ofsteam, Btu/lbm

a,b

=. steam constants as per following table:

F

Where:

Steam conditi on

a (Btu/lbm)

b (dimensionless)

Wet, <90% quality

291

11

Saturated, >90% quality

823

4.33

Superheated

831

4.33

g

=

gravitational constant = 32.2 ft/sec 2

P

=

static pressure at discharge, psig

=

[M (b - 1) 1 A b ][ 2J (ho - a)/g(2b - 1) ]1/2 - Pa

2-58


A

=

internaI area of dis charge pipe, in2

Pa

=

atmospheric pressure = 14.7 psi M, V, P, A F

token at this location

--

W =Weight of entire assembly

Figure 2-37 The dynamic load factor (DLF) is used to account for the increased load caused by the sudden application of the dis charge force. (Note that DLFs are discussed in great detail in Sections 4 and 5 ofthese seminar notes.) For the purposes ofthis estimate, the DLF varies between 1.1 and 2.0, depending upon the rigidity of the valve installation and the opening time ofthe valve. If the piping system is relatively rigidly restrained, the DLF can he calculated by fin ding the natural period ofvibration ofthe valve installation, treating it as a single degreeof-freedom oscillator:

=

0.1846 [ W H3 / g E 1]112

T

=

natural period of vibration, sec

W

=

weight of relief valve installation, lb

H

=

distance, run pipe to center of outlet pipe (see Figure 2-36), in

g

=

gravitational constant = 386.4 in/sec2

E

=


1

=

moment ofinertia ofinlet pipe, in4

T Where:

Next, the ratio of the valve opening time, to, to the fundamental period of vibration of the valve installation, T, should be found. This ratio is then used to determine the DLF from the chart in Figure 2-37. (Note that in the event that the opening time is not known, a conservative value of 2.0 for the DLF should be used.)

2-59


2.2

2.0

\

1.8 u...

c5

~

1.6

\

1.4

\

1.2

0.1

\

r--- rl'- i'-

0.2

0.4 0.6 0.8 1.0

2.0

4.0 6.0 8.0 10

Ratia af volve apening time ta periad af vibration toiT

Figure 2-38

2-60

20

3


3.0

Modeling And Analysis Of The Piping System .......................................................... 1

3.1

Computer Representation Of Basic Elements ........................................................... 2

3.2

Piping Configuration Modeling Techniques ............................................................. 19

3.3

Expansion Joint Modeling And Evaluation ............................................................. 20 3.3.1

Expansion Joint Stiffnesses .......................................................................... 20

3.3.2 Evaluation of Expansion Joint Allowable Movements ................................ 23 3.3.3 Use of the ERATE Program .......................................................................... 25 3.3.4 Modeling ofUnbalanced Pressure Force ...................................................... 28 3.3.5 Modeling ofTie Rods ...................................................................................... 35 3.3.6 Expansion Joint Assemblies .......................................................................... 35 3.4

Piping Nozzle Evaluation .......................................................................................... 43 3.4.1 Equipment Nozzle Load Analysis ................................................................. 43 3.4.1.1 NEMA SM23 Standard for Steam Turbines ................................. 44 3.4.1.2 API 610 Standard for CentrifugaI Pumps ..................................... 49 3.4.1.3 API 617 Standard for CentrifugaI Compressors ........................... 54 3.4.1.4 API 661 Standard for Air Cooled Heat Exchangers ...................... 54 3.4.1.5 HEl Standard for Closed Feedwater Heaters ............................... 57 3.4.2 Calculation ofVessel Stresses Due to Nozzle Loads .................................... 59 3.4.2.1 Calculation ofVessel Stresses Due to Nozzle Loads ..................... 60 3.4.2.2 Running a Sample WRC 107 Calculation ...................................... 65 3.4.2.3 Evaluating Vessel Stresses ............................................................. 75 3.4.2.4 Completing the Sample Calculation .................................. :........... 83 3.4.3

Estimation ofVessel Nozzle Flexibilities ...................................................... 84 3.4.3.1 Use ofWRC Bulletin 297 ................................................................ 86 3.4.3.2 Modeling Nozzles for Flexibility Calculations ............................... 92

3.5

Restraint Modeling .................................................................................................... 97 3.5.1 Restraint Types .............................................................................................. 97 3.5.1.1 Anchor .............................................................................................. 97 3.5.1.2 Restraint .......................................................................................... 98 3.5.1.3 Spring Hanger ................................................................................. 99 3.5.1.4 Hanger ........................................................................................... 100 3.5.1.5 Support .......................................................................................... 100

1

3.5.1.6 Snubber .......................................................................................... 101 3.5.1.7 Sway Brace .................................................................................... 103 3.5.2

Non-linear Effects ........................................................................................ 103 3.5.2.1 Friction ........................................................................................... 103 3.5.2.2 One-Way Restraints ...................................................................... 104 3.5.2.3 Gaps ............................................................................................... 104 3.5.2.4 Large Rotation Restraints: ........................................................... 105 3.5.2.5 Bi-linear Stiffnesses ...................................................................... 107

3.5.3 Evaluation of Restraint Stiffness ................................................................ 108 3.5.3.1 Use of the Structural Steel Modeler ............................................. 112 3.5.4 Use of CNODES When Modeling Restraints ............................................. 116 3.6

Miscellaneous Topics ............................................................................................... 118 3.6.1

Consideration ofCold Spring ...................................................................... 118

3.6.2

Fiberglass Reinforced Plastic Pipe .............................................................. 122

3.6.3 Underground Pipe ........................................................................................ 124 3.6.3.1 Modeling Soil Restraint ................................................................ 126 3.6.3.2 Automated Underground Piping Modeler .................................... 128 3.6.4 Jacketed Pipe ............................................................................................... 130 3.6.5 Flange Leakage Analysis ............................................................................ 132 3.6.5.1 Equivalent Pressure Calculation ................................................. 132 3.6.5.1 Flange Leakage Analysis Module ................................................ 133

2


3.0 Modeling And Analysis Of The Piping System The first two sections ofthese seminar notes have served to give the user an overview of the requirements of pipe stress analysis. This section presents the task on a detailed level, presenting ideas for modeling of various piping configurations and explaining specific analyses which may be performed to evaluate individual piping components. Included in this section is information on the following subjects: 1

Computer representation ofbasic elements

2

Piping configuration modeling techniques

3

Expansion joint modeling and evaluation

4

-

Nozzle evaluation, including evaluation of equipment loads, determination of nozzle/vessel stresses, and estimation of nozzle/vessel connection flexibilities

5

-

Piping restraints/structural modeling

6

-

Miscellaneous topics (coldspring, underground pipe, plastic pipe,jacketed pipe, flange leakage analysis, etc.)

3-1


3.1 Computer Representation Of Basic Elements Pipe stress analysis computer software algorithms are based upon certain assumptions. These assumptions serve to make the computer model (and its corresponding analytical results) only an approximation of reality. In many cases this approximation may be sufficiently close to reality to fall within the tolerances, margins, and factors of safety of the problem to be an adequate representation, while in other cases the user may find itnecessary to refine the model through more detailedmodeling. This section describes the assumptions used in the computer algorithms in order that the user may more fully understand the limitations (and the potential work-arounds) of the system. The "stiffness method" algorithm, which is used to perform the actual analysis done by CAESAR II and other prominent pipe stress/structural computer programs is described in detail in Section 6 of the se seminar notes.

o @ ~

•

•

•

Arbitrary Cross Section

Pipe Cross Section

Structural Cross Section

"Stick" Member

Figure 3-1 Piping basic elements are modeled as centerline, or "stick" members. These elements are defined by two node points (one at the "from" end, and the other at the "to" end), each with fixed spatial coordinates and six degrees of freedom (three translational and three rotational). The elements are further defmed by a constant (non-varying along the element length) set of stiffness parameters (i.e., material and cross-sectional properties). Response of the elements under load is governed according to recognized strength of material

3-2


relationships (as described in Section 6 ofthese notes), subject to certain limiting assumptions. These assumptions, described on the followingpages, govern the relationship between the mathematical model in the computer and the actual pipe existing in the power plant or refrnery. AlI elements remain stable under load (local buckling of cross-sections is ignored):

1

48" 00

0.375" Wall

Local Buckling of Cross Section

Section A-A

Figure 3-2

2

-

Plane sections remain plane:

Center of Bending

D

B

Figure 3-3 The computer algorithm assumes that points A and B (of Figure 3-3) always lie on the same cross-sectional plane, whether in the deformed or the undeformed state.

3-3


J

L

1

~

:~

B

1

0 F

L'

Figure 3-4 In reality, the moment F x L (in Figure 3-4) does not produce a uniform "plane-

sections-rem ain-plane" bending load at the cross-section A-B, since it causes local warping. 3

-

Hooke's Law is applicable across all fibers of the cross-section:

Compressive Normal Stress

"---.. Tensile Normal Uniform Bending Stress

Pipe Shoe W/Saddle

(Normal stresses very linearly from the neutral axis)

Saddle inhibits uniform bending and extension along ail fibers at the cross section.

Figure 3-5 4

-

Hooke's Law is applicable throughout the entire load range:

7

1

Stress Distribution Remains Unear

... Not Plastic

Figure 3-6

3-4


5 -

Moments and forces applied to the beam are assumed to act about the neutral axis:

L

Should not be modeled as: (Unless the F * L moment can be assumed negligible)

Figure 3-7 6

-

Element cross-sections do not ovalize under load (except as adjusted for bend elements):

A

-0-

~CTLA-A

A

This ovalization will make the pipe more flexible, i.e. the pipe will bend easier. Ovalization of this type for straight pipes is not considered.

Figure 3-8

The stresses at the ovalized section are intensified due to: 1 2

reduction in section modulus, and -

added local plate bending in the top and bottom fibers.

3-5


7

Applied loads are not affected by the deformed state of the structure (P-delta effect):

Np

6. =

F - -..-

0.25 in

=

1000 lb

1 1 1 1 1 1 1 1 1 1

Figure 3-9 In reality, there will be an addition al moment applied to the system, equal to the load times its displaced distance from the neutral axis ofthe structure (i.e., 1000 pounds x 0.25 inches = 250 in-lb). The computer software models this load as strictly a force with no applied moment. 8

Rotational deformations of the system are assumed to be small:

Sterling From y

Sequentiel 90· Rotetions About Z, Y, end X-Axes About Y About X About Z y

y

y

,~, ,~, /~, /~, Sterling From y

Sequentiel 90· Rotetions About Y, X, end Z-Axes About Y

About X

y

y

About Z y

,~, /~, /~, ,~, Figure 3-10

3-6


Node point rotations are added vectorially by the computer software. This is not a valid representation of reality for large rotations, as demonstrated in Figure 3-10 for three 900 rotations. 9

Boundary conditions are assumed to respond in a linear fashion:

Non-Linear Restraint Response Different for Up and Down Loads (OneWay Restraint)

Linear Restraint Response Constant Throughout Load Range

Figure 3-11 The stiffness algorithm cannot solve for non-linear restraint conditions, such as one-directional restraints, bi-linear restraints (soil or bottomed out springs), friction, etc. However, CAESAR II does include a procedure which overcomes this limitation; see point 8 below. These limitations are of the most concern when modeling the following situations (pointers for increasing accuracy in each situation are also given): Large Diameter/thin wallpipingorducts: In thiscase,itisadvisabletominimize localized loadings by distributing them with pads or saddles, or do plate buckling analysis (preferably with finite element software) when the loads cannot be altered.

1

2

-

3

4

Localized stress conditions not explicitly covered by an SIF, i.e. a saddle: The portion of the pipe impacted by the saddle may be modeled as a rigid element, while saddle/piping local stresses may be estimated through the use of finite element analysis or through the use ofWelding Research Council Bulletins, such as 107 and 198. Pipe connections to thin walled vessels: The flexibility of the connection may be modeled by a flexible element (such as that generated using Welding Research Council Bulletin 297), while stresses in the pipe and vessel may be estimated through the use offinite element analysis orthrough the use ofWel ding Research Council Bulletins 107 and 297.

-

Highly corrosive systems (especially when subjected to cyclic loadings): Corrosion of a pipe results in an irregular cross-section which is usually modeled by using the uncorroded cross-section for load generation (weight and thermal forces), and the fully corroded cross- section for calculation ofthe section modul us

3-7


(stress calculation). Corrosion is much more dangerous under fatigue loadings due to the fact that it provides many more opportunities for crack initiation; in order to compensate, a low cyclic reduction factor should be used to match the allowable expansion stress range to the fatigue curve for a highly corroded material. 5

Elbows: Elbows ovalize significantly when subjected to bending loads. This can be accounted for by increasing the flexibility ofthe elbow element in the computer model and multiplying the calculated stress by a stress intensification factor (this is done automatically by most programs such as CAESAR II). Code defined "flexibility factors" for bends have been determined theoretically and verified experimentally.

t--_~dline

= "\

Flange .A" 1

r - rnean raJ

~

p:l

th: r

(From BS 806-1975)

~tion

cross

Figure 3-12 The flexibility and stress intensification factors of bends must be reviewed in those cases where ovalization is inhibited (such as when the elbow is stiffened by flanges or welded attachments). The piping codes provide correction factors for bends with one or two flanges, but omit geometries such as shown in Figure 3-13.

(A)

(8)

(C)

Figure 3-13 These attachments almost certainly affect the flexibility, and more importantly, the stress intensification factors for the bends. The factors for heavily stiffened bends, such as that shown in Figure 3-13(A), could he estimated using finite element analysis, or stiffness could he increased by modeling the elbows as

3-8


flanged, or simply as straight pieces ofpipe (with increased stress intensification factors applied). In less pronounced cases such as those shown in Figures 3-13 (B) and (C), deviations from the response of an unstiffened bend is usually ignored. 6

-

Loadings which produce stresses which are weIl outside of the code allowable ranges: These loads will tend to produce stresses weIl beyond the material yield stress, stresses in the buckling range, large dis placements resulting in significant P-delta loads, or large rotations Oeading to inaccurate results). This limits programs such as CAESAR II as accurate analysis tools throughout the full range of potential loadings. However accuracy is not affected for those loads which are ofmost interest to the engineer- code allowables are based upon the fact that the analysis being done assumes linear material response.

7

-

Non-linear boundary conditions: The effects of non-linear restraints must be simulated through an iterative process aimed at convergence of the non-linear restraints inlegitimate states-forexample, with the pipe liftedoffataone-way support (and with the support function removed from the analysis), or with the pipe sliding along a frictional restraint (and with an appropriate force applied along that line of action in the analysis). This process is activated (during static analysis) automatically when a non-linear effect is detected by CAESAR II.

8

Non-homogenous elements: As noted, piping elements are modeled as stick elements ofconstant cross-section and material properties. In certain cases, such as with reducers, which have a variable cross-section, this is not a valid representation. An element such as this is usually modeled as a single, or as a series of elements, each having average parameters. For example, a l2x8 standard wall reducer may be modeled as a 10-inch standard wall pipe Capproximately the average of the inlet and outlet pipes), or as two segments, with outer diameters and wall thicknesses interpolated hetween the two. When using those codes which define a stress intensification factor for reducers, one would have to he calculated and specifically applied at that location.

9

Rigid elements: Rigid elements, such as valves and flanges are most difficult to model due to the inability to represent their geometry, and their stress distribution with stick elements. Therefore, pipe stress software cannot be used to accurately determine the effects ofthe piping system on rigid elements. Analysis of these components is best left to fmite element analysis, test, or other recognized methods. However, the effects of the rigid elements on the piping system can he simulated by providing an element ofhigh relative stiffness in the model (it is always more important to adequately model relative stiffnesses than absolute stiffnesses when constructing a model). This is done by providing an element with sufficiently large cross-section, and having the defined weight of the rigid item. In CAESAR II, a rigid element is modeled as having: a) aninsulationfactorof1.75Ccompared tothematchingpipe), unless a zero weight rigid Ca modeling construct) b) fluid weight of the matching pipe added, unless a zero weight rigid c) the same inside diameter and 10 times the wall thickness ofthe matching pipe 3-9


3.2 Piping Configuration Modeling Techniques Piping may he modeled in varying detail, depending upon how much accuracy is required. This section looks at the various ways (providing corresponding degrees ofaccuracy) in which sample piping configurations might he modeled. Consider the following geometry, of a large diameter pipe supported hy a dual spring assemhly:

1

\

t

J 1\

Vl L

_

50" diameter

1/2" wall -.J ____ _ 4· trunnion support shown. (Typ. both sides)

V

Figure 3·14

Simplest Method:

50" diameter pipe Flexible Restraint K at centerline K =2 Kspring

Figure 3·15 Limitations: 1

-

local stress calculations not considered for 50" pipe

2

-

stiffness of trunnions not considered

3

-

torsional resistance due to the restraint pair is not considered (see Figure 3-16)

4

-

local flexihility of the shell of the 50" pipe is not considered

3-10


Figure3-1G More Accurate:

L

Rigid element - zero weight Kspring

Kspring

Figure 3-17 Limitations: 1

local stress calculations not considered for 50" pipe

2

stiffness of trunnions not considered

4

local flexibility of the shell of the 50" pipe is not considered

Most Accurate:

Ksheillocal (WRC 297)

Kshell local (WRC 297)

\

~-------il,...------____

\

/

Pipe element

/ 'modeling trunion Kspring

Rigid element - zero weight Length = 1/2 00

Figure 3-18

3-11


Limitations: 1

local flexibilities and stresses only as close as WRC 297 and WRC 107 bulletins (see discussion in this Section 3.4 of these seminar notes)

Looking at another configuration, a he avy-wall forged WYE fitting:

Figure 3-19

Simplest Model:

~

Incoming pipe with branch properties coded to intersection

~point. Apply SIF's forwelding tee here. Incoming pipe with header~ properties coded to ~ intersection point.


weight offorged fitting probably underestimated considerably

2

rigidity of forged fitting probably underestimated considerably

3

stress intensification factors may be too conservative

3-12


More Accurate:

Rigid elements whose cumulative weights equal that ot the torged titting and tlanges.

Figure 3-21

Limitations: 1

no provision for stress calculations in forging, but this isn't usually a problem, because ofthe extra heavy wall of the fitting would ensure that the connecting pipe would probably fail first. Any questions regarding load capacity should probably be directed to the fitting manufacturer

Most Accurate:

Rigid element modeling tlange.

Figure 3-22

3-13


Comments: 1 2 3

the flexibility ofthis model will he more accurate (but only marginally so for a heavy fitting) -

stresses (unintensified) will be computed at the crotch; however, there will be sorne unknown intensification factor existing at the crotch this model probably does not yield any significant im provement overthe previous one

One of the most common types ofpipe support is shown in Figure 3-23:

6" dia. std. wall stanchion

Figure 3-23

Simplest Model:

·V" direction restraint applied at the bend point shown.

Figure 3-24

3-14


Limitations: 1

flexibility of the stanchion is not included in the model

2

-

the point of application of the stanchion is not at the correct location on the bend curvature

3

-

pipe may lift off of (or lock up with) modeled support due to thermal expansion between centerline of horizontal run and point of application on riser

4

-

stiffening effect on bend of stanchion not considered

5

-

local stresses at stanchion not considered

More Accurate:

~ Pipe between

·C·

t

nodes Cand D with properties of stanchion

Figure 3-25

Limitations: 1

stanchion doesn't act at the proper point on the bend curvature

2

-

stiffening effect on bend of stanchion not considered

3

-

local stresses at stanchion not considered

3-15


Most Accurate:

41.4 deg. For Long Radius Bend 48.2 deg. For Short Radius Bend

R(1 - cosa )--....-4

L LL U1' -SiOO}

el

1

Use WRC107 To Calculate Local Stresses

Aa

1

1

Ct

:'-------

1

Pipe Element Between Points "B" and ·C". Restraint at Stanchion Node "C"


points A and B aren't exactly at the same location (this can be resolved using CAESAR n's "OFFSETS" feature, but other pipe stress software may have a difficult time with this)

2

-

modeling the stiffening effect of the stanchion on the bend through the use of a single flange bend is an approximate solution

3

-

local stresses at the stanchion are only as accurate as WRC 107 bulletin

A few configurations which illustrate solutions to potentially tricky modeling situations follow below: The distance L in Figure 3-27 may become important if the gap on the guide closes and there is a horizontal restraint force which will cause a torsional moment to exist in both members.

3-16


Centerline of the pipe "stick" model

L Centerline of the structural 'stick" model

Figure 3-27 Because the elbow in Figure 3-28 connects directly to the equipment flange and the equipment flange is anchored, the stiffness ofthe model in this local region is very high. If the stanchion connects at A and the equipment centerline is at B, the differential thermal growth of the elbow between those points could put enormously high loads on both the stanchion and the equipment model. This is also in reality, a difficult problem to design for. Unless the user is willing to put a spring at the stanchion location, the differential thermal growth in this small area might result in large nozzle loads.

A Rotating Equipment Centerline

B

Figure 3-28 In the Figure 3-29, a small, but heavy process monitor and actuator is mounted on the line. The rigidity, weight, and moment due to the offset is best modeled using a weightless rigid element going from the centerline of the pipe out to the center of gravity of the process monitor, at which point a small rigid element with the weight ofthe equipment should be modeled. The rigidity ofthe body of the monitor (within the pipeline) should be modeled as

3-17


a rigid as weIl. (Note that some engineers may prefer to model the effects ofthis equipment by applying a force equal to the weight and a moment equal to the weight times offset at the centerline ofthe pipe. This approach, although acceptable for static analysis, is absolutely incorrect for dynamic analysis, and should therefore be avoided since it cannot be promised that no dynamic analysis will he conducted on a system in the future.)

ç Pump Figure 3-29 In Figure 3-30, the large 18 inch line comes directly from a flue-gas furnace, passes through a small exchanger and enters a waste heat boiler. This is a very stiffsystem relative to the vessel connections. Therefore, instead ofmodeling the connections as rigid anchors (which would give the same relative stiffness to the restraints and to the piping), WRC Bulletin 297 should he used to estimate and model the nozzle flexibilities. This method will provide the best approximation of the distribution of the piping loads to the vessels.

Soiler

--~~--~--~~

~ ~umace

Soiler and fumace nozzle flexibilities? (This is a very tight. stiff system)

Figure 3-30

3-18


In Figure 3-31, rectangular ducting connects the two separators, which are rigid relative to the ductwork. In order to size each spring for i ts share ofthe distributed weight ofthe whole assembly plus the connected piping, it is best to simulate the stiffness of the duct through the use of an equivalent structural member or piping element.

User defined cross section in structural

Separator

~{!Sprin9

stee'

Pi'"' ' ' '

"Ri9idElem7 ! Separators modeled as pipe

Location

Figure 3-31 An angle valve could be modeled as shown in Figure 3-32. It may be necessary to model it as three rigid elements if the weigh t of the operator is significant in com parison to the valve body.

Figure 3-32 The following sections of these seminar notes provide more detailed methods for modeling and analyzing specifie components of the piping model.

3-19


3.3 Expansion Joint Modeling And Evaluation Expansion joints are used when it is necessary to provide a large amount offlexibility in a small space. Expansion joints are constructed out of sheet metal, which, after rolling and welding to for a cylinder, has convolutions (also called corrugations) formed in it via either hydraulic pressure or rolling. Expansion joints may vary in terms of the number and type of convolutions, the material, the number of plies, all ofwhich effect the pressure capacity, the stiffness, and the allowable movement.

Figure 3-33 For the most part, these details are taken care ofby the expansion joint manufacturer. A typical expansion joint piping design proceeds: The decision is made to use an expansion joint in the piping system. (In many design problems the joint is used to protect a sensitive piece of equipment from excessive nozzle loads.)

1

2

-

Based upon the design temperature and pressure, a standard expansion joint is selected from a manufacturer's catalog. The properties ofthat bellows are then inserted into the piping model.

3

-

Ifthe bellows reduces loads and stresses as intended then the range ofexpansion

movements on the bellows must be checked. For each bellows there is a limit to the cumulative axial, bending and lateral displacement that can be absorbed by the joint without excessively deforming the convolutions or causing fatigue failure. These limits are presented in different ways in different manufacturer' s catalogs, but are always functions of the number of applied cycles, bellows material properties and convolution shape. Where excessive displacement is a problem, increasing the number of convolutions can be the solution. 4

-

Once the bellows movement is within the allowable range of movements, the design is completed. A competent expansion joint manufacturer should be able to provide assistance throughout the design stage as required.

3.3.1 Expansion Joint Stiffnesses Each particular combination of material, thickness, and convolution geometry has a different axial spring rate (per convolution) associated with it. Bending and lateral convolution spring rates can be computed from the axial spring rate.

3-20


The behavior of a bellows under load is described by the following equations:

Where: F

=

axial force in each convolution (also the axial force throughout the entire bellows), lb

f

=

axial stiffness per convolution, lb/in

= NKax N

=

number of convolutions in the joint

Kax

=

total expansion joint axial stiffness, ib/in

ex

=

axial displacement per convolution, in

=

X/N

=

total axial dis placement of joint, in

Mr

=

bending moment in each convolution (also the bending moment supported by the entire bellows), in-lb

D

=

is the effective diameter of the joint (equal to the inside diameter plus the height of one convolution), in

er

=

axial displacement per convolution resulting from a rotation of the convolu tion,in

=

(rxD)/(2N)

r

=

bending rotation of single convolution, radians

v

= fD

x

Where:

ey /

(2 1)

Where:

v

=

shear force in each convolution (also the shear force supported by the entire bellows), lb

3-21


=

axial displacement per convolution resulting from a lateral deflection of the convolution.

=

3Dy/(NI)

y

=

totallateral displacement of the joint, in

1

=

length of the bellows, in

ey

These expressions can easily be converted into stiffness and flexibility coefficients:

=

F/x

Bending Flexibility: MIr

=

(1/8) (Kax) (D2)

Lateral Stiffness:

=

(3/2) (D2) (Kax) / (12)

Axial Stiffness:

Kax

V/y

These stiffness values are provided in most manufacturer's catalogs. In the event that the manufacturer only gives axial stiffness, the other two can be calculated once the effective diameter and length are known. (Note that torsional stiffnesses are not usually provided, since unprotected expansion joints are not designed to carry torsionalloads and may fail catastrophically if inadvertently exposed to even moderate torsional moments.)

Note however that the bending flexibility coefficient should not he used in any piping program. The bending stiffness that should be used is exactly four times the hending flexibility. This is because the so-called bending flexibility is calculated by applying a moment (M r ) to the free end of an expansion joint and observing its end rotation (9). A computer model, however, expects a bending stiffness to be the ratio of the applied moment to the angular rotation at the end of an expansion joint that is fixed against translation - i.e., a representation of guided cantilever. This angular stiffness for a guided cantilever expansion joint model is calculated as:

Mrfr

=

(Kax) (D2) /2

FLEXIBILITY

STIFFNESS

Figure 3-34 Some pipe stress programs only offer "point", or zero-Iength expansion joint models. (In CAESAR II the user can define "fmite length" or ''point'' expansion joints.) There is a difference in terms of how the two models are entered. As seen above, for finite length expansion joints, the lateral and bending stiffnesses are related by the equation:

3-22


Bending Stiffness

=

M!r = Kax x D2/2

=V/y x 12/3

=

Lateral Stiffness x 12/3

Because ofthis exact relation, and since the length is known, the user can only enter one of these two values. CAESAR II computes the other value using this equation. For a "point" expansion joint, the length is unknown, so aIl three stiffnesses must be definedfor the model.

Example: Consider an expansion joint with the following parameters: Nominal diameter = 4 in Effective Area = 19.6 in2 Kax (from manufacturer) = 316 lb/in

Bellows Length = 4.447 in The expansion joint stiffnesses are calculated as: M!r

=

(1/8) (Kax) (D2)

D

=

[4 x 19.6/ pi

M!r

=

(1/8) (4.9955 2 ) (316) = 985.7 in-lb/rad = 17.2 in-Ib/deg.

]1/2

= 4.9955 in

The bending stiffness to use in a piping program would be: 4 x M!r V/y

=

4 x 17.2 = 68.8 in-Ib/deg

=

(3/2) (D2) (Kax) / (12)

=

(3/2) (4.9955 2 ) (316) / (4.447 2 ) = 598.14 lb/in

3.3.2 Evaluation of Expansion Joint Allowable Movements Since the failure mode of expansion joints is fatigue, the relative expansion displacements hetween the start and end ofthe expansionjoint must be checked against the manufacturer's allowables. Note that the allowables provided will not be absolute values, but will he based upon a specifie number of cyclic applications. The manufacturer must always provide a fatigue curve or some other type of adjustment factor in order to determine the allowable displacement for a different number of cycles. Occasionally, the manufacturer provides allowable movements only for axial displacements. In this case, the equations given in Section 3.3.1 can be used to calculate an equivalent axial displacement from lateral and rotational displacements: Er

=

R D / 2 , or:

Er

=

0.00872665

eD

3-23


=

3DY/l

Er

=

total equivalent axial displacement due to rotation, in

R

=

total rotation on expansion joint, radians

D

=

effective diameter of expansion joint, in

e

=

total rotation on expansion joint, degrees

Ey

=

total equivalent axial displacement due to lateral displacement, in

y

=

totallateral displacement on expansion joint, in

1

=

length of expansion joint, in

Ey Where:

Therefore, movements on an expansion joint are acceptable if:

x + Er + Ey <= Xall, or:

x + 0.00872665 D e +

3 DY /1 <= Xall

Where:

x

=

actual axial displacement of expansion joint, in

Xall

=

allowable axial displacement of expansion joint, in

Example: Assume that the expansion joint used in the example from Section 3.3.1 must be checked for its range of applied expansion displacements. The following parameters apply: Number of Convolutions = 12 Manufacturer's allowable axial displacement (for 3000 load cycles) = 1.43 in The expansion joint runs between node points 120 and 125, with the axis of the bellows coincident with the global X-axis, in the stress analysis model. The range of expansion displacements (assume that fewer than 3000 cycles are expected) from the stress analysis output are shown below:

NODE

DX

DY

OZ

RX

RY

RZ

120

0.3

0.25

0.0

0.0

1.23

0.03

125

-0.1

0.12

0.0

0.0

-0.02

0.89

3-24


Now it is necessary to compute the relative dis placements between the ends ofthe expansion joint. This is often not a simple task since the rotations ofthe expansionjoint ends can cause lateral translations which don't produce axial deformation of the joint. In the event that displacements and rotations are small, and the expansion joint runs along a global axis, the relative dis placements of the expansion joint can he fairly closely approximated:

x e

= (DX1-DX2) = 0.3 - (-0.1) = 0.4 in =

[(RY1-RY2)2+(RZ1-RZ2)2]1/2=[(1.23-( -0.02»2+(0.03-0.89)2]1/2 = 1.52 0

Y

=

[(DY1-DY2)2+(DZ1-DZ2)2]1/2 = [(0.25-0.12)2+(0.0-0.0)2]1/2 = 0.13 in

Using the interaction formula, the range of expansion movements is checked as:

x + 0.00872665 D e +

3 DY / l <= Xalh or:

0.4 + 0.00872665(4.9955)1.52+3(4.9955)(0.13)/4.447 = 0.904 in <= 1.43 in

3.3.3 Use of the ERATE program For more complex configurations, the relative expansion joint end displacements can be evaluated using an auxiliary routine (accessible from the WRC 297/SIFlFlange option of CAESAR fi). Full instructions for the use ofthis routine are found in the CAESAR fi U ser's Manual. This routine requires 23 data items (with a single value per line), describing the orientation, length, and end displacements of the expansion joint; from this, it calculates the relative displacements. The required items are described below: 1

2 3 4 5 6 7 8 9 10 Il 12 13 14 15 16 17 18 19 20 21 22 23

-

Nodei Nodej N L D Xi Yi Zi X'~

y

J

Zj -

-

node number of "from" end of the expansion joint node number of "to" end of the expansion joint number of convolutions in the expansion joint flexible length of the expansion joint effective diameter of the expansion joint node i's X-coordinate node i's Y-coordinate node i's Z-coordinate node j's X-coordinate node j's Y-coordinate node j's Z-coordinate

DXi DYi DZi

RXi RYi RZi DX'J DYj

DZj RX'J RY'J RZj

3-25


The input for the current example is shown in Figure 3-35.

EJMA EXPANSION JOINT RATING Node Number for "FROM" end •.....•...•.•...•.•.. Node Number for "TO" end ••••....••...••.....•.. Number of Convolutions ..••••....••..••......••.. Flexible Joint Length ...••••....••..•••.... (in.) Effective Diameter ..•...••••....••...•..... (in.)

120.000 125.000 12.000 4.447 4.996

X Coordinate of "from" end .•....••........ (in.) Coordinate of "from" end .•....••...••... (in.) Z Coordinate of "from" end .•........•••... (in.)

.000 .000 .000

X Coordinate of "to" end ••.•....••..•••... (in.) Coordinate of "to" end ••••....••..•••... (in.) Z Coordinate of "to" end ••.....•••..•••... (in.)

4.447 .000 .000

X Displacement of "from" Displacement of "from" Z Displacement of "from" X Rotation of "from" end y Rotation of "from" end Z Rotati on of "from" end

end ...•••..•••... (in.) end ....••..•••... (in.) end ....••..•••... (in.) ••••....••..•••... (deg) ••••...•••..•••... (deg) ••••...••...•••... (deg)

.300 .250 .000 .000 1.230 .030

end ••...•••..•••... (in.) end ••....••..•••... (in.) end ••....••..•.•... (in.) ...•••....••..•••... (deg) ..••••....••..••.... (deg) .•••••.........••... (deg)

-.100 .120 .000 .000 - .020 .890

y

y

y

X Displacement of "toN Displacement of "to" Z Displacement of "to" X Rotation of "to" end y Rotation of "to" end Z Rotation of "to" end y

More

Compute Help Figure 3-35

3-26


Output for this problem is shown in Figure 3-36.

EJMA EXPANSION JOINT RATING

More

OUTPUT: AXIAL DISPLACEMENTS PER CORRIGATION Axial Axial Axial Axial

Displacement Displacement due to Lateral Displacement due to Rotation Displacement TOTAL

.033 .044

.006 .083

RELATIVE MOVEMENTS OF END "i" WITH RESPECT TO END (Local Joint Coordinate System) Relative Relative Relative Relative

Axial Displacement. "x" Lateral Displacement. My" Bending. "theta" (deg) Torsion (deg)

.401

.158 1.517 -.019

RELATIVE MOVEMENTS OF END "i" WITH RESPECT TO END (Global Piping Coordinate System) Relative Relative Relative Relative Relative Relative

"j"

X Displacement Y Displacement Z Displacement Rotation about X (deg) Rotation about Y (deg) Rotation about Z (deg)

"j"

-.399 -.132 .095 .000 1.250 .860

To Exit For Help

Input Output

More

Figure 3-36 Results - e(total) - are given in terms of equivalent axial displacement per convolution - in this case, 0.083 inches per convolution. The accuracy of the initial estimate (done without the benefit ofthe ERATE program) of an equivalent axial extension of 0.904 inches can be determined by multiplying 0.083 times 12 convolutions, for a total equivalent extension of 0.996 inches (an error of approximately 9%). It can he seen that the bulk of the error is due to the underestimation of the lateral displacement (see x, y, theta, tors - the total axial, lateral, rotational, and torsional displacements respectively on the bellows as a whole) on the expansion joint - the estimate was 0.13 inches, while ERATE calculated it asO.158inches. Thisisimportant,sincelateralloadingisverycritical. Additionally, without the use of a calculational aid such as ERATE, the lateral displacement will be the most difficult to estimate, especially as the rotational angle increases. Torsion on the expansion joint should be approximately zero. Ifnot, it is recommended that the acceptability of the bellows for the torsionalload be verified by contacting the expansion joint manufacturer.

3-27


3.3.4 Modeling of Unbalanced Pressure Force Normally pressure in a pipeline is absorbed by tension in the pipe walls:

Figure 3-37 12 inch nominal diameter pipe, standard wall (Di = 12 in), P = 250 psi Axial Tension = Area x Pressure = pi/4 di2p

= pi/4 (12 2) (250) = 28,274 lb Due to the axial flexibility of expansion joints, they are incapable of carrying this large axial load. Due to the larger internaI diameter of the convolutions of the expansion joint, the pressure thrust force is calculated from an effective diameter, which is:

Where: Deff

=

effective diameter ofbellows, in

Di

=

internaI diameter of expansion joint (or pipe), in

w

=

internaI height of one convolution, in

The unbalanced pressure force is therefore:

=

P (pi/4) De#

Fp

=

pressure thrust force, lb

P

=

system pressure, psi

Fp Where:

The pressure force is actually developed at the point where the pressure encounters the first metal area perpendicular to the axis ofthe expansion joint - for example, a capped end or a change in direction. The distribution of the pressure thrust loads in a number of configurations is shown in Figures 3-38 through 3-42.

3-28


Pressure thrust loads in untied, unanchored expansion joints:

~_.E~: ]10

"

End is blind flange

T

Tl

=

tension in bellows walls (tending to blow the bellows apart), lb

=

P (pi/4) DetF

=

tension in pipe wall

Figure3-3S

Pressure thrust loads in anchored expansion joint installation:

~,

" T2

Flanged end anchored

=

tension in bellows walls, lb

=

0

Tl

=

compression in pipe wall, lb

T2

=

totalload on anchor, lb

=

P (pi/4) DetF

T

Figure 3-39

3-29


Pressure thrust in tied expansion joint:

:'~~iF -- -Tl

T~~ Ts •

=

tension in bellows, lb

=

0

Tl

=

tension in pipe wall, lb

T3

=

tension in tie bars, lb

T

Tl

--9i Tl

Tl

Il \

Figure 3-40

Pressure thrust loads in rotating equipment without tie bars (ends anchored):

=


=

0

Tl

=

compression in pipe on either side of expansion joint, lb

T2

=

pressure load on impeller and on inside of elbow, lb

T4

= =

reaction load on pump base and on anchor, lb

T

P (pi/4) DetF

Figure 3-41

3-30


Pressure thrust loads on rotating equipment with tie bars:

T

Tl

T2

T3

=


=

0

=

tension in pipe on either side of expansion joint, lb

=

P (piJ4) Di 2

=

pressure load on impeller and on inside of elbow, lb

=

P (piJ4) Di 2

=

tie bar load, lb

=

P (piJ4) De~

Figure 3-42 Note that there are no reaction loads (T4) when tie bars are used. Most pipe stress programs such as CAESAR II automatically calculate the unbalanced pressure thrust load and simplify the model by applying the entire magnitude at either end of the bellows. In most cases, this is an adequate approximation of the actual situation. Greater modeling accuracy can be achieved by disabling the application of pressure load at the bellows (by defming an effective diameter as 0.0), and calculating and applyingthe thrust load manually to the model as so: Apply the force T - Tl at the ends of the bellows. Apply the force T2 - Tl at the locations identified by the T2 arrows. Expansionjoint, tie rod, and reaction loads T, T3, and T 4 will be calculated correctly by the program. In any event, even though the tension/compression in the pipe wall may not be completely accurate in the default computer model, the load tending to open the bellows will be; this is usually a much more critical detail to consider when designing a system to absorb the pressure thrust. It is left to the user to confirm that this is normally not a major design issue.

3-31


3.3.5 Modeling of lie Rads Ifrestraints are not used to absorb the unbalanced pressure load, it must be handled by tie rods. These are tension-carrying rods attached to either end of the expansion joint, which prevent the ends from pulling apart. Since tie bars assure that there is no pressure reaction from the expansion joint, tie rods can be modeled in two different ways: implicitly, by omitting both the tie-rods and the pressure load from the model, or

1 2

-

explicitly, by including both the tie-rods and the pressure load in the model.

Assuming that the tie rods absorb 100% of the load, the net effect ofboth ofthese models on the piping system are the same.

Implicit Model of Tie Rods: The first case noted above is obviously the simpler ofthe two. Omitting the tie rods is possible if there is no pressure load on the bellows; this can be omitted by defining an effective diameter equal to 0.0 for the expansion joint. Tie rods, besides absorbing the pressure load, also prevent extension and compression of the bellows under piping operating loads. Therefore, when the user leaves the tie bars and pressure thrust out ofthe model, it is also necessary to set the axial stiffness ofthe expansion joint to be essentially rigid (or actually to the total axial stiffness ofthe tie rods, which is AE/ 1). If the axialload on the expansion joint is tensile then the surrounding pipe is trying to stretch the tie bars even further. Ifthe axialload on the expansionjoint is compressive, then: 1

If the compression is less than the pressure thrust load, there is not a problem.

2

If the compression exceeds the pressure thrust load, then the tie rods will be in compression. The compression must be checked to ensure that it is not so great that it buckles the tie bars. If the tie rods are tension only (i.e., lock nuts are placed only on the outside ofthe expansion joint flanges) then some redesign is required, either: a) put nuts on the tie bars on both sides of the expansion joint flanges, or b) redesign the piping system so that the compressive load is not so great.

Explicit Model of Tie Rods: When explicitly modeling the tie rods, the pressure load is included in the model by defining an effective diameter for the expansion joint. The tie rods can be modeled by using a structural element (of the same cross-sectional area as the tie rods) to connect the two ends of the expansion joint. The structural element used could be a pipe, a rigid element, or a user-defined structural steel element.

3-32


In the event that a single pipe element is used to represent the tie rods, the best way to provide the same axial stiffness as the tie rods is to leave the diameter equal to the diameter of the attached piping and set the wall thickness approximately equal to:

=

(N/4) Drod2 / Do

N

=

number of tie bars

Drod

=

diameter or tie rod, in

Do

=

outer diameter of pipe element, in

t Where:

If a rigid element is used to model the tie rods, again the diameter should be set to that of the attached piping; the wall thickness should be set to: t

=

(N/40) Drod2 / Do

The rigid element should be given a weight equal to the total weight of aU of the tie rods, which, if made of steel, is approximately:

=

N (0.283) L (pi/4) Drod2

W

=

total weight of tie rods, lb

L

=

length of tie rods, in

W Where:

The tie rods are modeledin CAESAR II to resistonly axialloads through the use ofrestraints with "CNODEs" (other nodes in the system to which a restraint is connected). Consider the expansionjointltie rod assembly shown in Figure 3-43. The bellows element is modeled as running between the two node points 5 and 10. The tie rod element is then run from node point 5 to node point 20, using the same delta-coordinates as for the expansion joint. This puts node points 10 and 20 at a coincident location, without any actual attachment. The attachment is provided by placing a restraint at the far end of the tie rod (node point 20) in the direction of the expansionjoint axis, in this case the Y direction. Placement of a restraint here in this manner restrains node point 20 (the end of the tie rod) against a rigid point in space; this can be adjusted by defining the restraint node point 10 as a CNODE. This means that node point 20 is not restrained against a point in space, but rather that it cannot move in the global Y-direction relative to node point 10 - the end of the expansion joint - an effective representation of a tied expansion joint.

3-33


Tie bars designed to take compressive or tensile loads axially.

J!j) ~~ -

"""""::iilif1 '1

,~ ~

1

.... -

Figure 3-43 The tie rods should be set to the ambient temperature, if they are outside of the piping insulation, or to a temperature doser to the operating temperature ifthey are inside ofthe insulation. Tie rods may also be modeled in a more complex fashion, using multiple rigid elements, as shown in Figure 3-44. In this model, loosening of the nuts on the rods due to rotation of the expansion joint flanges will be simulated.

2025 3025---<1.. -~~.... 1025 4025

15

Figure 3-44 As noted, tie rods must be checked for potential buckling loads after the analysis is complete. Or alternatively, they may be designed to take tension only. This is done by placing locknuts only on the outside of the flange, as shown in Figure 3-45. In this case, the expansion joint is prevented from extending by the nut, while the flange can move freely during joint contraction. This configuration can be modeled in CAESAR n by using one-way restraints (or even gaps, ifappropriate) between the end of the tie rodelement and the CNODEs. For example, if the tie rod shown in Figure 3-42 was tension only, it would be modeled by placing a +Y restraint at node point 20, with 10 as the CNODE, indicatingthat the end of the tie rod cannot move down against the expansion joint (but can move up).

3-34


Possible free movement of tie bar end:.,:

• Rigid element from centerline of pipe to centerline of tie bars. TENSION ONLY TIE BARS

Figure 3-45 Complex expansionjointltie rod models are cumbersome to build and check, but where hot, large diameter tight piping systems are being analyzed they yield the most accurate model. This is especially true where tie rods are long and not designed for compression. In these cases a slight rotation of the expansion joint can put one side ofthe tie rods in compression and the other side in a greater tension.

3.3.6 Expansion Joint Assemblies Expansionjoints may be used in a number of different types of assemblies, based upon the application. Various assemblies are described and sample models thereof are shown. (Note that CAESAR II provides an expansion joint modeling feature which can automatically build many of the expansion joint assemblies shown here. It is accessed by pressing J at the piping input spread sheet.)

3-35


Ringe Joints: Hinge expansion joints are shown in Figure 3-46.

SINGLE HINGE EXPANSION JOINT

DOUBLE HINGED EXPANSION JOINT

;HingeArm\

~

/

r--/ l

[1[1

1

fi -

_ / Hinge Pi/"

0

][

Center Hinge Arms / Hinge Pin

ilU

"'-

'Hinge Pin

Figure 3-46 The hinges restrict angular rotation of the beIlows to a single plane, and may be used in a single- or double-hinged configuration, the latter ofwhich comes as a single unit. When using a pair of single hinged joints, the joints should be placed as far apart as possible to reduce angular rotations as weIl as forces and moments. In most cases, the hinges are designed to pass through the full pressure thrust load, so there is no need for tie rods. In some cases, the hinge connections may be slotted to permit axial displacement of the bellows, however, then the pressure thrust must be absorbed by adjacent anchors. A typical hinge application is shown in Figure 3-47. Note that the piping system requires bending in one plane only.

1

..

J:'''~::~ 1 .. ;..( .. 1

_ - - -- -- -- -- --- _.... PG

EOU"MfNT

'L 1

1 1

"", 1

• 7' ~

7""

,', !~~ ,

1

..

~---

'----.;;;r DIA

Figure 3-47

3-36

~ lA


A computer model for a single hinge expansion joint is shown in Figure 3-48.

A

z

Hinge axis

X

Rigid elements, weight of each equal to 1/2 total hinge assembly weight.

d&~.~--~~==~.C=~~-4~ Remember: Hinges are almost always used in pairs.

40

44

45 46

47

50

Figure 3-48 A zero-Iength expansion joint is placed between node points 45 and 46. The axial, lateral and torsional stiffnesses ofthe expansion are set to rigid (lE 12), with the hending stiffness taken from the manufacturer's catalog (or calculated from Deff2 Kax / 2. The effective diameter for the pressure thrust may be set to zero. The bending restriction is modeled by placing a rotational restraint, in the direction perpendicular to the axis of the pipe and the hinge, at node point 45, with a CNODE at 46. Halfofthe weight of the hinge hardware is assigned to each of the rigid elements, as it may he important for hanger design and/or equipment loadings.

Gimbal Joints: Gimbal expansion joints are shown in Figure 3-49.

~HingePin

l} gTI

U/Gimbal Ring

r--~~,,!11'i nngepln l ,

1

~

( " •

h

1

1

•

\..1-

1 1 1 1

-i 1 )

-

.

V

1

1 1\'--1 '1

.

L:J

r-+P

1 1

Il

11 ..l' Il S=r i IlflfJ i ~ U 1

-

1 Ir

1 Il.

~

11

1.1

t--

H

1 1 1

141. 1!

1015-1020 Bellows

t o'I' U

1 L.....I

25-30 Gimbal 1035-1040 Bellows

I

ANGULAR AND TRANSVERSE

Figure 3-49

3-37


Gimbals are designed to permit angular rotation in any plane. The hinges and gimbal ring are capable of absorbing axial pressure or vacuum loads, dead weight ofadjacent piping, and torsional moments. Like hinges, if gimbals are used in pairs, they should be located as far apart as possible to maximize absorbed dis placement, and reduce rotation and forces and moments. A typical gimbaljoint application is shown in Figure 3-50. Note that the piping system requires bending in two planes.

~ss .....

PG

Figure 3-50 A computer model for a gimbaljoint, such as the first one shown in Figure 3-49 is very simple to build - one simply defines an expansion joint with rigid axial, lateral, and torsional stiffnesses, and a bending stiffness equal to that ofthe actual expansionjoint used, since the gimbal is free to bend in all directions. (Rigid elements with weights equal to the weight of the gimbal assembly may be included as weIl.) Sometimes, however, a gimbal may be used in conjunction with hinges, as shown in the diagram ofthe angular/transversejoint in Figure 3-49. A computer model for something like this is more complex to build - one solution is shown in Figure 3-51. ~20

3~~

1··---4· •.---~. 2530

1

1040 45

1020

Figure 3-51 The hinges/gimbal are modeled as the rigid series of elements running from 10 to 45 along the top ofthe figure, while the bellows and spool piece are modeled as the elements running along the bottom of the figure.

3-38


Element 10 to 15 is a rigid element, having a length equal to the distance from the face of the inlet flange to the axis of the first hinge, with a weight equal to approximately onequarter of the total weight of the hinge/gimbal hardware (note that fluid and insulation weigh t is automatically added to non-weightless rigid elements, so that should be considered when assigning a weight to these elements). The hinge (element 15 to 20) is modeled as a zero-Iength expansion joint with rigid (lE12) axial, lateral, and torsional stiffnesses, and a bending stiffness of 1 (which is effectively zero). The hinge is restricted to one-directional rotation by restraining node point 15 rotationally about the X-axis, with a CNODE of20. Element 20 to 25 is another rigid element, having a length equal to the distance from the axis of the first hinge to the mid-point ofthe gimbal, and again a weight equal to one-quarter of the total hardware weight. The gimbal (element 25 to 30) is a zero-Iength expansionjoint with rigid axial, lateral, and torsional stiffnesses, and a bending stiffness of1 (unrestricted by any restraints). Element 30 to 35 is a third rigid element, having a length equal to the distance from the mid-point of the gimbal to the axis of the second hinge (again with onequarter of the total hardware weight). The second hinge (element 35 to 40) is modeled in the same way as the first, except that the rotational restraint applied at node point 35 is about the Y-axis, with a CNODE of 40. The final rigid element, from node point 40 to 45, has a length equal to the distance from the axis ofthe second hinge to the face of the outlet flange, and provides the fmal quarter of the hardware weight. Since neither the hinges nor gimbals are internally pressurized, the expansion joints which are used to model them should be given effective diameters of zero as weIL The expansion joints and spool piece will be modeled from node point 10 to 45 as weIl, indicating that the centerlines of the two assemblies are coincident, but connections are present only at the end points. Elements 10 to 1015 and 1040 to 45 are rigid elements with the length and weight ofthe two end flanges. Elements 1015 to 1020 and 1035 to 1040 are finite length expansion joints modeled with the exact properties of the actual bellows used (including effective diameter). Element 1020 to 1035 is modeled as a normal pipe element, representing the spool piece between the two expansion joints.

Universal Joints: A univers al expansion joint is shown in Figure 3-52. Universaljoints consist oftwo unrestricted expansion joints flanking a spool piece. They are usually used to absorb large lateral movements in any direction. By increasing the length of the center pipe the amount of lateral displacement absorbed can be increased, with a corresponding reduction in the lateral forces and hending moments. In most cases univers al joints are tied to prevent the pressure from blowing apart the assembly. (When a univers al expansion joint must absorb axial movement other than its own axial growth, an untied univers al should he used - in that case, adjacent restraints must be designed to handle the pressure thrust.) A center support may be provided on the tie rods to help support the weight of the center piece, to provide limit stops for the displacement, and/or to reduce the length of compressive rods (and the corresponding tendency of the rods to buckle).

3-39


UNIVERSAL EXPANSION JOINT

Figure 3-52 A simple model of a universal expansion joint is shown in Figure 3-53.

Zero length expansion joints with axial, lateral and torsiorial stiffnesses rigid, and no pressure thrust. Ambjent Temperatyre 1

~

!

~~

.006

~

Expansion joint stiffnesses inserted between the zero length elements: 5-1005 & 6-1006

Figure 3-53

3-40

a


Since the tie rods isolate the pressure loads and thermal axial growth of the joint from the remainder of the system, the assembly can be simply modeled by calling the effective diameter of the bellows zero, the axial stiffness of the bellows rigid, and the temperature of the assembly ambient. Elements 4 to 5 and 1006 to 7 are rigid elements (with the weight of the tie rods, etc.), elements 5 to 1005 and 6 to 1006 are expansion joints with effective diameters ofzero, axial stiffness ofrigid, and bending and lateral stiffnesses as determined from the manufacturer. Element 1005 to 1006 is simply a pipe element representing the spool piece. The totallength of the elements from node point 4 to node point 7 should be the same as the length of the tie rods, and, as noted above, the assembly should all be set to ambient temperature. More complex models involve entering as accurately as possible the bellows, tie rods, and all supporting mechanisms. These models are very cumbersome to build but will give the most accurate representation of the loads, movements, and other conditions in and around the joint. More complex univers al joint models are shown in Figure 3-54.

J(x

z

1002

/i 1~==I=*lH'~---r

:J"

Use double acting restraint with gap and connecting nOde~

Gap to limit lateral 1 deflection 1.5"

1036

=

L

1030 1003

X

1003

Figure 3-54

3-41


Other expansion joint configurations may he modeled by modeling various combinations of hellows, hinges, gimbals, tie rods, limit stops, andother hardware as shown in these figures. A more accurate (and more likely correct) representation ofthe real configuration can usually he achieved with a more complex model. When modeling the assembly, total hardware weight must be considere d, including internaI or external sleeves and bellows end connection details, in addition to the items noted above.

3-42


3.4 Piping Nozzle Evaluation Often piping configurations that are acceptable by stress criteria are limited by the loads which they place on connected equipment and vessels. Piping loads on the nozzles of equipment such as pumps, compressors, turbines, and heat exchangers may have the tendency to deform or overstress equipment casings, overload hearings, or cause shaft binding. Normally manufacturers should provide allowable nozzle loadings to which their equipment may be subjected, or they may reference industry standards, such as NEMA SM-23 (Steam Turbines), API 610 (CentrifugaI Pumps), API 617 (Centrifugal Compressors), API 661 (Air Cooled Heat Exchangers), or HEl (for Closed Feedwater Heaters). These standards provide look-up tables or simple calculations which serve as a common reference for equipment vendor and engineer. Piping attached to vessels induces stresses in the vessel walls, in the form of membrane and bending stresses. These stresses must he evaluated against the requirements ofthe ASME Boiler and Pressure Vessel Code, Section VIII, Division 2. Calculation of stresses in a vessel wall is difficult without a finite element analysis; the hest means of doing a hand calculation is to use aids such as Welding Research Council (WRC) Bulletin 107 "Local Stresses in Spherical and Cylindrical Shells due to External Loadings". Piping attached tovessels may also tend to bend, buckle, orotherwise deform the vessel wall, permitting some displacement or rotation of the connection under load. Therefore, totally rigid restraint models may not be accurate representations ofpiping to vessel connections. Flexibilities of the connection should be estimated when possible; this can he done with the assistance ofliterature such as Welding Research Council Bulletin 297 "Local Stresses in Cylindrical Shells Due to External Loadings on Nozzles - Supplement to WRC Bulletin No. 107". The three types of possible analyses which may be done on nozzles - evaluation of equipment loads, calculation ofvessel stresses, and calculation ofpiping/vessel connection flexibili ties - are described in this section. 3.4.1 Equipment Nozzle Load Analysis The most accurate means of evaluating a piece of equipment for anticipated nozzle loads is to perform a test. In lieu of a test, the next best method may be a finite element analysis, if operability failure (as opposed to stress failure) can he accurately determined from the model. In the absence of either ofthese, the engineer can often specify that the equipment meet a recognized standard, which provides for evaluation ofnozzle loads. This standard, which may provide look-up tables or simple calculations, becomes a common reference hetween the manufacturer and the engineer - a promise that the equipment can stand at least a certain set ofloads, which the engineer can then ensure that the piping loads remain below. It should be noted that these loads are minimum loads - in most case, the standards do not provide a means of actually evaluating the capacities of individual pieces of equipment. Equipment can be modeled in the pipingproblem in a number ofways. The nozzles can be considered to he rigid anchors, or entire pieces of equipment can be built-up from an

3-43


assemblage ofrigid elements, with varying degrees of complexity. In either case the forces considered when evaluating the equipment are those forces which the pipe stress analysis shows are acting at the equipment connection. The load cases for which the nozzle loads are to be checked are the greater of those from the cold and the hot cases - that is, from the sustained and the operating load cases (except when cold spring is considered, in which case the cold case would be sustained plus the effects of cold spring). Typically suction, discharge and extraction lines are included in separate pipe stress models. Once all of the loadings on a particular piece of equipment have been computed, the equipment can be evaluated to determine whether these loads are acceptable (i.e., in accordance with the governing standard).

CAESAR II provides the ROT program, which may be accessed from the main menu, to automatically evaluate piping nozzle loads against the requirements of a number of these standards. Equipment (and standards) covered include: 1

-

Ste am Turbines Standard SM23

National Electrical Manufacturer's Association (NEMA)

2

-

CentrifugaI Pumps - American Petroleum Institute (API) Standard 610, 6th and 7th Editions

3

-

Centrifugal Compressors - API Standard 617

4

-

Air Cooled Heat Exchangers - API Standard 661

5

-

Closed Feedwater Heaters - Heat Exchange Institute (HEl) Standard

In order to use this program, the user is required to enter some description ofthe equipment (geometry, nozzle sizes, etc.) and the applied loads. Specific requirements ofthese standards (and the corresponding use of the ROT program) are described below. 3.4.1.1 NEMA SM23 Standard for Steam Turbines

NEMAStandard SM23 requires that ste am turbines be evaluated by two sets of simple force/ moment calculations. The two types of computations are used to satisfy: 1

-

individual nozzle allowables

2

-

cumulative equipment allowables

First, the loads on each individual suction, discharge, and extraction nozzle must satisfy the equation: 3F + M< 500D Where: F

=

resultant force on the particular nozzle, lb

3-44


M

=

resultant force on the particular nozzle, ft-lb

D

=

nominal pipe size of the connection, in

Ifthe loads on the individual nozzles are acceptable, the cumulative load on the equipment from all nozzles must be checked next. Cumulative equipment allowables require that forces and moments on aH connections he resolved at the discharge nozzle; the absolute magnitude ofeach of the force and moment components must then fall below a multiple of De, as: Fx <= 50 De Fy<= 125 De Fz <= 100 De

Mx <= 250 De My <= 125 De Mz <= 125 De Fe + M e /2 <= 125 De

Where: Fx

=

total X-force (from ail nozzles) on the equipment (where the X-axis is defined as being parallel to the equipment centerline), lb

Fy

=

total Y-force (from ail nozzles) on the equipment (where the Y-axis is coincident with the direction of gravity), lb

Fz

=

total Z-force (from all nozzles) on the equipment (where the Z-axis is defined by the right hand rule form the other two), lb

Fe

=

total resultant force acting on the equipment, lb

Mx

=

total X-moment (from forces and moments on aH nozzles) acting on the equipment, resolved about the discharge nozzle, ft-lb

My

= total Y-moment (from forces and moments on aH nozzles) acting on the equipment, resolved about the discharge nozzle, ft-lb

Mz

= total Z-moment (from forces and moments on aU nozzles) acting on the equipment, resolved about the discharge nozzle, ft-lb

Me

= total resultant moment acting on the equipment, resolved about the dis charge nozzle, ft-lb

De

=

diameter of an opening whose area is equal to the sum of the areas of aU individual equipment nozzles, in

3-45


An example of a NEMA SM23 analysis is shown in Figure 3-55. In this example, the turbine is modeled as a series ofrigid elements, so both the discharge and suction nozzle loads appear on the same pipe stress analysis output.

20

10

HEADER 'TIE-IN" 4"

1

2~-3

1

30~IL T~'~55

35:i<,50

TURBINE MODELED AS THREE RIGID ELEMENTS

45

............t

"\

1:--~ TURBINE ...............

Figure 3-55 The forces and moments acting on the elements connected to the turbine (excerpted from the CAESAR II output for this piping system) are shown below:

NODE

FX

FY

Fl

MX

MY

Ml

30

-108

-49

-93

73

188

603

35

108

67

93

162

-47

-481

50

-192

7

-11

369

-522

39

55

192

-63

11

78

117

-56

Since the sign conventions of these forces is that of forces acting on the elements, it is necessary to reverse the signs to get the forces and moments acting on the turbine nozzles at points 35 and 50, or:

NOZZLE LOADS AT TURBINE NODE

FX

FY

Fl

MX

MY

Ml

35

-108

-67

-93

-162

47

481

50

192

-7

11

-369

522

-39

3-46


When running the NEMA portion of the ROT program, there are two input spreadsheets upon which descriptive and load data must be entered, which in this case would be fIlled out as shown in Figure 3-56.

NEMA SMZ3 Input Data

EquipMent ID

NEMATl

Suction Nozzle Node NUMber •••••••.•.....••...••• Suction Nozzle NOMinal DiaMeter ••....••••..

35.0000 4.0000

Discharge Nozzle Node NuMher •••••••.....•...•••• Discharge Nozzle NOMinal DiaMeter ••.....••.
50.0000 8.00130

Factor for NEMA Allo~ables <1.85 for API 617) ••• EquipMent Centerline: 1-X. Z-Z •••....•••••..•••

1.013130

................

Extraction Nozzle ul Node NuMher Extraction Nozzle ul NOMinal DiaMeter .•.••.
................

Extraction Nozzle uZ Node NUMber Extraction Nozzle uZ NOMinal DiaMeter •..••.
PgUp/PgDn/HoMe/End-Change Page

[Escl-Exit

NEMA SMZ3 Input Data

EquipMent ID

[Ftl -Execute

= NEMATl

X Distance FrOM Discharge to Suction ...••. (in.) Distance FroM Discharge to Suction ...••• (in.) Z Distance FroM Discharge to Suction ...••• (in.)

-34.0000 .0000 15.0000

'i

X Force Acting on Suction Nozzle • .•.•....• (lb. ) Force Acting on Suction Nozzle • •.....••. (lb. ) Z Force Acting on Suction Nozzle · ••....••. (lb. )

-108.0000 -G7.000ll -93.0000

X MOMent Acting on Suction Nozzle ••.... (ft. lb • ) MOMent Acting on Suction Nozzle ...... (ft. lb. ) Z MOMent Action on Suction Nozzle ...... (ft.Ib.)

-lG2.0000 47.0000 481.0000

Force Acting on Discharge Nozzle · .....•• (lb. ) Force Acting on Discharge Nozzle · ....... (lb. ) Force Acting on Discharge Nozzle · .•••••• (lb. )

192.0000 -7.0000 11. 000ll

MOMent Acting on Discharge Nozzle · ..• (ft. lb. ) MOMent Acting on Discharge Nozzle · ... (ft. lb. ) MOMent Acting on Discharge Nozzle · ... (ft. lb. )

-3G9.0000 522.0000 -39.0000

'i

'i

X 't

Z X 'i

Z


Œsel-Exit

Figure 3-56

3-47

[Fil -Execute


The program then proceeds to check the loading on each nozzle, and then resolves the loads about the discharge and checks the cumulative loading on an equivalent nozzle, as described above. The two page NE MA output report for this particular example is shown in Figure 357. The results first show that the discharge nozzle passed (only reaching 30.4% of the allowable), the suction nozzle passed (reaching 49% ofthe allowable), and the cumulative load passed, with the highest summation load (Z-moment) reaching only 56% of the allowable. If the turbine had failed, the symbol **FAILED** would have appeared in the "STATUS" column next to the load combination that was excessive. Nozzle

Node

Diameter Diameter (Nom) (USED)

X

y

)

(i n. )

(i n. )

(i n. )

8.000 4.000

8.000 4.000

.000 -34.000

.000 .000

(i n.

DISCHARGE SUCTION

50 35

Distance From Discharge Nozzle Z (i n.

)

.000 15.000

Individual Nozzle Calculations Node

Nozzle

Components Resultants (lb. & ft.lb.) (lb. & ft.lb.) FX= FY= FZ=

192 -7 11

F=

MX= MY= MZ=

-369 522 -39

M=

640

FX= FY= FZ=

-108 -67 -93

F=

157

MX= MY= MZ=

-162 47 481

M=

MX(About dis. noz. )= MY(About dis. noz. )= MZ(About dis. noz. )=

83 -398 189

DISCHARGE

50

SUCTION

35

Values/Allowables 3F+M

1216

500*D(used)

4000

192 % of ALLOW.

=

30.40

3F+M

980

500*D(used)

2000

% of ALLOW.

49.00

509

Summation Calculations. Diameter Due to Equivalent Nozzle Areas. OC = Nozzle Loads Summations ( 1b. &

Allowables

8.94 in. % of ALLOW.

ft. 1b. )

SFX SFY SFZ FC(RSLT)

84 -74 -82 138

50*DC = 125*DC = 100*DC =

447 1118 894

18.79 6.62 9.17

SMX SMY SMZ MC(RSLT)

-447 170 631 792

250*DC = 125*DC = 125*DC =

2236 1118 1118

20.00 15.25 56.51

535

125*DC =

1118

47.85

=

FC + MC/2 =

Figure 3-57

3-48

Status


3.4.1.2 API 610 Standard for Centrifugai Pumps Analysis of pump nozzle loads is covered by API Standard 610 "CentrifugaI Pumps for General Refinery Service". The requirements of the 6th edition, issued in January of 1981, and those of the 7th edition, issued in February, 1989, can be evaluated using CAESAR II's ROT program. The requirements of the two editions differ slightly; those ofthe 6th edition are described first:

API Standard 610 basically presents reference tables ofnozzle loads which a specified size pump nozzle must be good for. The 6th edition of the standard provides separate allowables for two types of support/base plates: 1

heavy duty

2

standard

The "heavy duty" pump is typically a reinforced standard pump which can withstand higher piping loads. (The criteria necessary to meet the requirements of a heavy duty base plate/ support are discussed in paragraph 2.4.6 of API 610, 6th Edition.) AlI force components acting on each nozzle of a heavy dut y pump must be less than two times the value shown in Table 2 for that specific load component. For pumps with heavy duty base plates, this is the only requirement. A copy of Table 2 (from the 7th edition) is shown in Figure 3-58.

Table 2-Nozzle Loadings Note: Each value shown below indicates a range from minus that value to plus that value; for example, 160 indicates a range from -160 to +160, Nominal Size of Nozzle Flange (inches) ForcefMoment'

2

3

160 200 130 290

240 300 200 430

160 130 200 290

240 200 300 430

200 130 160 290

300 200

Each top nozzle

FX FY

FZ FR Each side nozzle FX FY

FZ FR Each end nozzle FX

FY FZ FR Each nozzle

MX MY MZ

340

MR

460

260 170

240

430 700 530 350 950

4

6

8

10

12

14

16

320 400

560 700

260 570

1200 1500 1000

1010

850 1100 700 1560

2200

1500 1800 1200 2600

1600 2000 1300 2900

190C 230( 150C 330(

560 460 700 1010

850 700 1100 1560

1200 1000 1500 2200

1500 1200 1800

1600 1300

190C 150C

2000

230C

2600

2900

330(

700 460 560 1010

1100 700 850 1560

1500 1000 1200 2200

1800 1200 1500 2600

2000

230C

1300 1600 2900

150C 190C 330(

1700 1300 870 2310

2600

3700

4500

2800

3400 2200

4700 3500

540C

1900 1300 3500

460

320 260 400

570 400

260 320 570 980

740 500 1330

1800 5000

6100

2300

6300

400C 270C 720C

• F = force, in pounds; M = moment, in foot-pounds; R = resultant. See Figures 1-5 for orientation 01 nozzle loads (X, Y, and Z).

Figure 3-58

3-49


The "standard" API 610 pump has a slightly more complex scheme for compliance: In no case shall any of the individual force or moment components on either the

1

suction or discharge nozzles exceed twice the allowable value for the component as shown in Table 2. 2

-

Providing that the first is satisfied, the pump passes ifthe resultant forces and moments on both the suction and discharge nozzles are within the allowables given in Table Cl, shown in Figure 3-59 (note that resultant allowables are included in Table 2 of the 7th edition).

Table C-1-Suggested A1lowable Resultant Forces and Moments (for Vendor's Standard Baseplates) Resultant ForceIMoment

Nominal Size of Nozzle FlInge (iDches)

2 430 690

3

4

640

860

1400 2000

6 I~ 3~

8

10

12-

2300 2700 2900 5200 6600 8200

NOTE: Fr = resultant force. iD pounds; Mr = resultant moment. in foot-pounds. - For lizes Jarger ibID 12 incbes. allowable resultants should he mutuaUy agreed upon by the purchuer IDd the vendor.

Figure 3-59 3

-

Ifthe pump passes the first requirement, but fails the second, the pump may still pass ifboth conditions described below are satisfied as weIl:

CONDITION 1 CRITERIA a) The ratio ofthe appliedresultantforce to the allowable resultantforce from Table Cl for each nozzle is less than or equal to 2. b) The ratio of the applied resultant moment to the allowable resultant moment from Table Cl for each nozzle is less than or equal to C; where C=2 for nozzles 6 in. and smaller, and C=(D+6)/D for nozzles larger than 6 in. c) For each nozzle, the sum of the force ratio and the moment ratio found in (a) and (b) above must be less than 2.

CONDITION 2 CRITERIA The summation ofthe forces and moments from both the suction and the dis charge nozzles, taken about the base point (which is the intersection of the shaft axis and the pedestal centerline), must he less than the sum ofthe force and moment resultant allowables for both the suction and discharge taken from Table Cl. The 7th edition ofthis standard modified the requirements to a small extent. The differences in the 7th edition are:

3-50


1

this edition does not recognize separate provisions for heavy duty baseplates; the methodology for evaluating any pump under this edition is the same as that for as standard base plate pump under the 6th edition

2

-

different allowables are provided for vertical in-line pumps that are supported only by the attached piping

3

-

there is no difference in the allowables for tension and compression loading on top nozzles

4

-

the allowables for larger diameter nozzles have been increased in some cases

5

-

Tables 2 and C-l have been combined into a single Table 2

An example of an API 610 analysis (if evaluating under the 6th edition, the pump is considered to have a standard baseplate) is shown in Figure 3-60. For this example, the suction and discharge piping were analyzed in separate models - the 8-inch suction piping was analyzed in ajob called ''1NLET3" and the 6-inch dis charge pipe was analyzed in ajob called "DISCH3". In both cases the pump nozzles were modeled as anchors. The forces and moments acting on these anchors, as printed in the CAESAR fi output report, are the nozzle forces and moments (with the same sign) that should he entered into the rotating equipment input spreadsheets. These reactions are shown below:

REACTION LOADS AT PUMP NODE FX 305 50

507 105

FY

FZ

MX

MY

MZ

-111

-231 490

3950 46

2861 -51

2189 -201

30

..-----=:::D CONTINUE ON WITH JOB "DISCH3"

c=-------. CONTINUE ON WITH JOB "INLETS"

~

SUCTlON~ ~~__ /

305 (ANCHOR)

1

y

50 (ANCHOR)

L

~~SE

1_'-_3--l-JJ

BASE POINT

3-51

POINT

Z~X 9'

PUMP FRONT VIEW (LOOKING ALONG 'X')

Figure 3-60

1--11'

PUMP SIDE VIEW (LOOKING ALONG NEGATIVE 'Z')


When running the API 610 portion of the ROT program, there are three input spreadsheets upon which descriptive and load data must be entered, which in this case would be filled out as shown in Figure 3-61.

API G1B Input Data

EquipMent ID = G10T1

Base Point Node NUMber ......................... . Base Plate Type: l-Standard. 2-Heavy Dut Y ...... .

100.0000 1.0000

Suction Nozzle Node NuMber ..................... . Suction Nozzle Type: I-Top. 2-Side. 3-End ...... . Suction Nozzle NOMinal DiaMeter ............
305.0000 3.0000 8.0000

Discharge Nozzle Node NUMber ................... . Discharge Nozzle Type: 1-Top. 2-Side. 3-End .... . Discharge Nozzle NOMinal DiaMeter ..........
50.0000 1.0000 5.0000

Factor for Table 2 Allowables .......•.••..•.... Factor for Table Cl Allowables ................. . EquipMent Centerline: 1-X. 2-2. 3-~ ........... .

1.0000

EquipMent ID

API G10 Input Data SUCTION

Nozzle Data Entry X.....
Nozzle Orientation Base Point to Nozzle

Cin.) 2 ..... Cin.)

-11.0000 .0000 .0000

~

X.....
-111. 0mm

Z ..•..
-231.0000

(ft.lb.) .. (ft.lb. ) Z.. (ft.lb.)

3350.0000 2851.0000 2189.0000

~ .....

Forces on Nozzle

MOMents on Nozzle

X..

~

~PI

EquipMent ID

G10 Input Data DISCHARGE

G10T1

507.0000

= G10T1

Nozzle Data Entry

Nozzle Orientation Base Point to Nozzle

X•••••
Forces on Nozzle

X.••.. (lb.) Y.....
MOMents on Nozzle

X•• (ft.lb.) y •• (ft.lb.)

Z ••
Figure 3-61

3-52

.0000 15.0000 -9.0000 105.0000 30.0000 490.0000 45.0000 -51.0000 -210.0000


The program then performs the compliance check, as described above. The two page API 610 output report for this particular example is shown in Figure 3-62. The results first show the status of the individual nozzle checks - each of the load components on both the discharge and the suction are within twice the Table 2 allowable; however, the resultant moment on the suction nozzle exceeds the allowable from Table C-l. Because ofthis failure, Conditions 1 and 2 must he checked. Since both of the se conditions are satisfied, the pump meets the requirements of API 610 - despite the one failure. INPUT DATA' BASE POINT NODE 1BB BASE PLATE TYPE STD. SUCTION TYPE END DISCHARGE TYPE TOP SHAFT PARALLEL TO X AXIS

-------------------------------------NOZZLE

NODE

DISCHARGE SUCTION

DIAMETER

SB 3B5

DISTANCE FROM BASE POINT ~

(in. )

X ( in. )

(in. )

Z (in. )

6.BBB 8.BBB

.BBB -l1.BBB

15.BBB .BBB

-9.BBB .BBB

-----------------,

LOAD COMPARlSON UITH THE ALLOUABLES IN TABLE 2 ( FACTOR =

-------NOZZLE

-------------NOZZLE LOADS Z>
NODE (

--------DISCHARGE

SB

a ft. lb.)

lb.

~=

FZ= MX= M'/= MZ=

SUCTION

3B5

% OF ALLOU.

FROM TABLE Z lb. a ft.lb.>

(

-----------

FX=

FX=

lBS 3B 498

UZB 7BB 9ZB

9.38 4.Z9 53.26

46 -51 -21B

34BB 26BB 1749

1.35 1.96 12.B7

14BB 17BB

23.B5 7.93 13.59

5ZBB 3BBB 26BB

75.96 75.29 B4.19

FZ=

587 -111 -231

MX= M'/= M2=

395B 2B&1 ZlB9

~=

1.BB

---------22BB

)

---------

STATUS

--------

RESULTANT LOAD COMPARISON WITH THE ALLOWABLES IN TABLE Cl ( FACTOR = NODE

NOZ2LE

NOZZLE LOADS

a

( lb. DISCHARGE SUCTION

ALLOWABLES % OF ALLOW. FROM TABLE Cl ft. lb.) ( lb. a ft. lb.>

SB

FR= MR=

5BZ Z2B

15BB 35BB

33.47 &.31

3B5

FR= MR=

56B 5345

Z3BB 52BB

24.78 IBZ.B1

Resultant applied MOMent on

1.B8)

STATUS

""FA 1 LED""

Suetion Nozzle

IS WITHIN ONE AND TWO TIMES THE ALLOWABLES FROM TABLE Cl. THE PUMP MA~ STILL SATISF~ API61B REQUIREMENTS IF CONDITIONS #1 AND #Z BELOW ARE MET. CONDITION #1 NOZZLE

NODE

DISCHARGE

SB

UALUE (Fa'Fr>+(Ma~Mr)

CONDITION =

Fa'Fr Ma'Nr SUCTION

3B5

(Fa'Fr)+(Ma~Nr>

Fa,yr Ma'Nr

=

.4B .33 .B6 1.28 .25 1.B3

< or = < or = < or =

STATUS

2.BB 2.BB 2. BB

< or = 2. BB < or = 2.BB <

or =

1.75

CONDITION #2 SUMMATION OF APPLIED FORCES MOMENTS AT THE BASE POINT

a

MAXIMUM APPLIED LIMIT AS DESCRIBED IN APPENDIX C

7121

125BB

Figure 3-62

3-53

STATUS


3.4.1.3 API 617 Standard for Centrifugai Compressors The requirements ofthis standard are identical to those ofNEMA SM23, except that all of the NEMA SM23 allowables are increased by 85%. Therefore it is not necessary to discuss this standard here.

3.4.1.4 API 661 Standard for Air Cooled Heat Exchangers The API 661 Standard for Air Cooled Heat Exchangers for General Refinery Service covers the allowable loads on the vertical, collinear nozzles found on most single- or multi-bundled air cooled heat exchangers. The types ofnozzles which may be checked are those which are labeled "9" in Figure 3-63.

-

r

uge"d 1. Tubesheet 2. Plu, sheet 3. Top and bouom plates 4. Endplate

~.

9. Noule

Tube

6. Pus partition

10. Side frame

7. Stiffener

II. Tube spacer 12. Tubesupponcr.....membcr

8. Plu,

13. Tube keeper 14. Vent I~. Drain 16. IMtrument connection

Figure 5-Typical Construction of a Tube Bundle with Plug Headers

Figure 3-63

API 661 has the following two requirements: 1

each nozzle in the corroded condition shall be capable of withstanding the following moments and forces (referenced as Figure 8 values):

(lb)

Nominal Diameter

FX

FY

FZ

MX

(ft-lb)

MY

MZ

1 2 3 4 6 8 10 12 14

100 150 150 500 750 1200 1500 2000 2500

150 200 250 400 750 2000 2000 2000 2500

100 150 300 500 600 850 1000 1250 1500

50 70 200 400 800 1100 1250 1500 1750

70 120 300 600 1500 3000 3000 3000 3500

50 70 200 400 1050 1500 2000 2500 3000

3-54


2

the sum of the forces and moments on each fixed header (i.e. each individual bundle) will be less than 1500 pounds transverse to the bundle, 2500 pounds axial to the bundle, and 3000 pounds axial to the nozzle centerline, while the allowed moments are 3000, 2000, and 4000 foot-pounds respectively (referenced as Table allowables)

The input to the API 661 portion ofthe ROT program is relatively self-explanatory; typical input screens are shown in Figure 3-64 and the output report is shown in Figure 3-65. API 661 Input Data

EquipMent ID

= H661

Inlet Nozzle Node NUMber .. ( Optional ) .......... . Inlet Nozzle NOMinal DiaMeter •...•...•••.••• (in.) Outlet Nozzle Node NUMber.( Optional ) ••..••...•. Outlet Nozzle NOMinal DiaMeter ...•.•..•.••.• (in.) Figure 8 Force/MoMent Multiplier .....••...•••••• Resultant Force/MoMent Multiplier ..•..••...•••••• Tube Bundle Direction ( 1-X, z-z ) .............. .

API 661 Input Data SUCTION

EquipMent ID

= H661

Nozzle Data Entry

Nozzle Orientation (in.)

18.0000

X..... (lb.)

Z .•••• (lh.)

100.0000 --302.0000 50.0000

X•• (ft.lb.) (ft.lb.) Z •• (ft. lb. )

203.0000 300.0000 2300.0000

'f . . . . .

Forces on Nozzle

'f •••••

MOMents on Nozzle

'f ••

Figure 3-64

3-55

(lb.)


API Standard 661 1987 2nd Edition

File Date TiMe

User Entered Description : Suction Nozzle Node NUMber Discharge Nozzle Node NUMber

5

2B

Suction Nozzle NOMinal DiaMeter = Discharge Nozzle NOMinal DiaMeter User Entered Figure 8 User Entered For~MoM. Tube Bundle Direction

8 8 .BB .BB

Allo~able Allo~able

X

Units For This Analysis

in •• lb •• ft.lb .• in.

Suction 't

Distance

H661 DEC 16.1992 11: 6 pM

Figure 8 Values

Force li MOMent Ratios

Status

18.121

X 't

Force Force 2 Force

1121121.121 -31212.121 5121.121

12121B. 2121121121. 85B.

.1218 -.15 .B6

PASSED PASSED PASSED

MOMent MOMent 2 MOMent

2B3.B 3BI2I.B Z3121121.B

1113121. 3BBB. 151313.

.18 .113 1.53

PASSED PASSED FAILED

Discharge

Figure 8 Values

X 't

't

Distance

Force li MOMent Ratios

Status

-18.121

Force Force Z Force

21213.13 23.13 -2ZZ.B

12BB. 2121121B. 8513.

-.26


MOMent MOMent Z MOMent

3BB.12I 3BI2IB.12I 3121Z.121

11BB. 3121121121. 15121121.

.27 1.BB .ZB


X 't

X 't

.17 .131

[Enterl-Continue Resultant

Force~MoMent

Check

Resultant

Table

Allo~able

Ratios

Status

X 't

Force Force Z Force

3B3.B -279.13 -172.13

25BB. 3I21BB. 15BB.

.12 .B9 .11


MOMent MOMent Z MOMent

911.13 33BI2I.12I 2756.5

2I21BB. 4BBB. 3BBB.

.46

.82 .92


X 't

Overall Loading status

MM

FAILED

MM.

[Enterl-Continue

Figure 3-65

3-56


3.4.1.5 HEl Standard for Closed Feedwater Heaters

The HEl Standard for Closed Feedwater Heaters provides a method for evaluating the allowable loads on shell type heat exchanger nozzles. This method is a simplification of the WRC Bulletin 107 methodology (discussed in Section 3.4.2 ofthese seminar notes), in which the allowable loads have been linearized to show the relationship between the maximum permitted radial force and the maximum permitted resultant bending moment. If this relationship is plotted (using the moment as the abscissa and the force as the ordinate) a straight line can be drawn between the maximum permitted force and the maximum permitted moment. Then, plotting the actual combination of applied force and moment, if these loads fall outside of the line, the nozzle is considered to fail, while ifit falls inside of the line, it is considered to pass. The maximum permitted force and moment is calculated from dimensionless parameters based upon the shell geometry; these parameters have been linearized for lookup by CAESAR II's ROT program. A sample input to the HEl portion of the ROT program is shown in Figure 3-66. Note that the program automatically calculates and includes the pressure thrust load (internaI area ofthe pipe times the internaI pressure) whenever the entered design pressure is greater than zero.

EquipMent ID = HEITl

HEl Nozzle/Uessel Input Data

Design Pressure (P) .................. (lb./sq.in.) Nozzle Outside DiaMeter .................... (in.) Shell Outside DiaMeter ....•....•........... (in. ) Shell Thickness (T) ...........•............. (in.) Material ~ield Strength (Sy) ......... (lb./sq.in.) Material Allowable stress (Sa) ....... (lb./sq.in.) MaxiMUM Radial Force .•.•.••..••...•.....••.. (lb.) MaxiMUM CircuMferential MOMent ..........• (ft.lb.) MaxiMUM Longitudinal MOMent .............. (ft.lb.)

[Esc]-Exit


150.0000 30.0000 75. 7500 .7500 31500.0000 17500.0000 Z0000.0000 8333.3330 1. 0000

[FU -Execute

Figure 3-66 The output report corresponding to this input is shown in Figure 3-67. Note that the nozzle failed for this application, since the load combination fell outside of the allowable load combination line.

3-57


Heat Exchange Institute (Nozzle Loads 84) File Date TiMe User Entered Description

HEITl DEC 16.1992 1B:44 pM

User Input Ualues Design Pressure Nozzle Outside DiaMeter Shell Outside DiaMeter Shell Thickness Material ~ield Strength Material Allowable stress MaxiMuM Radial Force MaxiMuM CircuMferential MOMent MaxiMUM Longitudinal MOMent Pressure Thrust Load

15B.BB 3B.BB 75.75 . 75 315BB.BB 175BB.BB 2BBBB.BB 8333.33 1. BB 9569B.95

IbJsq. in. in. in. in . IbJsq. in. IbJsq. in. lb. ft. lb. ft. lb. lb.

[Enter] - Continue

DiMensionless ParaMeters Selected Alpha Beta GaMMa Delta SigMa

44B.BB .35 5B.BB 34B.BB 11BB.BB

x

COMputed Force / MOMent Allowables: Frf MrCM MrlM MrM

= = =

53213.78 5B337.35 86B81.13 5B337.35

lb. ft. lb. ft. lb. ft. lb.

MrM

444B4. The Max allowed force for the input MOMents Note. applied force includes pressure thrust if P > B. B.

lb.

The nozzle FAILED in accordance with this analysis procedure. An "x" outside the shaded region shows failure. [Enter] - Continue

Figure 3-67

3-58


3.4.2 Calculation of Vessel Stresses Due to Nozzle Loads Loads from piping attached to vessels induce stresses in the vessel walls, in the form of membrane and bending stresses. These stresses normally must be evaluated against the requirements of the ASME Boiler and Pressure Vessel Code, Section VIII, Division 2. Accurate calculation of stresses in a vessel wall is difficult without a finite element analysis; the best means of doing a calculation otherwise is to use a reference which parameterizes results of finite element analyses. The most common reference of this type is Welding Research Council Bulletin 107. Section VIII Division 2 ofthe ASME Boiler and Pressure Vessel Code provides fairly detailed rules for allowed stress in nozzle junctions and vessels. A synopsis of the elastic code rules are outlined here in order to provide sorne rules ofthumb by which to analyze stresses in the vessel, at a junction: 1

-

Section AD-160.3 contains two conditions to determine if a fatigue analysis may be ignored for nozzles. The second ofthese, Condition BP is summarized below: a) The expected design number offull-range pressure cycles does not exceed the number of allowed cycles corresponding to an Sa value of 4Sm on the material fatigue curve, where Sm is the allowable stress intensity for the material at the operating temperature. b) The expected design range of pressure cycles other than startup or shutdown must be less than 1/4 the design pressure times (Sa/Sm), where Sa is the value obtained on the material fatigue curve for the specified number of significant pressure fluctuations. c) The vessel does not experience localized high stress due to heating. d) The full range of stress intensities due to mechanical loads (including piping reactions) does not exceed Sa from the fatigue curve for the expected number of load fluctuations.

2

-

If fatigue analysis is not required, then Appendix 4 states that the following limits must be satisfied: a) General pressure membrane stress intensity must be less than Sm. b) Primary membrane plus primary bending stress intensity must be less than 1.5Sm. c) Primary plus secondary stress intensity must be less than 3Sm.

Note that the 3S m limit applies to the range of stress intensity. The quantity 3Sm is defined as three times the average of the tabulated Sm values for the highest and lowest temperatures during the operation cycle. In the determination of the maximum primaryplus-secondary stress intensity range, it may be necessary to consider the superposition of cycles of various origins that pro duce a total range greater than the range of any of the individual cycles. The value of3Smmay vary with the specific cycle, or combination of cycles, being considered since the temperature extremes may be different in each case. 3-59


In pipe stress terminology, this can be approximated as:

1

The sum of the pressure stress intensity in the vessel and the local sustained stress intensity at the nozzle connection, computed using WRC 107, must be less than 1.5Sm.

2

The sum of the pressure stress intensity in the vessel, the local sustained stress intensity at the nozzle connection, and the local expansion stress intensity at the nozzle connection, computed usingWRC 107, must be less than 3S m (where Sm is the average of the Sm at the operating and installed temperature).

Because it is often difficult to include pressure in the localloading condition in a WRC 107 analysis, and because the area reinforcement requirements are supposed to take care of the pressure stress requirement at the intersection, it may sometimes be convenient to simplify the se requirements to the following: 1

The local sustained stress intensity at the nozzle connection, computed using WRC 107, should be less than 0.5S m.

2

The sum ofthe local sustained stress intensity at the nozzle connection and the local expansion stress intensity at the nozzle connection, computed using WRC 107, must be less than 2.0S m.

This is based upon the worst case assumption that the full value of Sm is used to satisfy the pressure stress; this leaves 0.5Sm to satisfy the local stresses from the sustained external loads. The same rationale can be applied to the second requirement as weIl, leaving 2.0Sm to satisfythe local stresses from the sustained and expansion externalloads. Ifthese reduced allowables are not satisfied then the engineer should review the magnitude of the pressure loading and revert back to considering it within the full local stress analysis.

3.4.2.1

Calculation of Vessel Stresses Due to Nozzle Loads

The Welding Research Council Bulletin No. 107 (WRC 107) has been used extensively since the mid 60's by design engineers to estimate local stresses in vessel/attachment junctions. Welding Research Council Bulletin 107 is a parameterization of the results of a set finite element analyses examining stresses in vessels due to loaded attachments. WRC 107 contains equations and non-dimensional curves (based upon parameters such as ratios ofthe nozzle to vessel diameter and the vessel diameter to vessel thickness) which are used to extract coefficients for the calculation of stresses in the vessel wall at the point of attachment. Note that WRC 107 computes stresses in the vessel shell at the nozzle/vessel interfacestresses in the nozzle wall (which in sorne cases can be higher than the stresses in the vessel wall) are not computed. Stresses in the nozzle wall may become greater than the stresses in the vessel wall as the tlT (nozzle to vessel thickness) ratio becomes less than one. WRC 107 may be used to analyze cylindrical or spherical vessel at attachments. The attachments may be round, square, or rectangular; the round and square attachments may be solid or hollow (i.e., a round hollow attachment represents a pipe or nozzle connection),

3-60


while only solid rectangular attachments (Lugs) are considered - the nozzle thickness does not effect the calculation. Appendix B to WRC 107 provides information on weldment stress intensification factors. Applications include vessel nozzles, pipejunctions, welded lugs and trunnions. The techniques introduced in this bulletin were updated and extended in WRC Bulletin 297, which may be used in a similar way as that explained here. The convention adopted by WRC 107 to define the applicable orientations ofthe applied loads and stresses for both spherical and cylindrical vessels are shown in Figure 3-68.

V1 (or ~ )

P

P

VL u

u

1

Mf

Mf

Stress positions A. B. C. 0 u - on outside surface 1- on inside surface

Stress positions A. B. C. 0 u - on outside surface 1- on inside surface

LOAD AND STRESS ORIENTATIONS ON CYLINDRICAL SHELLS

LOAD AND STRESS ORIENTATIONS ON SPHERICAL SHELLS

Figure 3-68 WRC Local Coordinates

The WRC 107 bulletin is used as follows. Based upon various dimensional ratios of the vessell nozzle configuration, the engineer selects 12 dimensionless parameters from as many different figures in the bulletin. These 12 parameters are used with local loads in 15 equations to calculate 80 different stresses - circumferential membrane, circumferential bending, longitudinal membrane, longitudinal ben ding, and shear stresses (in two directions) at each of eight locations in the vessel. These eight locations are the at the inner and outer edges (identified by the subscripts 1and u respectively) ofthe vessel, at the 00 ,90 0 , 1800 , and 270 0 azimuth (identified as Figure 3-68 locations A, D, B, and C respectively) around the nozzle.

3-61


It has also been a common practice to use WRC 107 to conservatively estimate vessel shell stress state at the edge of a reinforcing pad, if any. The vessel wall stress at the edge of a nozzle reinforcing pad can be obtained by considering a solid plug (solid inclusion), whose outside diameter is equal to the O.D. of the reinforcing pad, subjected to the same nozzle loading.

Before attempting to use WRC 107 to evaluate the stress state of any nozzle / vessel junction, one shall always make sure that the geometric restrictions limiting the application ofWRC 107 are not exceeded. These are dependent on the configuration and curves used (see AppendixA.2 ofWRC 107 for details). The CAESAR II WRC 107 module notifies the user when the bounds of the curves are exceeded. The WRC 107 method should NOT be used when the nozzle is very light or when dimensionless parameters fall outside the limits oftheir respective figures. The parameters in the WRC 107 figures should not be unreasonably exceeded. Output from the WRC 107 program includes the figure numbers for the curves accessed, the curve abscissa, and the values retrieved. The user is urged to check these outputs against the actual curve in WRC 107 to get a "feel" for the accuracy ofthe stresses calculated. For example, ifparameters for a particular problem are al ways near or past the end of the figures curve data, then the calculated stresses may not be reliable. The WRC 107 program can be actived from the Main Menu. The user will be prompted for a job name as shown in the figure below.

JobnaMe

Entr~

Enter AttachMent/Shell ID --} FILENAME, 6 Characters MaxiMuM [Enter]-Default<= EXP14

)

Mter the user selects WRC 107 option, the processor will request an input file name, this entry will also be used for the subsequent output files. The user is then presented with a list of the current units used by CAESAR II. Input for an existing WRC 107 calculation data will always be referred to in its original units setting. Ifthese units are not acceptable, a different unit set must be identified using the DATABASE configuration. The input data is accumulated by the processor in six spreadsheets. The first sheet is a title block, the second and third sheets colle ct the vessel and the nozzle (attachment) geometry data, respectively. The nozzle loading is specified on the last three spreadsheets, according to specific load cases, which are sustained, expansion and occasional cases. These loads are found in the CAE SAR II output restraint load summary under the corresponding load cases.

3-62


The WRC 107 specific input coordinate system has been incorporated into the program. The user only needs to define two vectors specified on the geometry data sheets. The first vector defines the direction of the centerline of the vessel where postitve is from "B" to "A". The second vector defines the direction of the piping/nozzle orientation, with the positive direction of this vector pointing from the nozzle connection towards the vessel centerline. Note, these two vectors must be perpendicular to each other. There is no need to convert the CAESAR II globalloads to the local WRC 107 orientation. Notice that the curves in WRC Bulletin 107 coyer essentially aIl applications ofnozzles in vessels or piping; however, should any of the interpolation parameters, i.e. Beta, etc. fall outside the limits of the available curves, sorne extrapolation of the WRC method must be used. The current default is to use the last value in the particular WRC table. If one wishes to control the extrapolation methodology interactively, you may do so by changing the WRC 107 default from "USE LAST CURVE VALUE" to "INTERACTIVE CONTROL" in the Computation Control option located inside the Configure-Setup module of the MAIN MENU. Once again, data beyond the limits of the bulletin curves should be treated with skepticism. The following page was reprinted from WRC Bulletin 107 to illustrate how the specified data is used to calculate the stress intensitites at the eight points on the vessel around the nozzle. In the event that a reinforcing pad is defined around the nozzle, the corresponding vessel stresses at the edge of the reinforcing pad will be calculated automatically using the rigid plug assumption discussed earlier.

3-63


_., •.....,...

1:. ~"L.

g ...........,. L._M_.. T............... _L.oo", • __ L ....

" z ___

J".

..,

IIC,: _ _ ift.11h

1I1.:' _ _ lftoolft.o III, 1", Ibo VC,="--~

=__

(J

VL::--I\.

v..... .......

................

(0.175)

~:..

::--

s..... c:..c.....l_ 4w ..:

2. G.....,

v.........e.......

+=--

.) .......... IM4. ""

.) -

T _ _ _ 1.

'.1:1 _ _ 1••

1.=

..........

-

- - R.,.

·IIOTI: bter.llr.nl ...., .. I• IC,,""'CI .1.1. Il,,,

- - 1 ••

teft ...

.-i_

CUINDRICAI.. SHELL

1) When t ; O. S c larqest absolute ~qnitude of either S - l/2 [Ox+O~:!: {(ex - Of)2 + 4t2"Jor I(o~ - 0,)2 + 4't Z 2) When 't '" 0, S

largest al/solute Jna9nitude of either

S - 0x' 0, or COx -

0,'

3-64


3.4.2.2 Running a Sample WRC 107 Calculation The example problem which follows goes through a comprehensive local stress analysis of a vessel/nozzle using WRC 107 criteria. y Positive Axes for WRC 107

l;:z WRC 107 Example Problem

Node Numbers in the Piping Model

Node Numbers Around Nozzle Junction

local stiffness of the vessellnserted between nodes 55 and 56

Node Numbers Around Nozzle Junction Figure 3-69· Node Numbers in the Piping Model

3-65


In the figure, the user may also notice that there are two nodes occupying the same space at the nozzle/vessel surface junction: nodes 55 and 56. An anchor at 55 with a connecting node at 56 could be used to model the local vessel flexibility as "rigid". The anchor could then be replaced with a WRC 297 local vessel flexibility model, and the job be rerun to get a good idea of the "range" of loads and displacements that exist in the system around the vessel nozzle. In either case, the various restraint loads (forces and moments due to sustained, expansion and occasional loads) can be obtained from the appropriate CAESAR II restraint report. These loads reflect the action of the piping on the vessel. The sustained and expansion restraint report of the "rigid" anchor model are shown in Figure 370. For purposes of illustration, both the global system loads and the corresponding local WRC 107loads are summarized in Figure 3-71.

CAESAR Il RESTRAINT REPORT CASE 2 (SUS) W+P1

Sustained~

FilE: EX107 DATE:JUN 19, 1987

WRC 107 EXAMPlE PROBlEM RIGID ANCHOR VESSEl MODEl

---------- Momemts(ft.lb.) ----------

;;------- Forc:lb.)---------FZ

MX

MY

MZ

55 70 5

-26. -26. 26.

-1389. -22216.

32. 32.

-65. 355.

127. 288.

4235.

O.

O.

-1296.

o.

O. O. O. O.

-32.

5 5 5

O. O. O. O. O.

O. O. O.

O. O. O. O.

O. O. O.

14233.

96.

O. O. O. O. O.

O.

-661.

-156.

O. O.

TYPE RIGID ANCHOR RIGID ANCHOR DISPL. REACTION DISPL. REACTION DISPL. REACTION DISPL. REACTION DISPL. REACTION DISPL. REACTION

Sustainedpiping loads on the vessel in global coownates

Expansion piping loads on the vessel in global coordinates CAESAR Il RESTRAINT REPORT CASE 3 (EXP) 03 (EXP) = 01-02

Expansion~ NODE 55 70 5 5 5

---------- Forces(lb.)--------FX

WRC 107 EXAMPlE PROBlEM RIGID ANCHOR VESSEl MODEl

FilE: EX107 DATE: JUN 19, 1987

---------- Momemts(ft.lb.) ----------

FY

FZ

MX

MY

MZ

TYPE

6573. 8573. -8573.

23715. 23715.

-5866. -5866.

31659. -44599.

-5414. -34744.

-52583. -282611.

O.

O. O.

-23715.

O. O.

O.

5866.

O. O. O.

O. O. O.

O. O. O.

Restraint Report from CAESAR II Figure 3-70

3-66

RIGID ANCHOR RIGID ANCHOR DISPL. REACTION DISPL. REACTION DlSPL. REACTION


SUMMARY OF RESTRAINT LOADS ON THE VESSEL (lb) Y

(lb) X

(lb) Z

(tUb) MX

(tUb) MY

(ft. lb) MZ

SUSTAINED

-26

-1389

32

-65

127

4235

EXPANSION

8573

23715

-5866

31659

-5414

-52583

WRC107LOCALCOMPONENTS FORCE P(+X)

FORCE VL(-Y)

FORCE VC(+Z)

MOMENT T(-X)

MOMENl MOMENT MC(+Y) ML(+Z)

SUSTAINED

-26

1389

32

65

127

4235

EXPANSION

8573

-23715

-5866

-31659

-5414

-52583

Global to Local Loads Figure 3-71

The total sustained axialload on the nozzle may not be reflected in the restraint report. A pressure thrust load will contribute an additional axial load to the nozzle. The pressure thrust force always tends to push the nozzle 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 (12 2 ) / 4

=

-31,128

The P load may be adjusted automatically for the input by CAESAR II's WRC 107 module, if the user so requests. The actual preparation of the WRC 107 calculation 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. The CAESAR II program performs this conversion automatically. The user will, however, have to identify the vectors defining the vessel as well as the nozzle centerline. The following figure is provided to illustrate the definition of the direction vectors of the vessel and the nozzle.

3-67


NOZZ_E DIRECTIOr--;VECTOR -..--

--e

B

.. - . A

VESSE~

DI~ECTION

VECTOR y

• ------. x Figure 3-72 Notice that in order to define a vessel direction vector, the user first needs to designate the output data points (A->D) as defined by the WRC 107 Bulletin. Note that the line between data points B and A defines the vessel centerline (except for nozzles on heads, where the vessel centerline will have to be defined along a direction which is perpendicular to that of the nozzle). Since, in the vessel/nozzle configuration shown, pointAis 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 nozzZe direction vector is aZways defined as the vector pointing {rom the vesseZ nozzZe connection to the centerZine ofvesseZ. The program first prompts the user for the entries of geometric data describing both the vessel and nozzle, followed by spreadsheets for loadings. The values of the geometric entries in this example are shown in the following printouts from the program.

3-68


URC 187 Analysis

Shell'AttachMent ID = EXP14 UESSEL DATA SHEET

Uessel Hode HUMber (Optional) •••••••••••••....•. Uessel Type: 1-Cylindrical, Z-Spherical •••.•.•••

&0.0000 1.0000

Uessel Outside DiaMeter (D) ••••••••.•••••••••••• Uessel Thickness (T) •.••••••.•••.••••.••..••••••

119.8250 0.&250

Uessel Centerline Direction Cosine (UX) Uessel Centerline Direction Cosine (UV) Uessel Centerline Direction Cosine (UZ)

-1. 0000

Cold S. I. AlloL.!able (SMC) •••.•.•.••••.••..••••.• Hot S. I. AlloL.!able (SMh) •..•••••.•..•...••• 1...

PgUp'PgDn'HoMe'End-Change Page

URC 187 Analysis

[Escl-Exit

20000.0000 20000.0000

[F1l -Execute

Shell'AttachMent ID

=

[?l-Help

EXP14

HOZZLE DATA SHEET Hozzle Hode HuMber •.•••••.••••••••.•••.•••••.••• Hozzle Type: I-Round HolloL.!, Z-Round Solid, 3-Square HolloL.!, 4-Square Solid, 5-Rectangular Solid ••••..••.•••.•••

55.0000

Hozzle Outside DiaMeter (d or Z*CI) ...•••.• (in.) Hozzle Thickness (t) •.•••.••••••••••••.••.• (in.)

12.7500 0.3750

Hozzle Centerline Direction Cosine (HX) Hozzle Centerline Direction Cosine (HV) Hozzle Centerline Direction Cosine (HZ)

1.0000

1.0000

1

ParaMeter for Rectangular AttachMent (CZ) •• (in.) Reinforcing Pad Uidth (U) •••••••••••••••••• (in.) Reinforcing Pad Thickness (Tl) •••.•••.•••.• (in.)

PgUp'PgDn'HoMe'End-Change Page

[Escl-Exit

[Fil -Execute

Vessel and Nozzle Data Figure 3-73

3-69

[?l-Help


The user may enter up to three sets of loadings representing Sustained (SUS), Expansion (EXP), and Occasional (OCC) load cases. The program automaticaIly performs the stress calculation ofeach of the load cases consecutively. In the present case, we only have to be concerned about the sustained and thermal expansion cases. The loads are shown in the foIlowing two screens. The user can elect to leave any input ceIls blank ifthey are found not applicable. After checking aIl the entries made, the user should press [Ft] to initiate the analysis.

URC lB7 Analysis

Shell/AttachMent ID = EXP14

GLOBAL FORCES/MOMENTS (SUS) Global Force Fx .••....•.•..•.•.......•..•.. (lb.) Global Force Fy .•.•....•.•.....•........... (lb. ) Global Force Fz .......•..•..•..•••......•.. (lb.)

-21).0fl00 -1389.0000 32.0000

Global MOMent Mx •.•....•.••.•..•.......• (ft. lb.) Global MOMent My ......•..•..•..•........ (:Ft. lb.) Global MOMent Mz •.........•.......•....• (:Ft. lb.)

-1)5.0000 127.0000 4235.0000

InternaI Pressure (P) ..............• (lb./sq.in.) Include Pressure Thrust: l-Ves/De:Fault, B-No ...•

275.0000 1.0000

1


URC lB7 Analysis

[Escl-Exit

[Fll -Execute

[1l-Help

Shell/AttachMent ID = EXP14

GLOBAL FORCES/MOMENTS (EXP) Global Force Fx ...•....•....•..•........... (lb.) Global Force Fy .•.........•........•....•.• (lb.) Global Force Fz .•......•..•....•.....•....• (lb.)

8573.0000 23715.0000 -581)1).0000

Global MOMent Mx ...•....•..........•.... (:Ft. lb.) Global MOMent My .......•....•..•.•..•... (:Ft. lb.) Global MOMent Mz ....•.•.....•..•........ (:Ft. lb.)

31G59.0000 -5414.0000 -52583.0000

1


[Escl-Exit

Nozzle Loads Figure 3-74 3-70

[Fll -Execute

[1l-Help


After the input echo, the parameters extracted from the WRC 107 figures are printed on the screen. This step is similar to determining the data hy hand. These non-dimensional values are comhined with the nozzle loads to calculate the two normal stresses and one shear stress. The stresses will he reported on the outer and inner vessel surfaces (upper & lower, respectively) of the four points A, B, C & D located around the nozzle. The program provides the normal and shear stresses and translates them into stress intensities which can he used for comparisons against material allowahles. The output of the stress computations are shown in the four pages shown in Figure 3-75. As the output shows, the largest expansion stress intensity (117475 psi) occurs at the outer surface of point B (Bu). WRC 107 Stress Ca1cu1ations Attachment/She11 ID = EXP14

Page =

Date = Mar 6, 1996 Time = 2:02 pm

1 of

Vessel Stresses @ Nozzle Junction DESCRIPTION: Vessel Node: 60 Vessel Type: Cyl indri cal THIS IS INPUT TITLE PAGE FOR CAESAR II VesselOri.: 0.00,-1.00, 0.00 APPLICATION GUIDE, EXAMPLE NO.14.
Dimensions

Vessel Mean Rad. Vessel Thickness Noz. Outside Rad. Nozzle Thickness

Rm= T = ro = t =

59.688 0.625 6.375 0.375

in·1 in·1 in·1 in·1 1 1

(SUSTAINED)

Axial Force P Circ. Sh. Force VC = Long. Sh. Force VL = Circ. Moment MC = Long. Moment ML= Tors. Moment MT=

Parameter(s) used in the Interpolation of Dimensionless Loads: Gamma =

95.50 Dimensionless Loads for Cylindrical Shells Beta

Curves read for N(PHI) M(PHI) N(PHI) M(PHI) N(PHI) M(PHI)

/ 1 / / / 1

( p/Rm ) ( P ) ( MC/(Rm**2 * * ( MC/(Rm ( ML/ (Rm**2 * * ( ML/(Rm

N(x) M(x) N(x) M(x) N(x) M(x)

/ / / / / 1

( p/Rm ) ( P ) ( MC/(Rm**2 ( MCI (Rm ( ML/(Rm**2 ( ML/(Rm

Figure

Value

Beta) Beta) Beta) Beta)

0.093 0.093 0.093 0.093 0.093 0.093

4C 2Cl 3A lA 3B lB

14.994 0.059 3.449 0.085 10.793 0.035


0.093 0.093 0.093 0.093 0.093 0.093

3C 1C1 4A 2A 4B 2B

12.082 0.097 5.631 0.045 3.511 0.051

Stress N(PHI) M(PHI) M(PHI)

Points C & D (March 1979) 0.093 / ( p/Rm ) 0.093 1 ( P ) * Beta) ) 1 ( ML/(Rm 0.093

3C 1C 1B1

12.082 0.094 0.035

N(x) M(x) M(x)

/ ( p/Rm ) / ( P ) 1 ( ML/ (Rm

4C 2C 2B1

14.994 0.060 0.052

* * * *

* Beta) )

0.093 0.093 0.093

Figure 3-75

3-71

-31128. 32. 1389. 127. 4235. 65.

lb. lb. lb. ft.lb. ft.lb. ft.lb.

4


Date = Mar 6, 1996 Time = 2:02 pm

WRC 107 Stress Calculations Attachment/Shell ID = EXP14

Page =

2 of

Vessel Stresses @ Nozzle Junction Stress values at (lb./sq.in.)

Type of Stress Circ. Circ. Circ. Circ. Circ. Circ.

Memb. Bend. Memb. Bend. Memb. Bend.

Loadl P P MC MC

-Pl

1

-Q

1

-Pl

1

-Q

1

ML -Pl ML -Q

1

Total Circ. Stress

1

1

Au

Al

Bu

Bl

Cu

Cl

Du

Dl

12510 12510 12510 12510 10081 10081 28242 -28242 28242 -28242 44865 -44865 -25 -25 0 0 0 0 0 0 0 -358 358 0 -2635 -2635 2635 2635 0 0 -4938 4938 4938 -4938 0 0

10081 10081 44865 ·44865 25 25 -358 358 0 0 0 0

33179 -13429

54563 -34451

55329 ·35117

10081 10081 10081 10081 12510 12510 46473 -46473 46473 -46473 28748 -28748 0 0 0 0 -41 -41 0 0 0 0 -190 190 857 -857 -857 857 0 0 -7325 7325 7325 -7325 0 0

12510 12510 28748 -28748 41 41 ·190 190 0 0 0 0

48372 -29924

41489 ·16387

1

Long. Long. Long. Long. Long. Long.


P P MC MC ML ML

-Pl

1

-Q

1

-Pl

1

-Q

1

-Pl

1

-Q

1

48325 -18035

1

Total Long. Stress

1

64736 -42860

41027 -16089

VC -Pl VL -Pl MT -Pl

2 0 4

2 0 4

-2 0 4

-2 0 4

0 -110 4

0 -110 4

110 4

0 110 4

Total Shear Stress

6

6

2

2

-106

-106

114

114

48372

29924

64736

42860

54563

34451

55329

35117

Shear Shear Shear

Stress Intensity

Figure 3-75 (Cont.)

3-72

4


Date = Mar 6, 1996 Time = 2: 02 pm

WRC 107 Stress Calculations Attachment/Shell ID = EXP14

Page

3 of

Vessel Stresses @ Nozzle Junction Vessel Node: 60 Vessel Type: Cylindrical VesselOri.: 0.00, ·1.00, 0.00

DESCRIPTION: THIS IS INPUT TITLE PAGE FOR CAESAR II APPLICATION GUIDE, EXAMPLE NO.14.

Nozzle Node: 55 Nozzle Type: Round-hollow Nozzle Ori.: 1.00, 0.00, 0.00

WRC 107 STRESS CALCULATION AND ASME SEC. VIII DIV.2 STRESS SUMMATIONS.

Dimensions Vessel Mean Rad. Vessel Thickness Noz. Outside Rad. Nozzle Thickness

Rm= T ro t

Nozzle Loads 59.688 0.625 6.375 0.375

in·1 in·1 in·1 in·1 1 1

(EXPANSION)

Axial Force P Circ. Sh. Force VC Long. Sh. Force VL Circ. Moment MC = Long. Moment ML Tors. Moment MT =

8573. -5866_ -23715. -5414. -52583. -31659.

Parameter(s) used in the Interpolation of Dimensionless Loads: 95.50

Gamma

Dimensionless Loads for Cylindrical Shells Beta

Curves read for

Figure

Value


0.093 0.093 0.093 0.093 0.093 0.093

4C 2C1 3A lA 3B lB

14.994 0.059 3.449 0.085 10.793 0.035


0.093 0.093 0.093 0.093 0.093 0.093

3C 1C1 4A 2A 4B 2B

12.082 0.097 5.631 0.045 3.511 0.051

Stress Points C & D (March 1979) 0.093 N(PHI) 1 ( p/Rm ) 0.093 M(PHI) 1 ( P ) M(PHI) 1 ( MLI (Rm * Beta) ) 0.093

3C 1C 1B1

12.082 0.094 0.035

0.093 0.093 0.093

4C 2C 2B1

14.994 0.060 0.052

N(PHI) M(PHI) N(PHI) M(PHI) N(PHI) M(PHI)

1 ( p/Rm ) 1 ( P ) 1 ( Mcl (Rm**2 * * 1 ( MCI (Rm 1 ( ML/(Rm**2 * * 1 ( MLI (Rm

N(x) M(x) N(x) M(x) N(x) M(x)

1 ( P/Rm ) 1 ( P ) 1 ( MC/(Rm**2 1 ( MC/(Rm 1 ( MLI (Rm**2 1 ( MLI (Rm

N(x) M(x) M(x)

1 ( p/Rm ) 1 ( P ) 1 ( MLI (Rm

* * * *

* Beta) )

Figure 3-75 (Cont.)

3-73

lb. lb. lb. ft.lb. ft.lb. ft.lb.

4


WRC 107 Stress Calculations Attachment!Shell ID = EXP14

Date Time

= Mar

6, 1996 2:02 pm

Page

4 of

Vessel Stresses @ Nozzle Junction Stress values at (lb. !sq. in.)

Type of Stress

Au

Al

Du

Dl

Circ. Memb. P 'Q 1 ·3445 ·3445 ·3445 ·3445 ·2776 ·2776 ·2776 7778 ·7778 7778 ·12356 12356 ·12356 Circ. Bend. P 'Q 1 ·7778 Circ. Memb. MC 'Q 1 0 0 0 0 1076 1076 -1076 0 0 0 15282 -15282 -15282 Circ. Bend. MC -Q 1 Circ. Memb. ML -Q 1 32728 32728 -32728 -32728 0 0
·2776 12356 -1076 15282 0

Loadl

Bu

BI

Cu

Cl

1

Total Circ. Stress Long. Long. Long. Long. Long. Long.


P P MC MC ML ML

-Q -Q -Q -Q -Q -Q

Total Long. Stress VC -Q VL -Q MT -Q

1

82823 -24257-105269

32923

1226

-4626 -31490

23786

-2776 -2776 -2776 12799 -12799 12799 0 0 0 0 0 10647 10647 -10647 -10647 90954 -90954 -90954 90954

-3445 -7917 1758 8120 0

-3445 7917 1758 -8120

-3445 -7917 -1758 -8120 0

-3445 7917 -1758 8120 0 0

86026 -70284 -117176

90330

-1484

-1890 -21240

10834

468

0 1894 -2380

-1894 -2380

0 -1894 -2380

-2776 -12799 0

-2380

-468 0 -2380

468 0 -2380

-2380

0 1894 -2380

Total Shear Stress

-2848

-2848

-1912

-1912

-486

-486

-4274

-4274

Stress Intensity

87691

70459 117475

90393

2879

4709

33038

25069

Shear Shear Shear

-468

Figure 3-75 (Cont.)

3-74

4


3.4.2.3 Evaluating Vessel Stresses The ASME Section VIII, Division 2 code provides for a fairly elaborate procedure to analyze the local stresses in vessels and nozzles. Only the elastic analysis approach will be discussed in this manual. The user should always refer to the applicable code if any of the limits described in this section are approached, if any unusual material, weld, or stress situation exists, or ifthere are non-linear concerns such as the material's operation in the creep range. The first step in the procedure is to determine if the elastic approach is satisfactory. Section AD-160 contains the exact method and basically states that if aIl of the followingconditions are met, then the elastic approach is sufficient and fatigue analysis need not be done: a) The expected design number of full-range pressure cycles does not exceed the number of allowed cycles corresponding to an Sa value of 3Sm(4S mfor non-integral attachments) on the material fatigue curve. The Sm is the allowable stress intensity for the material at the operating temperature. b) The expected design range of pressure cycles other than startup or shutdown must be less than 1/3 (114 for non-integral attachments) the design pressure times (S/Sm)' where Sais the value obtained on the material fatigue curve for the specified number of significant pressure fluctuations. c)

The vessel does not experience localized high stress due to heating.

d) The full range of stress intensities due to mechanical loads (including piping reactions) does not exceed Sa from the fatigue curve for the expected number ofload fluctuations. Once the user has decided that an elastic analysis will be satisfactory, the comprehensive approach as used in the CAESAR II local stress evaluation is appropriate. This method will be described in detail below, after a discussion of the Section VIII Div. 2 Requirements.

3-75


ASME Section VIII Division 2 • Elastic Analysis of Nozzle

Ideally in order to address the local allowable stress problem, the user should have the endurance curve for the material ofconstruction and complete design pressure / temperature loading information. If any of the elastic limits are approached, or ifthere is anything out of the ordinary about the nozzle/vessel connection design, the code should be carefully consulted before performing the local stress analysis. The material Sm table and the endurance curve for carbon steels are given in this section for illustration. Only values taken directly from the code should be used in design. There are essentially three criteria that must be satisfied before the stresses in the vessel wall due to nozzle loads can be considered within the allowables. These three criteria can be suIDmaried as: Pm
Where P ID, Ph Pb, and Q are the general primary membrane stress intensity, the local primary membrane stress intensity, the local primary bending stress intensity, and the total secondary stress intensity (membrane plus bending), respectively; and k, Smh, and Smavg are the occassional stress factor, the hot material allowable stress, and the average material allowable stress intensity (Smb + Sme) / 2. Due to the stress classification defined by Section VIII, Division 2 in the vicinity of nozzles, as given in the Table 4-120.1, the bending stress terms caused by any externalload moments or interal pressure in the vessel wall near a nozzle or other opening, should be classified as Q, or the secondary stresses, regardless of whether they were caused by sustained or expansion loads. This causes Pb to disappear, and leads to a much more detailed classification:

P - General primary membrane stress intensity (primarily due to internaI presID

P

sure); Local primary membrane stress intensity, which may include: Membrane stress due to internaI pressure; Local membrane stress due to applied sustained forces and moments.

Q -

Secondary stress intensity, which may include: Bending stress due to internal pressure; Bending stress due to applied sustained forces and moments; Membrane stress due to applied expansion forces; Bending stress due to applied expansion forces and moments Membrane stress due to applied expansion moments

3-76


Each of the stress terms defined in the above classifications contain three parts: two stress components in normal directions and one shear stress component. To combine these stresses, the following rules apply: 1) Compute the normal and shear components for each of the three stress intensities, i.e. Pm, Ph and Q; 2) Compute the stress intensity due to the Pm and compare it against kSmh; 3) Add the individual normal and shear stress components due to Pmand Pl; compute the resultant stress intensity and compare its value against 1.5k~; 4) Add the individual normal and shear stress components due to Pm, Ph and Q, compute the resultant stress intensity, and compare its value to against 3Smavg ' 5) If there is an occasionalload as well as a sustained load, these types may be repeated using a k value of 1.2. These criteria can be readily found from Figure 4-130.1 ofAppendix 4 ofASME Section VIII, Division 2 and the surrounding text. Note that the primary bending stress term, Pb, is not applicable to the shell stress evaluation, and therefore disappears from the Section VIII, Division 2 requirements. Under the same analogy, the peak stress limit may also be written as: Pl + Q + F < Sa where: F represents fatigue stresses. The above equation need not be satisfied, provided the elastic limit criteria of AD-160 is met based on the statement explicitly given in Section 5-100, which is cited below: "If the specified operation of the vessel meets all of the conditions of AD-160, no analysis for cyclic operation is required and it may be assumed that the peak stress limit discussed in 4135 has been satisfied by compliance with the applicable requirements for materials, design, fabrication, testing and inspection ofthis division."

3-77


1995 SECnON Il

T.... ZA

TAlLEZ-' SEcnON Ill. CLASS 1 AND SEtnON VIII, DIVISION 2 DESIGN STRESS INTENSITY VALUES S. FOR FERROUS MATERIALS

Lm.

NonIÎIIII

e..--

No. C

c

PI'DCIICI ....

Sar."p,W1d._

sNo.

..45

K01700 1(01700

ElA SlA A

1(02504 1(Q2504 1(02!O1

C C C-Si

WlcL_

S"'5)

Smls.DiIIe

S"'5) SA-1Clf>

7

C

Sar,_

8 q

C C C C

10 11

12 l3 14

Plaie

SA-215 SAo", SAoU3

C-Un

17 lB 19 20

C-Si

Pla 5m1S.DiIIe

5 ... 516 SA-524

5 .....71 5 .....71

C

WId..,.

e-w.-si

Wld.1I.

C

WId.JIÏ!Ie

C-Si C-Si

W1d. • •

C-U~Si

WId. • • WId.DiIIe

p-

5 .....72 5 .....7Z 5 .....7Z 5 .....72

21

C-U....si C-U....si C-$i ":-Si

c.inoJs

S"'U1 5 ...Z16

29

·';-Si

F_

SA411of>

JO

C-U....si C-Si C-Si C C-MII

25 2110 27

n 32 33 34

35 3& 37

38 39 40 41 42

Plaie" . . . . . . . . .

F.......

SA-)5O

Bar._

SAo352 SA-IIoIIoO S..... 75 SAo765

c.tDille

F. . . . .

C-M~Si

Pla Pla

C-Si C-UneSi C-4I....si C-Si C-u..-$i

C-Si

C~i

>2

C

KOZIOI K03001 KO:JOOI KOHOl KO)OOl

>2

55

KOllOO KOUOK02IOI KOZm

Il CAS5

cas AS5 B55 C55 E55

5"'"

SMrf>Z

F.......

c-.

KOZZOO KOZ2OO

SA-,,. SA-".

C~i

S"'515

in.

50 B

B A50 55

Wld.DiIIe Smls. ..... WId._ S...... _

C-U"

A

S ...ZI5 5 .....72 5 .....75

C-MII C-MII

lS

23 24

sar._

WId.*

5 .....75

SA-W

C

110

21 2Z

Pie Plaie

SIW ~

SM72

4

SmIS._

CImI

CGIIIlJ T_

UNS ND.

45

C

f>

10. .

AI., DIIi9I

S.....75 5A415

3

S

T.""

A WCA 1

KOZI01 K02OO1 K01100 KOZZOZ I(OZ_ K01701 1(0)502 J02!02 K03_

WCA

Ka_ J02504 J025o-

IloO 1

KOlO46

LFl

LCA

IloO IloO

WId. • • WId." WId.III11C

SA-SU. 5 .....71 S.....71 SA-671

C860 CCIIoO CEIIoO

WId. JIÏ!Ie WId._ WId.DiIIe

$M72 S.....72 S.....72

CIIoO EIIoO

B60

IloO

K02401 K02100 K02401 K02100 K02402 K02401 K02100 K02402

.,u

Example Material Table (For Values of Sm) Figure 3-76

3-78

P-ND.

'ND.-


PARl'D-MlOPIkIES

T..... 2A

TABLE2A SEC1ION III, CLAIS 1 AND SECTION VlIL DMSION Z DESIGN STRESS INTENSI1Y VALUES S. FOR FERROUS MATERIALS MIlL

T_

LiIR No.

S--.

.,..... MIL

5.......

AcIIIIIC-...... r.... LiIIIIIs

m

bI

4S 4S 45

22.5

NP

700

24

700 700

7DO

48 48 48

30

700!SPn 700!SPn

HP

70CI

50 50

2S Z7

7DOCSPn

50 50

27

10

27

11

55

27.5

700 700 700CSPn

12

55 55 55

700 700 700

700

700

NP

16

55 55

30 30 30 3D 30

700

7DO

Z

4 5

6 7

1 CJ

13

14 15

24

JO JO

CS·l CS-l

Gu G4,521 G4, 0. "

700

cs·a

G4,1iZ1

7DO 7DO

CS-l CS·! CS·l

GZI GZl G4.621 G4.G5,GI G21

NP

NP

NP

700

NP 700

CS-l

7DO

cs-z

IIP

17

5S

3D

70CI

II

55

NP

7DO 7DO

19

55

20

55

JO JO JO

700 700

NP NP

21

55

30

22

JO

2l

55 55

24

S5

3D 3D

700 700 700 700

NP NP NP NP

25

sa

J'

700CSPn

2. 27

SI

40

NP

fIG fIG fIG

30

JO

700 700 700

7DO 7DO 7DO 7DO

za

29 3D 31

32

33 ]4

3D

JO JO

fIG fIG fIG

30

fIG fIG

JO JO

700 700 700 700CSPn

GZI GZI G4.GZ1

CS·Z

G4.GZI G4

cs.z

GZI GZI

700 700

CS-z CS-z

700

19

fIG

32

40

fIG

41

fIG fIG

li li

700

N' NP

J2

700

NP

~.GZI

G4.&21

~

cs-z

n

42

CS-z

c:s.,

NP NP

3.

J2

G4.G5.GI ' •• G5.GI G4.Iii5.GI G4.65.GI

7DO

700 700 700 700 700

l2

' •• 621 GZl G4.65.GI G4.G5.GI

CS·Z CS-Z cs·z

fIG fIG fIG loCI

n

CS-2 CS.z

104.521 G4 104,&21 G4,&21 G4

700

36 37

35

CS·Z CS-z

700 7DO

NP

.....

CIIM No.

YIII-2

GI

,

,......

EzIInIII

(N' ...........

(Spr . . . . .~

HP NP NP

G4.GZ1

G4.GZl G4.1ii5, GI G4.65.'" G4.65. ... G4.Iii5.GI G•• G5.GI G4.65.GI

269

Example Material Table (For Values of Sm) Figure 3-76 (Cont.)

3-79


1995 SECTION D

T.... U.

TULEn SECTION m, CLASS 1 AND SEcnON VIII, DIVISION 2 DESIGN STRESS INTENSITYVALUES S. FOR FERROUS MATERIAU

DaiIJa sans I/IWIItr, !IIi (....... .., 1000 .. Clll..-1IIil.... MetIt Til....... "F, NIt ÛCIdI9 L.itIf No.

2

3 4

,.

5

7 8

., 10 11 12

13 14 15 le.

1. 17

19 20

21 2Z ZJ 24

Z5 2e. 27

21 29

:JO 31

32 J3 )4

35 36 )7

3. 39 40 41 42

-2010 100

150

200

2SO

300

4ClO

soo

1303

600

OSO

700

15.0 15.0 15.0

13.7 14.c. 15.0

12.9 13.7 13.7

12.1

11.1

14.2 14.2

12.9 12.9

11.'

10.9 11.6

11.9

ll.b

10.lI n.5 11.5

16••0 16..0 16..0

16.0 10.0 16..0

16.0 16.0 16..0

16.0 16.0 16..0

16.0 le•• O 16..0

14.8 14.1 14.8

t4.5 14.5 14.5

14.4 14•• 14.4

16.7 16.7 10.1 110.7 18.3

15.2 16.4 16..4 16..4 16..7

14.8 16.0 16..0 16..0 lU

14.3 15.4 15.4 15•• 15.7

13.5 14•• 14•• 14•• 14.9

12.3 13.3 13.3 13.J

13.6

12.1 13.1 13.1 13.1 13.3

12.0 13.0 13.0 13.0 13.2

18.3 lllo3 11.3 11.3 11.3

lU 18.3 11.3 lU lU

17.7 17.7 17.7 17.7 17.7

17.2 17.2 17.2 17.2 11.2

110.2 110.2 16..2 16.2 110.2

14.1 14.1 14.1 14.1 14.1

14.5 14.5 14.5 14.5 14.5

14.4 14.4 14.4 14.4 14.4

11.3 lU lU lU

11.3 11.3

lU 11.3

17.7 17.7 17.7 17.7

17.2 11.2 17.2 17.2

16..2 16..2 110.2 16..2

14.1 14.1 14.1 14.1

14.5 14.5 14.5 14.5

14.4 14.4 1••4 14.4

lU 18.3 lU lS.J

lll.J lU lU 18.3

17.7 17.7 17.7 17.7

17.2 17.2 17.2 17.2

16..2 110.2 16.2 16..2

14 .• 14.1 14.. 14"

14.5 14.5 14.5 14.5

14.4 14.4 14.4 14••

19.3 19.3 20.0 20.0 20.0

19.3 19.3 lU 11.3 11.3

19.3 19.3 17.7 17.7 17.7

19.3 19.3 17.2 17.2 17.2

19.3 19.3 16.2

lU lU

17.7 19.3 14.1 14.1 14..

17.4 19.3 1• .5 14.5 14.5

17.3 19.2 14.4 14.4 14.4

20.0 20.0 20.0 20.0 20.0

18.3 lU 11.3 18.3 18.3

11.7 17.1 17.7 17.7 17.7

17.2 17.2

lU lU

17.2 11.1 17.Z

110.2 110.2 110.2

14" 14.. 14•• 14.1 14.8

14.5 14.5 14.5 14.5 14.5

14.4 14.4 14.• 14.4 14.4

20.0 20.0 20-D 20-D

18.9 11.9 18.. 18.9 18.9

lU lU lU lU lU

17.3 17.3 17.3 17.3 17.3

15.8 U.I l5.I 15..

l5.5 lS.5 l5.5 lS.5

20.0

19.5 19.5 19.5 19.5 19..5

15.5

20.0 20.0 20.0

19.5 19.5 19.5

11.9 111.9 11.9

lU lU

17.3 17.3 17.3

15.1 l5.8 15.8

15.5 15.$ 15.5

lU

1.5.1

750

l.S.4

15.. 15.. 15.4 U.4 15.4

15.. 15.4

270

Example MateriaI Table (For VaIues of Sm) Figure 3-76 (Cont.)

3-80

800

850

900


_:

.UE· . . . .... IZIT_ ' ...... ' _ _ _ . _ ..... _ _ _ .. _ _ 1 Z I.._ U1S ._ .'l1Il11L

>

... i

z~

.,;:;

;

.~

::~

1 3:

.,>> Z

...ô '0

_..FIG. 5-110.1 DESIGN FATIGUE CUIIVU FOR CA_. LOW ALLOY. SERIES 4XX, KIION ALLOY STEELS AllO "110" TEIISILE STEELS FOR TEMPERATURES NOT EXCEEDING 7IIII'F

Example Fatigue Curve (For Values of Sa) Figure 3-77 Procedure to Perform Elastic Analyses of Nozzles The procedure for checking stresses in vessel shells using WRC 107 can be summarized as follows: Step 1 -

Check that no geometric limitations invalidate the use ofWRC 107;

Step 2 -

IfWRC 107 is applicable, check to see whether or not the elastic approach as outlined in Section VIII, Division 2, AD-160 is satisfactory;

Step 3 -

Compute the sustained, expansion and occasionalloads in the vessel shell due to the applied nozzle loads. Consider the local restraint configuration in order to determine whether or not the axial pressure thrust load (P * Ain) should be added to the sustained (and/or occasionalloads). If desired by the user, this thrust load will be automatically calculated and added to the applied loads.

Step 4 -

Calculate pressure stresses, Pm, on the vessel shell wall in both longitudinal and circumferential (hoop) directions for both sustained and occasional cases. Notice that two different pressure terms are required in carrying out the pressure stress calculations. P is the design pressure of the system (sustained), while Pvar is the DIFFERENCE between the peak pressure and the design pressure of the system, which will be used to qualify the vessel membrane stress under the occasionalload case. Note that the Pm stresses will be calculated automatically if a pressure value is enter by the user. 3-81


Step 5 -

Run WRC 107 to calculate the Ph and Q stresses as defined earlier. Note that the local stresses due to sustained, expansion and occasional loads can be computed simultaneously.

Step 6 -

Various stress components can be obtained from combining the stress intensities computed from applyingthe sustained, expansion and occasional loads. These stress intensities can then be used to carry out the stress summations and the results are used to determine acceptability of the local stresses in the vessel shell. Notice nowCAESAR II can provide the WRC 107 stress summation module in line with the stress calculation routines

Under the above procedure, the equations used in CAESAR II to qualify the various stress components can be summarized as follows: P m(SUS) < Smh P m(SUS + DCC) < 1.2Smh P m(SUS) + Pl(SUS) < 1.5Smh Pm(SUS + OCC) + Pl(SUS + DCC) < 1.5(1.2)Smh P m(SUS + DCC) + Pl(SUS + OCC) + Q(SUS + EXP + OCC) < 1.5(Smc + Smh)

3-82


3.4.2.4 Completing the Sample Calculation Once the stress intensities are computed, the user can elect to use 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 WRC 107 stress summation routine can be activated from the WRC 107 Menu. The stress summation will be performed automatically after the user identifies the problem name of the stress calculation. A sample output is given below. WRC 107 Stress Summations Attachment/Shell ID = EXP14

Date Time

= =

Mar 6, 1996 2:03 pm

Page = l of 1

Vessel Stress Summation @ Nozzle Junction

Type of Stress Intensity

Stress values at (lb. /sq. in.)

Au

Location

Al

Bu

B1

Cu

Cl

Du

Dl

Circ. Circ. Circ. Circ.

Pm Pl Q Q

(SUS) (SUS) (SUS) (EXP)

26125 26125 26125 26125 0 0 9875 9875 15145 15145 10056 10056 10106 10106 23304 ·23304 33180 ·33180 44507 -44507 45223 -45223 82823 -24257-105269 32923 1226 -4626 -31490 23786

Long. Long. Long. Long.

Pm Pl Q Q


0 0 0 9224 9224 10938 10938 39148 - 39148 53798 -53798 86026 -70284-117176 90330

Shear Shear Shear Shear

Pm Pl Q Q


2 4 -2848

2 4 -2848

-2 4 -1912

-2 4 -1912

0 -110 4 -486

-110 4 -486

110 4 -4274

110 4 -4274

S.1. Pm (SUS)

26125

26125

26125

26125

12994

12994

12994

12994

S.1. Pm+P1 (SUS)

36000

36000

41270

41270

25463

25463

25545

25545

143059 100299

52607

47992

55893

39087

34819

20533

S.1. Pm+P1+Q (TOTAL)

1

Max. S.1. S.1. Allowab1e (lb. / sq. in. )

Type of Stress Intensity S.1. Pm (SUS) S.1. Pm+Pl (SUS) S.1. Pm+P1+Q (TOTAL)

12994 12994 12994 12994 12469 12469 12551 12551 28558 -28558 28938 -28938 -1484 -1890 -21240 10834

1

26125 41270 143059

20000 30000 60000

Result

Fai1ed Failed Failed

Figure 3-78 Since the present nozzle loading will cause stress intensities that are not acceptable to the ASME Section VIII, Division 2 criteria, it will have to be corrected. One way of dealing with this type of situation is to adjust the nozzle loading at its source, while the other option might be to reinforce the nozzle connection on the vessel side either by increasing the vessel thickness or by adding a reinforcing pad. The same analysis procedure can be repeated until the final results become acceptable. Note that once a reinforcing pad is seZected, the program will automatically compute the stress state at the edge of the pad as well as at the edge of the nozzZe.

3-83


3.4.3 Estimation of Vessel Nozzle Flexibilities As noted previously, piping attached to vessels can bend or otherwise deform the vessel wall, permitting some displacement or rotation of the connection under load. Therefore, totally rigid restraint models may not be accurate representations of piping to vessel connections - for example, they will probably be highly conservative when calculating nozzle loads during the expansion loading case. Where possible, a more realistic stiffness for the connection should be estimated when possible. One means of doing this is to use the Welding Research Council Bulletin 297 "Local Stresses in Cylindrical Shells Due to External Loadings on Nozzles - Supplement to WRC Bulletin No. 107", which published in August 1984 as an update to WRC Bulletin 107. This update expanded upon the work done for the evaluation of stresses in two normally intersecting cylindrical shells, but also parameterized finite element analyses done to predict nozzle flexibilities. As note d, vessel flexibilities usually reduce loads and stresses in the piping system, so use ofthem is generally less conservative than using completely rigid intersections. However, in some configurations, where the use offlexibilities redistributes loads to rotating equipment, their use may be more conservative (and more realistic). An example is shown in Figure 3-80.

To Rack

ç

Pump Suction

Figure 3-80

With the connection at node point 5 modeled as rigid, the loads from the thermal growth of the rack piping are taken by the nozzle and kept off of the pump battery. However, if the nozzle is truly flexible, it will deform, loading up the pump flanges. It may be necessary to perform a sensitivity study ofthe model to shed light on the true condition ofthe system. The results ofmodeling a system with rigid and flexible nozzles may fall in one ofthree regions, as shown in Figure 3-81.

3-84


x - Rigid Nozzle Model o - Flexible Nozzle Model Stress

x

x

x

x

x x

o 00

Region 1

0

0

0

Region 2

Region 3

.

- Case Studies

Figure 3-81

In Region 1, stresses are high for both models - the model is insensitive to vessel flexibilities. A redesign of the piping or reinforcement at the intersection is needed. (This is mostly the case with smaller diameter piping and heavy vessels.) In Region 2, stresses are high in the rigid model and low with the flexible model, indicating that the job is very sensitive to local flexibility. In this case it is necessary to take a closer look at the intersection: 1

-

2

Are the dimensionless parameters weIl within the limits of the theory? If not, vessel calculations may be way off. Is the nozzle truly an isolated nozzle? Ifnot, stresses near the nozzle could be much higher.

3

-

In most cases ofthis type the WRC 297 stiffnesses are so much smaller than the rigid stiffness the user can adjust the WRC 297 results toward stiffer junctions, (i.e., greater wall thicknesses, smaller radii) without affecting the piping solutions. Manytimes even a WRC 297 junction 10 to 100 times stifferthan what is normally calculated will still produce similar results. In these cases the analyst can comfortably put more confidence in the WRC 297 solution providing (1) and (2) above are answered in the affirmative.

4

-

Is the local vessel model very sensitive to changes (i.e., if the "L" dimensions change, or if the reinforcing pad is left out do the stiffnesses change very much)? If so, then it is necessary to build a range of solutions to study the parameters that effect the model and try to extract the results that are most in line with the assumptions that surround each parameter being varied.

5

-

Are other stresses (i.e., those due to pressure) high? If other stresses are high then the room for error is small.

3-85


6

-

Is the material highly susceptible to cracking or corrosion? In this case, the room for error is also small.

In Region 3, stresses for both models are low. In this case there is probably no problem, need for concern, nor need for extensive time and energy to be devoted to modeling of the nozzles.

3.4.3.1 Use of WRC Bulletin 297 WRC 297 indicates that a typical nozzle flexibility is equivalent to the flexibility of 3 to 13 elbows concentrated as a point spring at the junction of the pipe with the vessel. It is the opinion of the developers that the results of the WRC 297 flexibility calculations are within one order of magnitude of the exact flexibility solution, where the exact solution is defined by a plate finite element analysis of the model geometry. The bulletin provides for calculation of flexibilities in the three directions shown in Figure 3-82. AlI other directions (i.e., torsional, longitudinal shear, and circumferential shear) are considered to be rigid.

Circumferential moment

-~f--~

Axial

Figure 3-82 WRC 297 limits configurations to the following conditions:

d/D < 0.5 dit> 20

20 < Dtr < 2500

3-86


Where: d

=

outside diameter of nozzle, in

D

=

mean diameter of vessel, in

t

=

thickness of nozzle, in

T

=

thickness of vessel, in

The particular checks on the nozzle diameter and thickness are not that critical when dealing with flexibilities (i.e., solid plugjunctions are not that much more or less flexible than hollow nozzle junctions). The effect ofreinforcing pads should be considered if the width of the pad is at least 1.65(DT/2)l/2. In that case, the vessel wall should be increased to Twall + Tpad. The WRC 297 nozzle stiffnesses come from three values extracted from the two curves shown in Figures 3-83 and 3-84. These values (a, MrJ(ET3 e ), and Mc/(ET3 e» are based upon the following geometric parameters: equivalent vessellength L A

=

L / (DT)1/2

=

(d/D)

thickness ratio

=

=

(D/T)l/2 T/t

Where: Ll, L2

=

unstiffened length of vessel on each side of nozzle, in

3-87


... ...

oc

~

6

7

'"1

,.

!!!

~ ;;

.6

~ Q

~

i

.3

i.. g

il ~

1S

1

9

8

Figure 3-83

3-88


."

~

g

4

5

fi

1

r !II

1 ~

§ ~

1 1 !!.

~

i-

Figure 3-84

Mter fin ding a , Mr/CET3 8 ), and Mc!CET3 8 ), the nozzle stiffnesses are calculated as: Kax

=

4.95 a E T2 (D2 A )-112 lb/in

KL

=

Mr/(ET3 8 ) x ET3 x 2pi/360 in-lb /deg

Re

=

Mc!CET3 8 ) x ET3 x 2pi/360 in-lb /deg

This can be illustrated with an example problem. The dimensions are shown in Figure 385 for a nozzle/vessel combination made of a material with a modulus of elasticity of 29E6 psi.

3-89


d - 18 in. t = 0.25 in.

-----+------~~~----~~--------~~

8

0 =48in. T = 0.25 in.

l

L2 130 in.

1

225 in.

Figure 3-85 The first step is to check the WRC 297 limitations in order to determine whether it can apply to this situation: d/D

=

dit

= 18 1 0.25 =

72> 20

DIT

=

=

dIT

= 18 1 0.25 =

18/47.75 =

47.75/0.25

0.377 < 0.5

191 < 2500

72> 5

Therefore this nozzle/vessel configuration meets the requirements of the WRC 297 bulletin. The stiffness calculations are: L

= 8L1L2 1 [L 11/2 + L2112]2 = 8(130)(225)/[130 1/2+225 1/2]2 = 335.7 in

A

= LI (DT)1I2 = 335.71 (47.75 x 0.25)112 = 97.2

À

= (d/D) (DIT)1/2 = (18/47.75) (47.75/0.25)1/2 = 5.21

T/t

= 0.25 1 0.25 = 1.0

From Figure 3-83,= 5.0 (based on A and A)

Kax

= 4.95 a E T2 (D2 A )-112 = 4.95 x 5.0 x 29E6 x 0.25 2 x (47.75 2 x 97.2)-112 = 95,290 lb/in

From Figure 3-84, the curves are off of the scale for the longitudinal moment, so KL can be assumed to be rigid.

3-90


From Figure 3-84, for A = 100, T/t = 1.0, and Mc!(ET3 e)

= À 5.21,

=

3.7.

=

Mc!(ET3 e ) x ET3 x 2pi/360

=

3.7 x 29E6 x 0.253 x 2pi/360

= 29,262 in-lb/deg

CAESAR II provides a feature, accessed from the Kaux menu of the input spreadsheet, whereby the user may define nozzle and vessel parameters for a WRC 297 flexibility calculation. If a nozzle has been defined in the input, during error checking CAESAR II automatically performs the WRC 297 calculations, including lookup of the stiffness parameters from the digitized curves. A zero-length element with the calculated stiffnesses is then automatically inserted into the piping model at the nozzle location between the pipe and the vessel. The CAESAR II input and stiffness calculations (produced during input error checking) for the Figure 3-85 example are shown in Figures 3-86 and 3-87 respectively.

WRC 297

N022LE/VESSEL

FLEXIBILIT~

SPREADSHEET

Nozzle Node NUMber ............................. . Vessel Node NUMber (Optional) .................. . Nozzle Nozzle Vessel Vessel Vessel

Outside DiaMeter ................... Wall Thickness ...•.....•........... Outside DiaMeter ................... Wall Thickness ....................• Reinforcing Pad Thickness .........

(in.) (in.) (in.) (in.) (in.)

Dist. to stiffeners or head ............... (in.) Dist. to opposite side stiffeners or head . (in.) Vessel centerline direction vector X ....... . Vessel centerline direction vector ~ ....... . Vessel centerline direction vector 2 ....... . Vessel TeMperature (Optional) ........... (Deg. F) Vessel Material # (Optional) ............ (1-17)

F1 - (or Esc) to Exit

Figure 3-86

3-91

5.000

18.000 .250 48.000 .250


WRC NOZZLE DATA FOR NODE TERMINOLOG~:

Dl - distance to stiffner or head DZ - distance to opposite side stiffner or head L - unsupported length of cylindrical shell, L = 8CD1)CDZ) / [ sqtCD1) + sqtCDZ) ] ** Z D - Mean diaMeter of vessel T - wall thickness of vessel d - outside diaMeter of nozzle t - wall thickness of nozzle & - capital "laMbda", & = L / sqtc DT ) j - "laMbda", j = Cd/D) * sqt( D/T> Dl L = D = d &

CAUTIONS:

URC 297

5

130.000 335.699 47.750 18.000 97.161

DZ T/t = T = t j

ZZ5.000 1.000 .Z50 .Z50 5.Z10

LaMbda Cj) Greater than Z.0 produces approx. rigid junctions for longitudinal bending. The following cOMbinations can produce rigid axial junctions: j>1.6 &=10 / j>Z.5 &=Z0 / j>4.0 &=50 / j>6.Z &=>100

N022LE CALCULATIONS

URC N022LE NODE

=

UESSEL DMean(in.)= N022LE O.D. (in.)=

5 UESSEL THK. (in.)= N022LE THK. (in.)=

47.75e 18.eee

.25e .25e

AXIAL TRANSLATIONAL STIFFNESS (Ib./in.)= 173528 LONGITUDINAL BENDING STIFFNESS Cin.Ib./deg)= leeeeeeeeeeee CIRCUMFERENTIAL BENDING STIFFNESS (in.Ib./deg)= 27353 ANGLE BETUEEN N022LE LENGTH (L) (in.) = CAPITAL LAMBDA =

& UESSEL CENTERLINES(deg)= 335.699 97.161

THICKNESS RATIO SMALL LAMBDA

ge.eeee = =

1.eee 5.210

TO CONTINUE
TO PRINT

Figure 3-87 3.4.3.2 Modeling Nozzles for Flexibility Calculations

N ozzle flexibilities are most often modeled by describing two separate entities - the piping system and the vessel, and then connecting them with the zero-Iength WRC 297 flexibility element. Therefore there will be two node points at the exact same location in the space the node representing the surface ofthe vessel, and the node representing the ènd ofthe pipe. This usually means using a rigid element to represent the radius of the vessel.

3-92


Various examples of how to model nozzle to vessel connections are shown in Figures 3-88, 3-89, and 3-90. Figure 3-88 represents the simplest model. In this case, there are no thermal movements imposed from the vessel, and the overall stiffness ofthevessel is insignificant when compared to the localized flexibility. Accordingly, the vessel proper is not modeled, but is represented by an anchor with the WRC 297 stiffnesses.

~

A Il

:'\. ~

1

/1-------1mr-------""""'\

~\l

W

AA

~V'v

1

6

Pipe element running /away from vessel

)

.,

610

Point located on vessel surface

WRC 297 Flexibilities

WRC 297 Flexibilities

Figure 3-88 In this model, the user defines a vessel flexibility at node point 5 (with no vessel node point defined). This causes CAESAR II to insert a flexibility element between node point 5 and an infinitely rigid point in space (effectively a flexible anchor). Element 5 to 10 is defined as a normal pipe element. Note: The user should not put any anchors, restraints or non zero displacements at any WRC 297 nozzle or vessel nodes. CAESAR II inserts all the necessary restraints and/or anchors into the model to define the nozzle/vessel connection (i.e., 3 translational restraints and 3 rotational restraints). A more complex model is shown in Figure 3-89. In this case the radius ofthe vessel is modeled as a weightless rigid element. This incorporates the effect of the radial expansion of the vessel on the piping system; also, other thermal expansion of the vessel may be entered as imposed displacements at node point 5. This model, as does the previous one, still assumes that the overall stiffness of the vessel is insignificant when compared to the localized flexibility.

Vertical movement of vessel at the nozzle's elevation ..

ct

•

20 1 Figure 3-89 3-93


In this model, element 5 to 10 is coded as a weightless rigid, with the length equal to the outer radius of the vessel, and the material and temperature that ofthe vessel. Thermal growth at the centerline of the vessel at the elevation of the nozzle is calculated and imposed at node point 5. A nozzle is defined at node point 15, with a vessel node of 10, so a WRC 297 flexibility element is automatically inserted between these two nodes. Element 15 to 20 is defined as a normal pipe element. An even more complex model is shown in Figure 3-90. In this case the entire vessel is modeled using pipe and weightless rigid elements, and anchored at its base. This automatically incorporates the effect of aIl thermal expansion of the vessel, as weIl as its overall stiffness, on the piping system.

20 25 & 30 at same point in space at vessel surface

1 15-===. •

~_

25 30

35

10

5

Figure 3-90

In this model, the vessel is built of three pipe elements going from node point 5 to 20. The diameter, thickness, material, and temperature of the elements are the same ofthose of the portion of the vessel which they represent (for example, element 5 to 10 may represent an ambient skirt). An anchor is coded at node point 5 to represent the base anchorage of the vessel. Element 15 to 25 is a weightless rigid, with the length equal to the outer radius of the vessel, and the material and temperature that of the vessel- this provides the radial growth of the vessel. A nozzle is defined at node point 30, with a vessel node of25, so a WRC 297 flexibility element is inserted here. Element 30 to 35 is entered as a normal pipe element. Flexibilities of configurations not explicitly covered by WRC 297 may be approximated using the bulletin - the results, although not completely accurate, will probably be more accurate than using rigid connections. For example, using WRC 297 for a pipe attachment to the head of a vessel may yield adequate results when choosing values ofD, LI, or L2 as shown in Figure 3-91. Because this is an extension ofthe WRC 297 application, the user may wish to perform a sensitivity study of the effects of selecting different values of these dimensions.

3-94


-f f-d

Tray Elevation

D (?)

Figure 3-91 In the event that the nozzle is not normal to the vessel surface, WRC 297 may still give For example, one me ans ofmodeling a hillside nozzle is shown in Figure 3-92. Element 30 to 35 should be modeled as a weightless rigid, with a length equal to the outer radius ofthe vessel, and the outer diameter, wall thickness, material, and temperature equal to those ofthe vessel. WRC 297 flexibilities are calculated and inserted between node points 35 and 40. Element 40 to 45 is a 1/4" long pipe element (with the same properties of the nozzle) extending along the same line as element 30 to 35. Element 45 to 50 is the pipe nozzle connection itself. Note: The idea of the 1/4" long nozzle element is to properly orient the local vessel flexibilities with respect to the axial, longitudinal and transverse vessel directions. reasonab~e results.

3-95


_---+-1'"

X

""'-45

~d,t

z

50

Figure 3-92

This same idea can be used to model a lateral branch connection, as shown in Figure 3-93. In this case, the WRC 297 flexibilities are inserted between node points 110 and 115, and element 115 to 120 is the 1/4" long element used to orient the local axes.

105

Figure 3-93

3-96


3.5 Restraint Modeling One ofthe most important parts of any finite element analysis, be it piping, structural, plate or volumetric, is the accurate description of the boundary conditions. This is also one of the areas of the model most ripe for potential errors, due to the large variety of restraint configurations available, and possible user myopia. It is extremely important that aIl degrees of freedom actually restrained by a pipe restraint be coded in as restrained in the pipe stress analysis model; it is equally important that only those degrees of freedom actually restrained be coded in as restrained. Furthermore, with today's modern programs, it is also often necessary to indicate the potential non-linear effects (friction, gaps closing, liftoff, change in stiffness, etc.) of a restraint as it goes through a state change as weIl. Finally, the analyst must also consider the effects on the system of explicit modeling of restraint stiffnesses as weIl.

3.5.1 Restraint Types Typically encountered supportlrestraint types (the names used here are approximate since terminology varies in the industry) are described on the following pages, along with a simplistically described means ofmodeling them. It should be noted that what appears as a single restraint may in actuality consist of a combination of more than one of the types described here.

3.5.1.1 Anchor An anchor is a restraint which fully restrains the pipe; i.e., in aIl degrees offreedom - three translations and three rotations. An anchor is anywhere these degrees of freedom are restrained - it may be explicitly built for restraint purpose, it may be a by-product of a connection to an equipment nozzle, or it may be due to a construction practice such as grouting a wall penetration through which the pipe travels. Examples of anchors are shown in Figure 3-94. Anchors may be defined in CAESAR II by entering a type of Anchor, coding in six individual restraints (X, Y, Z, RX., RY, and RZ), or using a WRC 297 nozzle stiffness (see Section 3.4.3 of these seminar notes).

'~ 'Atlached ta Pump F~ange

Figure 3-94

3-97


3.5.1.2 Restraint The term restraint, although technicaIly applicable to any of these boundary conditions of the piping system, coIloquiaIly is confined to those restraints which have a relatively rigid stiffness, and restrain the pipe in a direction other than sim ply in the downward direction (i.e., a hanger, support, etc.). Restraints may act in more than one degree of freedom (but less than six, which would be an anchor). Restraints may be classified yet further in terms oftheir direction - for example, axial restraint, or guide (a restraint in both directions of the horizontal plane when applied to a riser, or in the lateral direction when applied to a horizontal run), or in terms oftheir function (for example, limit stop, where a gap is closed as the pipe moves, before the restraint kicks in). Restraints may consist almost entirely of manufacturer's hardware (i.e., clamp and sway strut assemblies), or may be built up of structural steel. Restraints are coded in CAESAR II as various combinations of restrained degrees offreedom (X, Y, Z, RX, RY, and RZ). Restraints in a skewed direction can be defined with direction cosines (or direction vectors) - i.e., 0.707, 0, 0.707 is equivalent to 1,0,1. Friction can be activated by defining a Mu value (coefficient of sliding friction) for the restraint (Mu values for steel on steel typicaIly range on the order of 0.3). Friction is represented by applying a force equal to Mu times the restraint load on the pipe along the direction of movement. Examples ofrestraints are shown in Figures 3-95 through 3-97. Figure 3-95 shows a simple restraint, consisting of a sway strut. A sway strut, due to the pins at the brackets, aIlows the pipe to move lateraIly (to the strut). The clamp aIlows the pipe to rotate in any direction. Therefore, the strut restrains only in one degree offreedom (i.e., Z). The radius of gyration of the strut is sufficient to prevent buckling, so the strut restrains in both directions of the degree offreedom. Since the pipe does not rub against anything as it moves, friction is not modeled.

Figure 3-95 Figure 3-96 shows three restraints built up of structural steel. In (a), the restraint is only in a single degree offreedom (Y), but in two directions (both up and down). In (b), lateral restraint is added as weIl, giving a two degree offreedom restraint. In (c), lugs are welded to the pipe to provide axial restraint, creating a restraint in three degrees of freedom (not an anchor, since the pipe is free to rotate about three axes). In these cases (although the third case is moot), friction should be modeled, since the pipe would rub against the restraint ifit moves.

3-98


'==

() (al

( bl

------

(c)

il

~

(1

Se<;t/On A·A

Figure 3-96 Figure 3-97 shows a restraint in four degrees offreedom - X, Z, RX, and RZ, since the lugs prevent lateral translation of the pipe, and the double lugs resist any moments about the Xor Z-axes. This restraint should also be modeled using friction.

Plpe---+

Concrete Eltlatlng wall aleeve

Figure 3-97

3.5.1.3 Spring Hanger Spring hangers may be variable or constant spring hangers. Springs are engineered hardware, which permit movement of the pipe (by providing less than rigid restraint) and apply an upward force to the pipe. Variable springs are modeled as a combination of a vertical, two-way restraint with a stiffness equal to that of the spring constant, and an external force (equal to the spring pre-Ioad) acting in the +Y direction. Constant springs are modeled as only an external force (equal to the spring pre-Ioad) acting in the +Y direction, since the spring constant is effectively zero. Modeling of spring hangers are extensively discussed in Section 2.4 of these seminar notes, so they are not discussed further here. 3-99


3.5.1.4 Hanger The most common type of restraint is that which rigidly resists weight; in other words it only acts against the downward movement of the pipe. Because it is the most common type, it merits two names - hanger and support. A hanger is specifically a rigid restraint, resisting only downward pipe movement, with a point of attachment above the pipe, that is, using a rod hanger assembly of some type. Typical hangers are shown in Figure 3-98. These are coded in CAESAR II as +Y restraints (where - or + preceding the restraint direction indicates one-way restraint), indicating that the restraint acts in the +Y direction against the pipe. Since the pipe has no structure to rub against, friction is not modeled.

o

"

'0. 0

.

"

Figure 3-98

3.5.1.5 Support Supports are those restraints which rigidly resist only the downward movement of the pipe, with the point of attachment below the pipe, that is, either with the pipe resting on a structure, or with an intervening saddle or trunnion. Typical supports are shown in Figure 3-99. These are coded in CAESAR II as +Y restraints, indicating that the restraint acts in the +Y direction against the pipe. Under certain circumstances, it may be desirable to model the trunnion or saddle with a rigid construction element (in order to consider the offset ofthe friction load, for example). Since the pipe slides against the structure, friction should be modeled (unless a roller, as in Figure 3-99f, or a teflon slide pad, is used).

3-100


(a)

( b)

(c)

(d)

If)

(e)

Types of sliding supports: (a) insulation protection saddle; (b) trunnion; (c), (d), (e) various types of saddles; (f) roller support.

Figure 3-99

3.5.1.6 Snubber A snubber is a type of engineered hardware which, due to its internaI mechanisms, offers no resistance to slowIy applied (i.e., static, such as weight or thermal) Ioads, but Iocks up and acts as a restraint when loads are applied quickly (i.e., dynamic loads such as earthquake or fluid hammer). There are two types ofsnubbers - hydraulic and mechanical- which are shown in Figures 3-100 and 3-101 respectively. Snubbers are modeled in CAESAR II by defining the restraint as XSNB, YSNB, or ZSNB (or with direction cosines). Any restraint defined as such is made inactive duringthe weight and thermalload cases, but activated for uniform load, wind, or any dynamic loads. Snubbers are discussed further in Section 5 of these seminar notes.

3-101


- ReseNoir Pivot-

,. . . Snubber valve

- Clevispin Piston rod eye

Figure 3-100

Step-up gearing

Figure 3-101

3-102


3.5.1.7 Sway Brace A sway brace (shown in Figure 3-102) is a specific type of engineered hardware, which appears to be a variable spring attached to a sway strut. It is usually used to tune the natural frequency of a system away from a resonance, by altering the stiffness term in the equation of dynamic motion. The sway brace is modeled as a restraint in a single degree offreedom, with a stiffness equal to the spring rate ofthe sway brace's variable spring. Since the sway brace is normally attached with a clamp, no friction would be modeled.

•

•

1 1 Figure 3-102

3.5.2 Non-linear Effects A restraint is said to be non-linear ifits response is not constant throughout its entire load range. Examples of non-linear effects are friction, one-way restraint, gaps, large rotation restraints, and bi-linear stiffnesses. A system with non-linear effects cannot be solved explicitly by the structural algorithm used; CAESAR II solves them through iteration until an acceptable solution is found. Problems with a large number ofnon-linear restraints may take quite a while to converge, or may not converge at aIl. In this case, the user must judiciously review the system in order to determine which ofthese effects may be excluded, or which convergence tolerances may be changed. Various non-linear effects are discussed below.

3.5.2.1 Friction Friction may be present whenever the pipe can move, and in the process of moving, rub against a restraint. As noted earlier, the maximum magnitude of the force resisting sliding is equal to Mu times the restraint force, where Mu is the dynamic (i.e., sliding) coefficient of friction. The frictional force is applied along the direction of the pipe movement. This is a non-linear effect, and requires an iterative solution, since applying a friction load to the piping system alters the results, changing the load at the restraint and the direction of pipe movement. Therefore, friction will have to be calculated again, and re-applied to the system, again changing the results. This continues until the frictionalload is unchanged,

3-103


within a tolerance (which can be adjusted by the user). In the event that there are numerous frictional restraints, this process may take a while (or may even not converge). Friction restraints have two states: sliding and not sliding. The restraint is not sliding if the force tending to move it is less than F times Mu. The restraint is sliding ifthe force tending to move it exceeds F times Mu. Friction is activated at a restraint in CAESAR II by entering a non-zero Mu value.

3.5.2.2 One-Way Restraints One-way restraints are restraints which are active only along half of the range of a specified degree offreedom. The most common application ofthese are hangers and supports, which prevent the pipe from moving down, but permit it to move up. Another example of a one-way restraint is a limit stop bumper, where a pipe expands axially until it bumps up against a restraint. Another example may be an accidentaI restraint - such as where the pipe may have been installed touching a column or penetration on one side, but not the other. One-way restraints obviously have two states - active or not active. The restraint is active if it is acting on the pipe, and inactive if the pipe has moved away from it. Analysis of oneway restraints require the following type ofiterative solution: the analysis is done with the restraint active. If the restraint load is in the direction of the one-way restraint, the analysis is completed. If the load is in the opposite direction of the restraint, the restraint is removed from the model, and the analysis is redone. Then the pipe's movement at that location is checked - if it is away from the restraint, the analysis is complete. If it is towards the restraint, the restraint is re-inserted into the model and the process continues. Again, ifthere are more than one one-way restraints, it may be a lengthy process. One-way restraints are coded in CAESAR II by placing a sign (+ or -) in front of the degree offreedom being restrained. The sign convention is that direction in which the restraint acts on the piping system (i.e., a +Y support acts upward on the pipe, so the pipe cannot move down).

3.5.2.3 Gaps Gap restraints are those restraints which do not touch the pipe upon installation (and therefore are not active), but are close enough to the pipe that they may become active during the normal movement of the pipe. The gaps may have been installed intentionally (such as in a limit stop), unintentionally (such as when a structural steel support is not fit up closely to the pipe), or simply by chance (such as when a pipe runs through a wall penetration). Examples of gap restraints are shown in Figure 3-103.

3-104


y

z-l

x , 2.5"

, 0.275"

Figure 3-103 Gap restraints have two states - closed and active or open and inactive. Analysis ofthese restraints require an iterative solution similar to that of one-way restraints: the analysis is initiaUy done without the restraint. The pipe's movement at that location is checked - if it is insufficient (or in the wrong direction) to close the gap, the analysis is complete. Ifit is sufficient to close the gap, an imposed displacement is put on the pipe to close the gap, the restraint is reinserted into the model, and the analysis is redone. The results are then rechecked - ifthe load on the restraint is in the right direction (tending to further close the closed gap), the analysis is complete. Ifit is in the direction where it is trying to reopen the gap, the imposed displacement and the restraint are removed and the system is reanalyzed. This process is repeated until the status of aU gaps converge. Gap restraints may be entered with friction, which further complicates the non-linear solution (since the friction wiU act at the gap restraint once the gap at the closes and the restraint begins to work, but not otherwise). In the event that gaps are unequal on either side of the pipe (as in the penetration in Figure 3-103), the gap restraints may be entered as multiple one-way restraints, each with unique gaps.

3.5.2.4 Large Rotation Restraints: A large rotation restraint is a restraint which rotates away from its originalline of action due to a pipe movement in a direction orthogonal to the originalline of action. This may occur when the restraint is provided by a rod or a strut and the orthogonal displacement is large. Since the length of the rod must remain constant, the rotation from the originalline of action is:

e

=

sin- 1 (il / L)

3-105


Where:

e

=

angle of rotation from originalline of action

~

=

orthogonal displacement of pipe, in

L

=

length of rod, in

Since the length of the rod must remain constant, the pipe will be forced to travel through the arc defined by the rod rotating about its point of attachment to the structure. This is illustrated in Figure 3-104. Note that the effect of the swing is much more pronounced when using short rods.

o

Thermal

Elevation change forced on the pipe.

Figure 3-104

3-106


These types of restraints are solved by initially doing the analysis with the restraints acting along their as-installed line of action. Mer the analysis is complete, the movement of the pipe at that location is noted. Based upon the movement of the pipe orthogonal to the line of action of the restraint, a new position on the arc of the restraint is calculated. The component ofthat position representing forced displacement along the line of action of the pipe is then used as an imposed displacement for the next analysis. Mer this analysis, the position of the pipe is checked again, the new position on the arc is recalculated, and the process is repeated, until the line of action of the restraint (and the position of the pipe) is unchanged (within a tolerance, which can be set by the user) from one analysis to the next. Large rotation restraints are entered in CAESAR II through the use of the term XROD, YROD, or ZROD. The direction in which the rod runs from the pipe is designated by the sign (+ or -, where no sign is taken to be the same as +). This type of support can be used to model supports where load bearing is sensitive to the angle of the line of action.

3.5.2.5 Bi-linear Stiffnesses Restraints with bi-linear stiffnesses maintain a constant stiffness up until a specified loading is achieved, at which point the stiffness changes. These can be used to represent bottomedout springs (through the normal range of the spring the stiffness is the spring constant, but once the spring compresses completely, the support essentially becomes a rigid hanger); soil supports (clay normally has constant stiffness until its ultimate load bearing capacity is reached, at which point the clay liquefies and offers virtually no resistance); and plastic hinges (a model where pipe or restraint yields throughout the cross-section). An example of a pipe in a soil trench, along with a graphic representation of the restraint response is shown in Figure 3-105.

PIPE IN A TRENCH:

Lx Ultimate Lateral Load = 120,000 lb. Estimated Lateral Stiffness = 60,000 Ib./in.

Ultimate Vertical Load = 240,000 lb. Estimated Vertical Stiffness = 533,333 Ib./in. Vertical bilinear soil spring model:

Lateral bilinear soil spring model: Ultimate Load

Force

Force

-----4=======+ Y

-----Jf-----. X K1 Ultimate Load (120,000)

Ultimate Load (240,000)

Figure 3-105 3-107


The iterative technique used for these restraints is explained below. The analysis is first do ne using a linear restraint with the first stiffness (KI)' The restraint load (F) is then checked, ifit is less than the transition load (Fy), then the analysis is complete. Ifnot, sorne modification is done to the model. First, a CNODE is added at the end of the restraint. A displacement equal to Fy/KI and a force equal to Fy are imposed on the CNODE, with a corresponding -Fy placed on the restrained point. The stiffness of the restraint is then changed to the second input stiffness K2, and the system is reanalyzed. The restraint load is then re-examined, and the process is repeated until the status of aIl restraints converges. Bi-linear restraints are entered in CAESAR II through the use of the designation X2, Y2, Z2, RX2, RY2, or RZ2. The user is also required to enter the two stiffnesses, as weIl as the ultimate, or transition, load. This type ofrestraint is discussed further in Section 3.6.2 of these seminar notes.

3.5.3 Evaluation of Restraint Stiffness In Section 2.3.3 ofthese seminar notes the significant effect that restraint stiffnesses may have on the piping system was described. Assuming restraints are rigid (when they are far from rigid) can lead to grossly conservative designs, possibly resulting in the unnecessary modification of the system. However, assuming that restraints are rigid cannot always be counted on to pro duce conservative solutions in a complex piping system, either. It is therefore often necessary to refine pipe stress analyses through the use ofestimated restraint stiffness. Section 3.4.3 provided one means of doing this, using techniques for estimating an accurate stiffness for piping to vessel anchors. If the stiffness is not entered by the user, CAESAR II defaults restraints to rigid - i.e., to have a stiffness (translational or rotational) of 1.0E12 lb/in or in-Ib/deg. In many cases, this is an extreme overestimate. For example, the stiffness of the "typical" restraint shown in Figure 3-106, can be conservatively estimated to be the stiffness ofthe bending of the W4X13 wide flange and the extension of the 1-inch diameter rod under load.

1"CP/,

Rad

Figure 3-106

3-108


The stiffness of the support is calculated as the inverse of the displacement of the support under a one pound load at the pipe, or (for this particular case):

Where: k

=

stiffness of restraint, lb/in

LI

=

length of wide flange, in

El

=

modulus of elasticity ofwide flange material, psi

1

=

moment of inertia of wide flange, in4

L2

=

length of rod, in

A

=

area of rod, in2

E2

=

modulus of elasticity of rod material, psi

For this example, the stiffness is: k

=

1 / [36 3/(3 x 29E6 x 11.3) + 36/(29E6 x pi/4)]

=

20,392 lb/in

This stiffness is far from 1E12. However the impact ofmodeling a restraint such as this as rigid depends on the characteristics ofthe piping system, since the true goal is to simulate a reasonable estimate of the relative stiffness ofrestraints to pipe. It is best to be concerned about modeling stiffnesses at restraints which obviously do not appear to be "rigid" relative to the pipe. Examples of anchors that should probably be modeled with a less than rigid stiffness are shown in Figures 3-107 and 3-108.

3-109


+ - The vessel nozzle is probably a very poor anchor here.

The vessel nozzle is probably a very rigid anchor here.

1

k------?~~======::-

Figure 3-107

Ta have called this point an anchor, and left the structure out of the model would not have been very accu rate.

Figure 3-108 The basic question that should be posed when reviewing restraints for possible modeling of the stiffness is "What is the relative stiffness of the restraint compared to the pipe and the loads being applied?" Influential factors include:

3-110


1

Large, hot, thick walled pipe usually has high axial stiffness and large loads, implying that impacted restraints may be less than rigid in comparison.

2

Alternatively, for small bore pipe, a restraint ofvirtually any stiffness probably appears to be rigid in comparison - when analyzing a 2-inch diameter line, results may not change noticeably even if the restraint stiffnesses are entered as 10,000 lb/in or 1E12 lblin.

3

Restraints in structures or buildings located high above grade may be affected by the flexibility of the building frame.

4

It is often difficult to build structures which can "rigidly" restrain moments. This is due both to the low torsional resistance of open sections, and due to the fact that structural members cannot be loaded in the strong axis for aIl three rotational directions. Performing the analysis using more realistic, flexible moment restraints will most likely cause the restraint to rotate slightly, transferring the moment load elsewhere in the system, and potentially dissipating it.

5

Connections at the top oftall towers may be flexible due to the lateral flexibility of the structure.

6

Soil anchors are cumulatively stiff, but locally very weak. It is common to continue buried pipe models for several hundred feet after the pipe has become buried. It is only after this distance that a "cumulative" (virtual) anchor is considered to have been formed. Waterhammer loads at buried elbows can cause pipes to "kick out" of the ground. In this case the local soil strength is smaIl, and the localload is very high. (Where this is considered a problem, concrete blocks and supports are poured around elbows that are part oflong elbow-elbow pairs.)

7

It is unrealistic to selectively model in restraint stiffnesses. For example, given three identical restraints in a row, if the center one has its true stiffness modeled and the other two do not, the analysis will force load away from the restraint at the center and toward those on the ends (which will not be the case in reality). It must be remembered when trying for accurate relative stiffness that the stiffnesses must be accurate not only relative to the pipe, but also relative to other supports.

8

In general, it is safe to say that the results of most analyses will increase in accuracy as the modeled restraint stiffnesses increase in accuracy.

Restraint stiffness can be calculated manuaIly, as was done for the restraint shown in Figure 3-106 and simply entered into CAESAR II when coding in the restraint. An alternative is to construct a partial model of the restraint as part of the piping system, using either pipe elements or structural members. For example, the flexibility oftall vessels can be modeled with pipe elements down to the foundation. In the case of structural steel models, often including only a small part of the model is sufficient to improve the behavior of the restraint. Complex structural steel models may be incorporated in to the piping system using the structural steel modeler, accessible from CAESAR II's main menu.

3-111

eOADE Pipe Stress Analysis Seminar Notes

3.5.3.1 Use of the Structural Steel Modeler Elements of structural steel members may be incorporated into the model of the piping system through the use of the structural steel modeler. These elements may vary from a single member (such as a saddle, welded to the pipe), to a complete restraint structure, to a complete model of the building steel. The steel may be modeled in order to include the stiffness ofthe structure in the analysis, or inorderto calculate the loads on the steel for stress analysis ofthese members (which can be done through the AISe unity check, accessible from CAESAR II's main menu). In general, constructing a structural model is very similar to constructing a piping modelthe same information is required: geometric layout, element cross sections, material parameters, boundary conditions, and loading. Internally, the calculations made are identical for the structure as for the pipe. The only major differences (aside from the actual process ofproblem coding) are: 1

Almost aIl joints between piping elements are assumed to be fixed connections (i.e., aIl three forces and aIl three moments are transferred between adjacent piping elements). In steel structures, the connections may transfer only selected loads between adjacent elements, depending upon the actualjoint construction. For example, the clip anglejoint shown in Figure 3-109 is one ofthe most common connection types - it is assumed to transmit forces, but no moments. Moment connections require welding of the flanges of the wide flange, since moments are carried in the flanges and shear is carried in the web. Therefore the user must be careful to accurately model the internaI connections between members. (Note that the default connection provides full fixity.)

I~ 1

1

B-llI~

1

1

1

1 1

F"mIITI!§;3

1

!,~

~~-/~!

Figure 3-109

3-112

r


2

Structural members perform very similarly to pipe elements except that the structural elements are not symmetric about their member axis; in fact, structural elements are usually weak for loading in one direction and strong in another. This means that the local orientation ofthe element is very important. Element orientation is specified through the use of the ANGLE parameter, which specifies the angle which the element is rotated away from its "standard" orientation. In CAESAR II's structural modeler, "standard" orientation is defined as follows: a) for elements running in the horizontal plane (beams), the element's weak axis coincides with the global Y-axis b) for elements running in the vertical plane (columns), the element's strong axis coincides with the global Z-axis c) for elements running in a skewed plane (bracing), the projection of the element's weak axis on the vertical plane coincides with the global Y-axis

Correct orientation of elements can be easily checked by using the structural modeler's PLOT commando CAESAR II's structural modeler uses keyword input, which may be entered interactively or through a file. Commonly used keywords are listed below:

HELP DEL EDIM EDIT ERR FIX LIST LOAD MATID PLOT QQUIT QUIT SECID STAT UNIF

Gives help Repeats last command Deletes input Defines structural element Leaves interactive for full screen edit mode Error checks input Defines restraint conditions Lists input Defines concentrated loads Defines material properties Plots structure Exits model building mode, does no error checking, does not build analysis files Exits model building mode, does error checking, builds analysis files Defines cross sectional properties Gives current model status Defines uniform loads

Where more data is required for a command, the program prompts the user for it. The program includes over 900 standard steel shapes, the properties of which may be accessed byname. Three databases are available-the 1977 AISC, 1989 AISC, and the 1991 German (DIN) standards. The user may enter the specifie cross-sectional parameters for non-standard shapes.

3-113


A sample problem, showing co ding of a pipe rack, is shown in Figure 3-110. The accompanying input illustrates the use the most common keywords to define a model.

1015 SECID = 1, W8X1 0 ;Define column cross section as #1 SECID = 2, W6X9 ;Define beam cross section as #2 MATID = 1, YM=29E6 POIS=.3 G=11 E6 DENS = .283 DEFAULT SECID = 1 EDIM 1005 1010 DY=12-0 EDIM 1010 1015 DY=12-0 EDIM 1035 1030 DY=12-0 EDIM 1030 1025 DY=12-0 DEFAULT SECID = 2 EDIM 1015 1020 DX =5-0 EDIM 10201025 DX=5-0 EDIM 1010 1030 DX=10-0 FIX 1005 ALL FIX 1035 ALL

.-----~~-----.

1025

12'-0" ;Define ail columns 1010

1030

;Define ail beams

12'-0"

-

W8x10

L

L

1005

X

1-

10'-0"

1035

-1

Figure 3-110 Note that a no de point (1020) must be placed along a structural member whenever an intersection with the piping system is to occur. The structure is included in the piping problem by entering the file name through an option of the Kaux menu. The pipe is then attached to the structure by usingthe attachment no de point on thestructure as the restraint CNODE. Many structures (such as building frames or continuous racks) have a high degree of repeatability. The user can take advantage of this through the use of no de and element generation commands. For example, the large structure shown in Figure 3-111 can be created using generation commands. The process is to first define one corner node, then fill a single column line of nodes, then sweep the line of nodes out into an area pattern, and finally sweep the area pattern up into a volume pattern of nodes, each step of which takes a single commando Elements can then be generated in a similar manner.

3-114


188

@Fill VoluMe

5

4

~3 @Fill Line

2

_____ ,)

(!) First

Node

Figure 3-111

The entire structure can be entered using only 14 commands to define the geometry, crosssectional properties, material properties, and boundary conditions: SECID 1 W14Xl20 SECID 2 WI0Xl2

COLUMNS BEAMS

MATID 1 YM=30E6 POIS=0.3 G=11E6 DENS=0.283

DEFINE MAT'L PARAMETERS

NODE NGEN NGEN NGEN

FIRST CORNER NODE LINE OF NODES 1 THRU 5 LINE INTO AREA PATTERN AREA PATTERN INTO VOLUME

1 (1) LAST=5 NODEINC=l DX=10-0 1 TO 5 LAST=30 NODEINC=5 DZ=15-0 1 TO 30 LAST=180 NODEINC=30 DY=22-0

EGEN 1 TO 31 LAST=60 DEFAULT SECID=2 EGEN EGEN EGEN EGEN

31 TO (151) 31 TO (271)

GENINC=30 GENLAST=180

GENERATE ALL COLUMNS DEFAULT SECTION TO BEAMS

32 LAST=35 GENINC=30 GENLAST=155 : GENERATE ONE X-Y SIDE : GENERATE OTHER X-Y SIDES (170) LAST=180 GENINC=5 36 INC=5 LAST=56 GENINC=30 GENLAST=176 : ONE Z-Y SIDE REMAINING Z-Y SIDES (295) LAST=180 GENINC=l

FIX 1 TO 30 ALL

: FIX ALL BASE NODES

3-115


3.5.4 Use of CNODES When Modeling Restraints As discussed elsewhere in these seminar notes, CNODES (short for connecting nodes) are used with a restraint whenever a point in the piping system is retrained against another point in the system. The CNODE point may be physically on a vessel, a steel structure, or another piece of pipe; or it may be a fictitious node which can be used to model a wide range of system behaviors, such as settlement. For example, pipe on pipe supports can only be modeled using CNODES. One example is shown in Figure 3-112. In that figure, no de point 30 rests on a saddle, which fits with an arc (unwelded) to the pipe at node point 225. Because the saddle is welded to the branch line and partially wraps around the pipe at node point 225, node point 30 cannot move down or in the Z-direction, but can move up and can slide in the X-direction -relative to node point 225. Ifthis restraint is simply coded as a +Y and Z restraint, with Mu=0.3, at node point 30, this would indicate restraint against a rigid point in space. This is obviously incorrect; the model can be rectified by placing a CNODE of225 on both the +Y and Z restraints. This will permit node point 30 to move freely relative to rigid points in space, but provides restraint relative to node point 225. A further example is shown in Figure 3-113. Node point 65 is supported from the overhead pipe at node point 195. The rod shown can handle tension only. Therefore node point 65 can move freely relative to rigid points in space, but is constrained against moving down relative to node point 195. The restraint is therefore specified as a +Y restraint at node point 65, with a CNODE of 195. Note that it doesn't matter which node is the connecting node as long as the proper relationship between the nodes is maintained; i.e., the restraint could be specified equally correctly as a -y restraint at node point 195, with a CNODE of 65.

saddle free to slide mu = 0.3

Figure 3-112

3-116


Figure 3-113 CNODES can also be used to accurately model the effect ofsettlement. In Figure 3-114, the pipe rests on the ground at node point 95. The ground at this point is expected to settle in the -y direction; the pipe may or may not settle with it dependingon the stiffness ofthe nearby pipe. This configuration can be modeled with a +Y restraint at node point 95, with a CNODE of1095 (friction is optional). The settlement displacement of -0.325 inches in the Y-direction is then imposed at node point 1095. This allows node point 95 to either lift-off, or to settle, according to the configuration of the piping system as a whole .

..-J---...;.----=---t Figure 3-114

3-117

0.325'


3.6 Miseellaneous Topies This section contains discussions of special modeling situations, such as cold spring, plastic pipe, underground pipe, jacketed pipe, and flange leakage analysis.

3.6.1 Consideration of Cold Spring Cold spring is the process of offsetting (or pre-Ioading) the piping system with displacement loads (usually accompli shed by cutting short or long the pipe runs between two anchors) for the purpose ofreducing the absolute expansion load on the system. Cold spring is used to: 1

-

hasten the thermal shakedown of the system in fewer operating cycles

2

-

reduce the magnitude ofloads on equipment and restraints, since often, only a single application of a large load is sufficient to damage these elements

Note that no credit can be taken for cold spring in the stress calculations, since the expansion stress provisions of the piping codes require the evaluation of the stress range, which is unaffected by cold spring (except possibility in the presence of non-linear boundary conditions, as discussed below). The cold spring merely adjusts the stress mean, but not the range. Many engineers avoid cold spring due to the difficulty of maintaining accurate records throughout the operating life of the unit. Future analysts attempting to make field repairs or modifications may not necessarily know about (and therefore include in the analysis) the cold spring specification. Due to the difficulty of properly installing a cold sprung system, most piping codes recommend that only 2/3 of the specified cold spring be used for the equipment load calculations. An example of how to calculate the amount of cold spring necessary to reduce equipment loads is provided in Figure 3-115. In the example shown, the pipe expands between the anchor and the equipment, placing excessive thermalloads on the nozzle. The ide a is to calculate the total thermal expansion which the pipe wishes to make between the two pumps, and then to offset the pipe by approximately half ofthat amount through the use of cut short elements.

3-118


10'-0

1

1

1

1

~.I

~ 5'-6

EQUIPMENT ANCHOR

t

2'-6

3'-9

J

Figure 3-115

The cold spring amount is calculated as:

Where: Ci

=

length of cold spring in direction i (where i is X, Y, or Z), in

Li

totallength of pipe subject to expansion in direction i, in

a

= =

dT

=

change in temperature, oF

mean thermal expansion coefficient of material between ambient and operating temperature, in/in/oF

For the case in Figure 3-115, assume that the operating temperature is 1170oF, ambient temperature is 70oF, and the coefficient of me an coefficient of thermal expansion between the two for the material is 7E-6 in/in/oF. In that case: Lx

=

(13)(12) + 8 + (5)(12) + 6 + (3)(12) + 9 =

431 in

Ly

=

(5)(12) + (10)(12)+ (2)(12) + 6

=

210 in

3-119


Lz

=

(30)(12) + (23)(12) + 7

=

643 in

Cx

=

(112) (431) (7E-6) (1170 - 70)

=

1.659 in

Cy

=

(1/2) (210) (7E-6) (1170 - 70)

=

0.809 in

Cz

=

(112) (643) (7E-6) (1170 - 70)

=

2.475 in

Therefore, one of the pipe runs in the X-direction should be cut short by approximately 1-5/ 8 inches, one of the runs in the Y-direction should be cut short by approximately 3/4 inches, and one of the runs in the Z-direction should be cut short by approximately 2-112 inches, as shown in the figure. Note that the (1/2) in the equation for the cold spring amount is used such that the mean stress is zero. In some cases it is desirable to have the operating load on the equipment as close to zero as possible. In this latter case the (1/2) should be omitted. The maximum stress magnitude will not change from a system without cold spring, but will now exist in the cold case rather than the hot. AlI pipe stress programs provide very specific methods of modeling cold spring. As of this writing (Version 3.18), CAESAR II provides two methods of specifying cold spring. (This is scheduled to change with Version 3.20 of the program, when cold spring will be more easily manipulated as a separate loading case.) In the first method, elements may be specified as being made of cut short or cut long materials. Cut short de scribes a cold sprung section ofpipe fabricated short by the amount of the cold spring, requiring an initial tensile load to close the final joint. Cut long describes a cold sprung section ofpipe fabricated long by the amount of cold spring, requiring an initial compressive load to close the final joint. The software models cut shorts and cut longs by applying end forces to the elements sufficient to reduce their length to zero (from the defined length) or increase their length to the defined length (from zero) respectively. (It should be remembered to make the lengths ofthese cold spring elements only 2/3 oftheir actuallengths to implement the code recommendations.) This is effectively what occurs during application of cold spring. The end forces applied to the elements are then included in the basic loading case F (for force), whereby they can be included in various load combinations. The drawback to this method occurs when other forces are present, such as applied external forces or spring hanger loads. In this case, the cold spring forces cannot be segregated from these other forces in the basic load case F. Therefore, the second method of modeling cold spring is more appropriate - using a second (or third) thermal case to represent the effects of cold spring. In this way the effects of cold spring can be isolated from all other loadings through the specification of the extra thermal case. This is done as follows: 1

-

model the system as normal, but use at least one element with a length and direction corresponding to the specified cold spring (the same as in the first method, but make it of the same material as the pipe, not of a special cut short or long material)

2

-

apply the normal operating temperatures to all elements ofthe model as thermal load case Tl - this represents the expansion of the system during operation

3-120


3

create thermalload case T2 representing only the effects of the cold spring - for this case: a) all non-cold spring elements are given a temperature equal to ambient b) all cut short elements are given an alpha value (instead of a temperature) of -0.6667, representing a shrinkage of 2/3 of its defined length c) all cut long elements are given an alpha value of 0.6667, representing an expansion of 2/3 of its defined length

Note that in order to enter an alpha value on the order of 0.6667, the alpha tolerance value of the CAESAR II setup file will probably have to be changed. For more information on changing the alpha tolerance, and modeling cold spring in general, the user is referred to the CAESAR II User's Manual. When analyzing a system with cold spring, a different set of load cases should be run. Assuming that the cold spring load case is T2 (as described in method 2 above), the following load cases probably constitute a good recommendation:

Load Case 1- P+W+F+T2 (OPE) - This is effectively a "cold operating" case - i.e., it represents the piping system in the cold condition, but includes both primary (P+W+F) and secondary (T2) loads, so it cannot be used for stress purposes. The reactions from this load case should be used for checking the restraint and equipment loads. Load Case 2 - P+W+F (SUS) - This is a sustained case from the point ofview that only primary loads are considered, and should therefore be used for checking the system sustained stresses. However, ifthere are non-linear effects such as one-way restraints, gaps, etc. present in the system, the restraint configuration should be examined to verify that it is a true representation of the restraint status during hot or cold operation. Load Case 3 - P+W+F +Tl +T2 (OPE) - This is the hot operating case, representing the piping system after thermal expansion. It is not used for stress purposes, but again the reactions from this load case are used for checking the restraint and equipment loads (they should be checked for the maximum loads from the cold or hot operating case). Load Case 4 - D1- D3 (EXP). This is the algebraic difference between, or the range of loading through which the pipe goes when heating up between, the cold and the hot cases. Therefore this is the expansion case, and is used to check the expansion stress requirements ofthe codes. Note that for completely linear systems, the expansion range (i.e., the difference between load case 3 and load case 1) is Tl, eliminating the effect of the cold spring. Care must be exercised when running cold spring and hanger design simultaneously. Cold spring in vertical runs of pipe adjacent to hanger design locations can cause inordinate weight loads to appear at the hanger positions. Cold spring effects should be omitted from the restrained weight run and included in the hanger operating run.

3-121


3.6.2 Fiberglass Reinforced Plastic Pipe The characteristic ofwrapped fiberglass reinforced plastic (FRP) pipe which distinguishes it from other piping materials is that it is orthotropic - the material has different properties in the longitudinal and hoop directions - and can therefore not be modeled by the standard CAESAR II pipe element. This is because of the method by which this type of pipe is made. For wrapped FRP pipe, the wall thickness is built up by wrapping layers of glass and matrix at several pre-specified angles about the pipe axis pipe. Twenty to thirty layers, wrapped at different angles, are usually used to obtain the desired pressure carrying capacity and bending strength. Because the layers are at different angles, and because the glass/matrix sheets are only capable ofaxialload carrying capacity, the resulting pipe has different strength characteristics in the hoop and longitudinal directions. For example, ifthe pipe was only wrapped at 90 degrees to the pipe axis, the resulting pipe could only contain hoop pressure stresses, with essentially no resistance against longitudinal pressure or bending loads. Because there is no standard angle ofwrap, the pipe properties must be gotten from the manufacturer. Obtaining these properties is fairly straight forward, as there are typical relationships that are used to calculate global properties based upon the local characteristics of the glass and matrix, given the angle and the degree of wrap.

Morrix Loyers (Exaggercieci)

Res!"

MCTrix

Figure 3-116

Note that today a non-wrapped form ofFRP pipe is also available. In this type ofpipe, the glass, in very small pieces, is enclosed in the matrix, and then sprayed into a piping mold. This method of construction provides for essentially isotropie properties, as the glass fibers are oriented at random in the matrix. In this instance, the standard CAESAR II pipe element can be used. Other plastics, such as PVC, also exhibit isotropie properties, and can therefore be modeled by the standard element as weIl. TheCAESAR II plastic pipe element is based upon a model ofwrapped fiberglass reinforced plastic pipe. When this element is requested, the following addition al material properties are required: G

shear modulus of elasticity , which is not necessarily related to the tensile modulus of elasticity (as for isotropie materials), psi

Eaxial

axial elastic modulus (different from the hoop elastic modulus), psi

3-122


Eaxial/Ehoop

* Vb/a

-

ratio of the axial elastic modulus over the hoop elastic modulus times Poisson's ratio of a strain in the axial direction resulting from a stress in the hoop direction, dimensionless coefficient of thermal expansion, inlinloF

CAESAR II prompts for these values whenever material #20 (plastic pipe) is selected. Unless otherwise entered, the following default values are used: Shear Modulus

=

80,000 psi

Axial Elastic Modulus

=

3.2E6 psi

=

0.15273

Thermal Expansion Coefficient =

12.0E-6 inlinloF

When analyzing plastic pipe, one primarily looks for points where the pipe is undersupported both horizontally and verticaIly. Piping designers used to working with steel pipe tend to under-support plastic pipe because they often use rules ofthumb and a "design eye" for the much stronger steel pipe. As a result supports near vertical risers may be placed too far from the verticalleg, causing excessive bending, and leakage at weakjoints. Horizontal supports should be provided liberally because they are inexpensive and lightly loaded, and because they prevent the pipe from buckling or moving into a position that is potentially dangerous. In fact, any horizontally unsupported line can "walk" its way off of supports, into neighboring lines, etc., if the designer is not careful. These types of problems seem to be exacerbated when working with plastic pipe. PracticaIly, the pressure stresses in plastic pipe should be considered before any flexibility analysis is done (this is consistent with the way any other pipe stress analysis is done). These pressures determine the required thickness of the pipe, and the degree ofwrap. N ext the pipe stress analysis should be done. There are few explicit piping codes or allowable stresses available for plastic pipe; it is up to the user to determine the appropriate flexibility and stress intensification factors, load combinations, and allowable stresses. CAESAR II models flexibility factors for plastic pipe elbows as 1.0, since the hoop modulus is generally considerably higher than the axial modulus, thus resisting cross sectional ovalization. Intersections and curved fittings are generally assumed to be approximately three times as thick as the matching pipe. When this is done an SIF of2.3, a value recommended by CibaGeigy for plastic pipe systems, is typically used. If the user has better stress intensification factor data, those values may be specified at individual fittings. It is conservative, and a practical approach, to combine aIl simultaneous loadings together to determine the maximum stress in the pipe. This includes the effects ofweight, pressure, and thermal effects. (When pressure is specified in plastic pipe, CAESAR II always activates the Bourdon pressure effect. This accounts for the displacements due to pressure elongation of the pipe, which can be significant in plastic pipe.) Preferably, the analysis results should be reviewed with the plastic pipe manufacturer to verify that the model is accurate and that the pipe supplied is capable ofwithstanding the stresses, pipe forces and

3-123


moments, and restraint reaction loads. Otherwise, the resulting operating stresses (both code and bending stresses) can then be compared to the maximum allowable stresses as specified by the pipe manufacturer. In some cases, the manufacturer does not provide allowable stresses, but rather, maximum allowable pressures and maximum recommended weight spans. In this case, the user can convert this information to allowable stresses by proceeding through the steps outlined below: 1

-

The user should look up in the manufacturer's specifications the maximum pressure and allowable weight spanforplastic pipe (oftheappropriate diameter), filled with water.

2

-

N ext a plastic pipe model comprised of straight pipe elements resting on vertical supports should be built. The model should include at least six equally spaced supports, with a node point placed at the midpoint of the middle span. The distance between supports should be equal to the maximum allowed pipe span obtained from the manufacturer's data.

3

-

Making sure that the modeled pipe is filled with water and pressurized at the maximum allowable pressure (obtained from the manufacturer's data), a weight plus pressure analysis on the pipe should be run.

4

-

The large st code stress and the large st bending stress found on the three node points of the middle span should then be used as the limits to the operating bending and code stresses found in the analysis of the actual system.

The real benefit of analysis of plastic pipe is that it helps to eliminate poorly supported systems that will eventually leak, or will cause distortion problems with the line. Whereas hot steel pipe can be easily over-supported, plastic pipe typically cannot. The tendency is to under-support it, or to support it incorrectly, producing large thermal moments at intermediate elbows. Both of these design flaws should be discovered easily with a stress analysis of the system.

3.6.3 Underground Pipe Analyzing an underground piping system presents circumstances quite different from those encountered when analyzing plant piping. The major problem is the accurate modeling, using point restraints, of the continuous effects of soil-pipe interaction. The pipe-soil interaction, which resists piping movements and forces, are divided into two categories frictional forces, which must be overcome by pipe sliding against the soil, and pressure forces, which are caused by the pipe pushing against the soil. Axial friction force is calculated as the product ofthe soil-pipe friction coefficient and the total normal force acting around the pipe. An example of a pipe buried in a trench is shown in Figure 3-117. In this case, the normal force acting on the pipe surface can be simplified as a top force W (for cases where the soil coyer ranges from one to three times the pipe diameter, this is approximately equal to the weight ofthe soil above the pipe), and a bottom force, equal to the weight of the pipe plus the soil above it. Therefore the frictional force offering axial

3-124


restraint on the pipe is estimated as:

=

Mu (W + W + Wp)

f

=

axial friction force, lb/in

Mu

=

coefficient of friction between pipe and soil

p

=

Su / 600 (for clay)

Su

=

Undrained shear strength

=

density ofbackfiIl soil, Ib/in3

D

=

outside diameter of pipe, in

H

=

height of soil coyer to top of pipe, in

Wp

=

weight of pipe and contents, lb/in

f

= Mu (2p

D H + Wp)

Where:

w

a. Trenched pIpe

b. Soil pressure

c. Idealized model

Figure 3-117

The soil density and friction coefficient can be obtained from soil tests performed along the pipe route. Where data is not available, the foIlowing values are typical: Silt - 0.3

Sand - 0.4

Gravel - 0.5

Clay = 0.6 to 2.4

Pressure forces are caused when the pipe tries to move lateraIly, pushing against the soil. Three different lateral soil forces normaIly encountered in piping analysis are shown in Figure 3-118. Each lateral restraint response can be idealized, as shown in Figure 3-118d, into two response stages: elastic, where the resisting force is proportional to the pipe displacement, and plastic, where resistance remains constant regardless of displacement. This type of restraint can be modeled using a bi-linear restraint, specifying an elastic stiffness, an ultimate load (for the elastic to plastic transition), and a plastic stiffness (which should be near zero). The soil will not only restrain the pipe against movements, but through couples, against rotations as weIl.

3-125


Displacement _ b. Downward

a. Upward

c. Sideward

d. Force displacement

Figure 3-118

The stiffness of the soil restraint (in lb/in per inch of pipe) can be calculated as the product of the soil modulus of elasticity times the outer diameter of the pipe. In the event that the soil modulus is not known, the stiffness can be determined from other soil parameters. For example, for sidewards motion, a passive pressure is created at the front surface of the pipe. According to L. C. Peng's paper "Stress Analysis Methods for Underground Pipelines" (published in the May 1978 edition of Pipe Line Industry), the ultimate load and idealized elastic response can be estimated from various soil parameters, as: U

=

k

= 33.336 x p (H + D) tan2(45+

1/2 x p (H + D)2 tan 2(45+

Where: U

=

ultimate load, lb (per inch of pipe)

=

soil friction angle, degrees

k

=

soil elastic stiffness, lb/in (per inch of pipe)

= eD e

=

soil modulus of elasticity, psi

In order to sufficiently simulate the continuous effect of soil restraint, it is often necessary to break underground piping into a finer mesh than would be necessary for plant piping.

3.6.3.1 Modeling Soil Restraint The following basic outline can be used for modeling buried pipe: 1

At least 200-300 feet of the buried pipe should be included in order to model the effect of a cumulative anchor

2

N odes (and associated restraints) should not bespaced further than 20 diameters apart for pipe greater than 12 inch in diameter, or greater than 30 diameters apart for 12 inch and under diameters.

3-126


3

Contributory area should be calculated for each no de as:

Where: A

=

contributory area, in2

Li

=

length ofthe pipe "i", in

Di

=

outside diameter of pipe "i", in

n

=

number of pipes framing into the node

For example, the contributory area for node 20 of the 12" nominal diameter pipe shown in Figure 3-119 is calculated as: A

=

1/2 (10 x 12 x 12.75 + 30 x 12 x 12.75)

15

= 3060 in2

25

20

10

Figure 3-119 4

-

The subgrade modulus of elasticity for the type of soil is then determined, preferably from actual soil tests. In lieu ofbetter information, the following data (taken from Joseph E. Bowles, "Foundation Analysis and Design", 3rd Edition, 1982) is available:

Sail type

Subgrade modulus Ckips/ft 3 )

loose sand medium dense sand dense sand clay and dense sand silty and dense sand clay qu < 4 ksf clay qu < 800 ksf clay qu > 1600 ksf

30-100 60-500 400-800 200-500 250-700 75-150 150-300 >300

3-127


(where qu is the unconfined compressive strength of the soil) If otherwise not known, the subgrade modulus, e, can be calculated from k = e

=

eD = 33.336 x r (H + D)2 tan2(45+ (33.336 / D) x r (H + D) tan 2(45+

<1>

<1>

/2), or:

/2)

5

-

The effective soil restraint stiffness for each node is calculated by multiplyingthe contributory area for each node times the soil subgrade modulus.

6

-

Lateral restraints (and possibly rotational restraints representing restraint force couples) with the calculated stiffnesses should then be inserted into the piping model at the appropriate nodes.

7

-

Next the density of the pipe should be set to zero, since the weight ofburied pipe is uniformly supported along its length. Weight loads in buried pipe do not cause deflections, stresses, or forces in the pipe. (Note that this step should be skipped when doing dynamic analysis of underground pipe, since the mass distribution is important in the dynamic analysis.)

8

-

Any axial stops in the form oflarge flanges or concrete anchors, designed to resist the thermal expansion of buried pipe, should then be coded into the model. If these are present, it is recommended that the restraint stiffnesses calculated above be reduced by approximately 25% in order to yield more conservative anchor loads.

3.6.3.2 Automated Underground Piping Modeler

CAESAR II provides an automatic underground pipe modeler, accessed from the main menu, which markedly simplifies this modelingprocess. The underground modeler provides two services to the user: If soil properties are not known, or if a good mathematical model of the soil is not available, CAESAR II provides a default soil model that may be used to approximate "typical" soil support characteristics.

1

2

-

Given soil support stiffnesses either from user input or from the default model, CAESAR II distributes the buried restraint stiffnesses over the buried part of the piping system. This is probably the most useful part of the buried pipe modeler. Properly breaking down the model into a finer element mesh and distributing restraints over the piping system is a very time consuming task to do accurately by hand, which the buried pipe modeler can do in seconds. The distribution of restraint stiffnesses over lateral bearing lengths, transition lengths, and over axial bearing lengths is described in detail in the CAESAR II user's manual.

Sel dom are soil properties known very accurately. Often there is absolutely no quantitative data available on the soil at the site. In these situations, the default soil model will probably provide as good an estimate of the actual soil properties as any. This model is based on a

3-128


combination of data available from a variety of sources (many from research on driven piles). After evaluating this data, it was determined that there was insufficient accuracy to differentiate between the horizontal and vertical soil restraint stiffnesses. Intuitively, it seems plausible that the downward stiffness of a buried pipe will be greater than both the lateral and upward stiffnesses, and the lateral stiffnesses will be greater than the upward stiffness, which will vary according to the buried depth. When there is good soil data and a good soil model available, it should be used in place ofCAESAR II's default soil model. These numbers can be input directlyto the buried pipe modeler. This improved model can take into account the differences among lateral, upward, and downward distributed soil stiffnesses. The underground piping modeler provides the user the opportunity to enter the following soil properties:

FRICTION COEFFICIENT SOI L DENSITY BURIED DEPTH TO TOP OF PIPE FRICTION ANGLE UNDRAINED SHEAR STRENGTH OVERBURDEN COMPACTION MULTIPLIER THERMAL EXPANSION COEFFICIENT TEMPERATURE CHANGE (Installed to operating) Note that the overburden compaction multiplier is a degree of compaction of the backfill, and is calculated by multiplying the Proctor Number (a measure ofbackfill efficiency defined in most soils textbooks as a ratio ofunit weights) by 8. The underground pipe modeler defaults to a value of the overburden compaction multiplier of8; this results in somewhat conservative restraint stiffnesses. Under common practice, this value is often reduced to somewhere in the range from 5 to 7. From this data, the underground pipe modeler calculates the axial friction loads, the transverse ultimate load and elastic stiffness, and automatically inserts the appropriate restraints, adding additional nodes as necessary. A diagram of a pipe for which soil restraints were generated by the underground piping modeler is shown in Figure 3-120.

3-129


Figure 3-120

3.6.4 Jacketed Pipe Jacketed piping can be idealized as two piping systems running along coincident Iines in space. The two piping systems are oftwo different sizes, so the smaller runs inside of the Iarger. The internaI pipe contains the piped fluid, while the outer pipe (the jacket) is used for protection or to carry a heated fluid to warm that in the inner pipe. An example ofjacketed pipe is shown in Figure 3-121.

3-130


t

Internai Support Between Core

/

Welded swage used as / e n d c ap .

L

Shrink fit or welded end cap.

l

RIGID RaD SUPPORT

~

and Jacket. (Spider)

X

Do (CORE) Di (JACKET)

' \ BLaCK FITTING

Ep •

(5) - Core pipe Il ange face.

éé

~@-I-e l''::'L.----O ~ ~

(10) - End Cap connection between core and jacket.

~

(15) - Point of flexible internai support between core and jacket. (20) - Node for the action of the rigid rod and for the monitoring of the displacements between the core and the jacket. This displacement should never be greater than (Di(Jacket)-Do(Core) ) / 2.

Figure 3-121

Jacketed piping systems are modeled by running the jacket elements directly on top of the core elements where the two are concentric. InternaI supports (spiders) offer negligible resistance to relative bending and axial displacement, so rigid restraints should be placed between the inner and outer pipe (for example between node points 15 and 1015) only in the local Y-and Z-directions. The end caps connecting the core to the jacket ofthe pipe are usually much stiffer than either the core or the jacket. For this reason at end cap locations (node

3-131


points 10 and 25), the inner and outer pipes should share the same node point (i.e., node points 1010 and 1025 should not be used) - this ensures that the rotations and displacements are identical for the two pipes at these locations. . The +y support acting on the jacket at node point 1020 does not cause any restraint to be inserted between nodes 20 and 1020. Node 20 is included in the model so that interference with the outside diameter can be checked at the 20-1020 cross section. Should there be concerns about interference, a restraint with a gap equal to the clearance between the inner and outer pipes can be entered. If a load develops at that restraint, this indicates an interference. The specific modeling process is fairly simple. The inner pipe is modeled first, with the user taking care to place a node point at each location where there is an internaI spacer support. Next the inner pipe is duplicated using the element block copy feature (accessed with the List hot key from the input spreadsheet). The entire run of the inner pipe should be copied, with a suitable node increment to ensure that no nodes are duplicated between the two copies. The second copy becomes the outer pipe. It is necessary to first change the diameter and wall thickness (and possibly the fluid density and temperature) of the pipe on the first screen of the outer pipe; these changes propagate through. N ext the user must go through and change the bend radii of each of the elbows in the outer pipe. The first and last node numbers of the outer pipe should then be changed to the same node numbers as those of the first and last point of the inner point - this serves to connect the inner to the outer pipe, a fact that can be confirmed by using CAESAR II's PLOT option. Finally, the internaI supports are modeled by placing guides (and vertical supports on horizontal runs) at each of the support points on the inner pipe, with CNODES to the corresponding points on the outer pipe. Any pipe restraints are then placed on the outer pipe only.

3.6.5 Flange Leakage Analysis AB noted in Section 2.1.3 ofthese seminar notes, pressure design offlanges is a complex task - flange design becomes even more difficult when piping loads are considered.

3.6.5.1 Equivalent Pressure Calculation Traditionally, analysis of flanges under piping loads has been done by converting piping loads to an "equivalent pressure", which can then be added to the actual system design pressure, the sum ofwhich is then compared to the ANSI Standard B16.5 "Pipe Flanges and Flanged Fittings" allowable pressures at temperature (see Section 2.1.3). Piping loads are converted to equivalent pressure by applyingthe pipe forces and moments over an equivalent gasket area and an equivalent gasket section modulus respectively, as follows: Ptotal =

P + Pe

Pe

4 x F / (pi G2) + 16 x M / (pi G3)

=

3-132


Where: Ptotal

=

"total pressure", to be compared to ANSI B16.5 allowables, psi

P

=

system design pressure, psi

Pe

=

equivalent pressure due to pipe loads, psi

F

=

axial piping force on flange, lb

G

=

diameter at gasket load reaction, in

M

=

resultant piping bending moment on flange, in-lb

This method is widely recognized to give highly conservative results, indicating failure in flanges which actually have quite a bit ofreserve strength. However, it has long been used due to the lack of any easy method for evaluating flanges under piping loads. One alternative is to perform a flange analysis according to the requirements of Section VIII, Division I of the ASME Boiler and Pressure Vessel Code. However, these calculations are only intended to assure that the flange will not be overstresssed by the necessary joint tightening loads that are required for a leak tight joint. The problem with these stress calculation methods is that they hold piping loads, which usually cause leakage failure in flanges, to standards for pressure and tightening loads, which usually have a different failure mode - stress failure. Stress calculation methods fail to address the deformation of the flange and its effect on gasket compression and leakage. The basic problem offlange leakage is a complex one not readily availing itselfto analysis. Facing selection, gasket type, operating temperature, and initial fitup loads are aIl factors that are either difficult or impractical to evaluate analytically.

3.6.5.1 Flange Leakage Analysis Module CAESAR II provides a tool which simplifies this analysis, a flange analysis module accessible from the WRC 297, SIFS, FLANGES submenu - which allows the user to evaluate leakage offlanges under load - a more realistic analysis process than performing an equivalent pressure analysis. The flange analysis module also automates the ASME B&PVC Section VIII, Division I flange stress calculations. The CAESAR II flange leakage model assumes that the user has aIready selected the gasket, has a flange design, and has analyzed the piping flexibility to compute the forces and moments exerted by the piping on the flange (possibly including the effects due to fitup tolerance, which can be done using restraints with CNODES and imposed forces or displacements). The ASME codes eliminate sorne of the decisions involving leakage by the publication of the gasket "m" factor. The "m" factor is the leak pressure ratio - the ratio of the pressure on the gasket required to prevent leakage, to the line pressure, times a factor of safety. These values are currently the subject of close scrutiny by many organizations, but the existing values have been use with a reasonably successful design history. It is with the "m" factor

3-133


that the CAESAR II flange leakage calculation starts and upon which it depends. It is recommended that the user aim for an "m" factor greater than 1.0. This should provide a safety factor greater than 2.0, and is consistent with other safety factors used in pipe stress analysis. If the flange analysis module predicts an "m" factor less than 1.0, then the loads on the flange should probably be reduced. The flange modeler determines the initial pressure on the gasket due to the tightening of the bolts, and the loss of pressure on the gasket due to the line pressure and the forces and moments acting on the flange. If the resulting pressure on the gasket (i.e., the initial minus alllosses) exceeds the gasket factor "m" times the line pressure, then the flange is considered to be "safe". There are a great many different types offlanges, facings, an gaskets. For the purpose ofthe flange modeler, an of these were generalized into a single model for leakage. Once this was done, the critical variables affecting leakage were retained in the analytical model, and the unnecessary variables were eliminated. It was determined that the deformation of the annular plate forming the flange, in conjunction with the deformation ofthe bolts and gasket, when subjected to bending, pressure, and axial forces were the critical variables to be evaluated. Various simplified elastic models were tested and a final model agreed upon that most closely correlated the results from finite element analyses of several typical flange configurations subject to bending and axial loads. Loads on the gasket were predicted within 15% for standardly dimensioned flanges, and other calculated values were within similar tolerances. The modeler also confirmed leakage of numerous flanges in actual plant applications as weIl. The basic flange deformation modes assumed to contribute most significantly to the unloading of the gasket are shown in Figure 3-122.

3-134


"

/lloment DiltribuUon

~

/

Gaaket

BoIt. Load

a:::::D

.---------rv:-r-2"7"""":lv,~1-r-"7"""":l7~1"'""'7i ;; Guket SWfne•• and ReacUon

PA

~

BoIt SUffness .... !if}

BoIt. Load

ADgular

RotaUon

])ue

To Moment

\

AD Remain ellsenUally

vertical for bendiDg and Dial type loadll

\

\ \

BoIt Force

Figure 3-122

3-135


The limitations of the model are that: 1

-

The gasket reaction and stiffness are concentrated at a point load at the center of the gasket loading area.

2

The boIt reaction and stiffness are concentrated at a single point and is assumed to be uniformly distributed around the annular plate which models the flange.

3

The pipe/hub interface is assumed to be flexible enough to allow rotation at the flange inner diameter at the point around the circumference where the bending moments produce a maximum stress in the pipe, so that the absolute rotation at the flange inner diameter is zero.

4

-

The gasket is assumed to be fairly stiff, so that the flange rotational stiffness is of the same order of magnitude as the gasket stiffness.

These analyticallimitations imply other more practical "usage" limits: 1

-

Full face gaskets cannot be modeled.

2

-

Leakage at self-energizing gaskets cannot be predicted.

3

-

Leakage for flanges with ring-type joints cannot be predicted.

4

-

Shear load effects on leakage are ignored.

5

-

The effect of the hub and pipe wall are not variable, and so are considered only approximately.

6

-

Leakage analysis for joints made up of flexible gaskets should not be attempted, since the effect ofvery flexible gaskets on leakage tends to be a function offactors other than the flexibility of the annular flange plate and boIts.

Complete instructions for the operation of the flange leakage analysis module are provided in the CAESAR II U ser's Manual. A brief description if given here. The data screen for the module is broken into two sections. The first section contains the input required for the leakage calculations, while the second section contains the additional input required for making the ASME Section VIII, Division 1 stress calculations. The flange leakage module permits (much of the data is optional) entry of the following information: Flange dimensions:

FLANGE INSIDE DIAMETER FLANGE THICKNESS Dimensional data for standard flanges (along with their bolts and gaskets) can be accessed from a built in database in the program.

3-136


BoIt data:

BOlT CIRClE DIAMETER NUMBER OF BOlTS BOlT DIAMETER BOlT INITIAL TIGHTENING STRESS The boIt initial tightening stress can be estimated as:

= 12 x T / (K d)

S Where: S

=

preload stress, psi

T

=

boIt torque, ft-lb

K

=

nut factor (as per the Standard Handbook of Machine Design, Figure 3-123)

d

=

nominal diameter of boIt, in

23.30

ST ANDARD HANDBOOK OF MACHINE DESIGN

TABLE 23·5 Nut Factors

NUl factor

Source

Lubricant or coating on the fastener 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 1. 12. 13. 14. 15.

1 5 1 1 2 2 2 2 3 3 3 3 4 4

Cadmium plate Zinc plate Black oxide Baked on PTFE Molydisulfide paste Machine oil Camaba wax (5% emulsion) 60 Spindle oil As rcccived steel fasteners . Molydisulfide grease Phosphate and oil Copper·based anti seize compound As received steel fasteners Plated fasteners Grease. oil. or wax

4

Reponed mean

Reponed range·

0.194-0.246 0.332 0.163-0.194 0.092-0.112 0.155 0.21 0.148 0.22 0.20 0.137 0.19 0.132 0.20 0.15 0.12

0.153~.328 0.262~.398 0.I09~.279

0.064-0.142 0:14-0.17 0.20-0.225 0.12-0.165 0.21-0.23 0.158-0.267 0.10-0.16 0.15-0.23 0.08-0.23 0.161-0.267

Figure 3-123 If not entered, the program defauIts to a boIt prestress of 45000/d 1l2 psi, a typical rule of thumb applied when field tightening bolts.

3-137


Gasket data:

EFFECTIVE GASKET DIAMETER UNCOMPRESSED GASKET DIAMETER UNCOMPRESSED GASKET THICKNESS EFFECTIVE GASKET WIDTH EFFECTIVE GASKET MODULUS LEAK PRESSURE RATIO Typical values for the effective gasket modulus are between 300,000 and 400,000 psi for spiral wound gaskets. The greater the modulus, the greater the tendency for the flange to leak. Therefore errors on the high side will tend to be more conservative. The leak pressure ratio is the "m" factor discussed above; the required value for each type ofgasket is given in Table 2-5.1 of the ASME Section VIII, Division 1 code. It is also accessible with the program HELP facility.

Load data:

EXTERNALLY APPLIED MOMENT EXTERNALLY APPLIED FORCE PRESSURE For the optional ASME Section VIII, Division 1 stress calculations, the following additional data is requested:

Flange dimensions:

FLANGE TYPE FLANGE OUTSIDE DIAMETER SMALL END HUB THICKNESS LARGE END HUB THICKNESS HUB LENGTH As above, dimensional data for standard flanges can be accessed from a built in database in the program. Any of eight standard flange types recognized by Section VIII can be selected from a graphie representation.

Operating and loading data:

DESIGN TEMPERATURE GASKET SEATING STRESS Permissible gasket seating stresses are provided in the program HELP facility.

3-138


Material allowables:

FLANGE STRESS ALLOWABLES BOLT STRESS ALLOWABLES STRESS ALLOWABLE MULTIPLIERS Allowable stresses for both flanges and bolts may be accessed from the program material database. Certain codes permit increases in the stress allowables in certain circumstances. The program HELP facility identifies those codes and the corresponding multipliers which maybe used. The program provides output in terms of Safety Factors - values of less than one usually indicate trouble. Sample input and output screens are shown in Figures 3-124 and 3-125 respectively.

3-139


FLANGE LEAKAGE/STRESS CALCULATIONS Flange Inside Dia~eter ........................... (in.) Flange Thickness ................................. (in.) Flange Rating COptionaU ............................. .

39.569 4.969

Boit Circle Dia~eter ............................. (in.) Itu~ber of Bolts ...................................... . Boit Dia~eter .................................... (in.) Boit Initial Tightening Stress ............ (lb./sq.in.)

38.599 32.999 1.599

Effectiue Gasket Dia~eter (G) .................... (in.) Unco~pressed Gasket Thickness .................... (in.) Effectiue Gasket Width (b) ....................... (in.) Leak Pressure Ratio (~) .............................. . Effectiue Gasket Modulus .................. (lb./sq.in.)

33.888 .963 .396 2.759 399999.999

Externally Applied Mo~ent ..... (optional) .... (in.lb.) Externally Applied Force ...... (optional) ....... (lb.) Pressure .................................. (l b ./sq . in. )

.999 .999 499.999

The following inputs are required only if the user wishes to perfor~ stress cales as per Sect VIII Diu. 1 To Exit

For Help

Inputt

Outpllt!

More J.

FLANGE LEAKAGE/STRESS CALCULATIONS Flange Inside Dia~eter ........................... (in.) Flange Thickness ................................. (in.) Flange Rating COptionaU ............................. .

3a.56a 4.969

Boit Circle Dia~eter ............................. (in.) Itu~ber of Bo 1ts ...................................... . Boit Dia~eter .................................... (in. ) Boit Initial Tightening Stress ............ (lb./sq.in.)

38.599 32.999 1.599

Effectiue Gasket Dia~eter (G) .................... (in.) Unco~pressed Gasket Thickness .................... (in.) Effectiue Gasket Width (b) ....................... (in.) Leak Pressure Ratio (~) .............................. . Effectiue Gasket Modulus .................. (lb./sq.in.)

33.888 .963 .396 2.759 399999.a99

Externally Applied Mo~ent ..... (optional) .... (in.lb.) Externally Applied Force ...... (optional) ....... (lb.) Pressure .................................. ( l b ./sq . in. )

.999 .999 499.999

The following inputs are required only if the user wishes to perfor~ stress cales as per Sect VIII Diu. 1 To Exit

For Help

Inputt

Figure 3-124

3-140

Output!

More J.


FLANGE

LEA~AGE/STRESS

More t

CALCULATIONS

ASnE Model Seating (Stress).....................

2.74

ASME SECT VIII DIV 1 STRESS MODEL -----------------------CALCULATED STRESSES (lb./sq.in.) Longitudinal Hub .. Radial Flange ..... Tangential Flange . MaxilllulII Auerage .. . Bolting .......... .

OPERATIHG 3285.

ALLOW 26259. 17599. 17599. 17599. 25999.

3685.

971. 3485. 9619.

SEATIHG 5685. 6379. 1688. 6932. 2683.

ALLOW 26259. 17599. 17599. 17599. 25999.

"*" Indicates Failure for an itelll.

STRESS SAFETY FACTOR: (If less than one then joint failure is predicted.) (Allowed/Actual) OPERATIHG SEATIHG 7.99 4.62 Longitudinal Hub ... . Radial Flange ...... . 4.75 2.74 To Exit

FLANGE

For Help

LEA~AGE/STRESS

Inputl

<~eypad>

More l

Output!

More t

CALCIJLATIONS

CALCULATED STRESSES (lb./sq.in.) Longitudinal Hub .. Radial Flange ..... Tangential Flange . MaxilllulII Auerage .. . Bolting .......... .

OPERATIHG 3285. 3685. 971. 3485. 9619.

ALLOW 26259. 17599. 17599. 17599. 25899.

SEATIHG ALLOW 5685. 26259. 6379. 17599. 1680. 17599. 6832. 17589. 25999. 2683.

"*" Indicates Failure for an itelll.

STRESS SAFETY FACTOR: (If less than one then joint failure is predicted.) (Allowed/Actual) OPERATIHG SEATIHG Longitudinal Hub ... . 7.99 4.62 Radial Flange ...... . 4.75 2.74 Tangential Flange .. . 18.92 19.41 MaxilllulII Auerage .... . 5.92 2.99 2.69 9.32 Bolting ............ . To Exit

1 2 3 4

For Help

STRESS

Longitudinal Hub Radial Flange Tangentiel Flange "axinun Average

5 Bolting

Inputt

<~eypad>

OPEHATINC 3285. 3685. 971. 3485. 9619.

RLLOURBLE 26258. 17588.

17588. 17588.

25888.

Output! SEATING ALLOUABLE 5685. 6379. 16BB. 6832. 2683.

5 .•..........._..._ ........

3

Note
show

the index ~he

valüë·s···c;n···i:ïie···t'ïangë·· .... ·.. ··· .............................................. .

general location

Dr

the stress.

Figure 3-125 3-141

26258. 17588.

Ib./sq.in

2588B.

Ib./sq.in

Ib./sq.in 1758B. Ib./sq.in 17588. Ib./sq.in

COADE Pipe Stress Analysis Seminar Notes Section 4 Table of Contents 4.0 Introduction to Dynamic Theory ............................................................................... 1 4.1

Characteristics ofDynamic Loads ............................................................................ 2 4.1.1 Variation ofLoad Versus Time - Load Types ............................................. 2 4.1.2 System Response Time Versus Timing ofLoad Change .............................. 9 4.1.3 Lack of System Equilibrium ........................................................................ 14 4.1.4 Movement of Dynamically Loaded System ................................................. 14 4.1.5 Relation of Induced System Loads to Applied Loads .................................. 14

4.2

The Dynamic Equation of Motion ........................................................................... 17 4.2.1 Evaluation of a Single Degree-of-Freedom System .................................... 17 4.2.2 The Dynamic Load Factor ........................................................................... 18

4.3

Evaluation ofMulti-Degree-of-Freedom Systems .................................................. 22 4.3.1

Modal Analysis ............................................................................................. 22

4.3.2 Modal Response Multipliers - Participation Factors and DLFs ............... 25 4.4

Eigensolver Algorithm ............................................................................................. 28

4.5

Accuracy of The Dynamic Model ............................................................................. 36 4.5.1

Mass Point Spacing ...................................................................................... 36

4.6 Types of Analysis ..................................................................................................... 45 4.6.1 Time History Analysis ................................................................................. 45 4.6.2 Seismic Spectrum Analysis .......................................................................... 45 4.6.2.1 Generation of the Response Spectrum .......................................... 46 4.6.2.2 Application to a Multi-Degree-ofFreedom System ...................... 57 4.6.2.3 Modal Results Summation Methods ............................................. 62 4.6.2.4 Combination of Spatial Components ............................................ 64 4.6.2.5 Missing Mass Correction ............................................................... 71 4.6.3 Force Spectrum Analysis (for Impulse Loadings) ....................................... 71 4.6.3.1 Generation of the Response Spectrum .......................................... 72 4.6.3.2 Application to a Multi-Degree-ofFreedom System ...................... 73 4.6.3.3 Summation of Responses .............................................................. 74 4.6.4 Harmonie Analysis ....................................................................................... 74


4.0 Introduction to Dynamic Theory Most of the loads discussed to this point in these seminar notes have been static loads (or in sorne cases, dynamic loads modeled as static loads). Static loads are those which are applied slowly enough that the piping system has time to react and internally distribute the loads, thusremaininginequilibrium. Onceinequilibrium,allforcesandmomentsareresolved(i.e. the sum of the forces and moments are zero), so the pipe doesn't move. With a dynamic load - a load which changes quickly with time - the piping system may not have time to internally distribute the loads, so forces and moments are not always resolved - resulting in unbalanced loads, and therefore pipe movement. Since the sum of the forces and moments are not necessarily equal to zero, the internally induced loads can he different - either higher or lower - than the applied loads.

4-1


4.1 Characteristics of Dynamic Loads Summarizing the major characteristics of static versus dynamic loads: Static Loads 1

2

-

3

Dynamic Loads

load varies slowly, or does not vary, with time (weight, thermal expansion, settlement, spring loads, etc.)

1 -

load varies quickly with time (earthquake, fluid hammer, vibration, relief valve, etc.)

system (internal forces and restraint loads) always has time to fully react to the applied load

2

-

system (internal forces and restraint loads) may not have time to fully react to the applied load before it changes

system is always in equilibrium (sum of forces and moments on system are zero)

3

-

system is not in equilibrium (sum of forces and moments on system are not zero)

4

-

with no unbalanced forces, system remains at rest

4

-

with unbalanced forces, system moves, according to F = MA

5

-

induced system reactions (internaI forces and restraint loads) are equal to applied loads

5

-

induced system reactions (internaI forces and restraint loads) are not equal to applied loads, and may be much higher or muchlower

The implications ofthese points are discussed in the following sections. 4.1.1 Variation of Load Versus Time -

Load Types

The force vs. time profiles of the dynamic loads most often encountered during the design of piping are usually one ofthree types - random, harmonic, and impulse. These profiles, and the load types having them, are described below. Random: With this type ofprofile, the loadchanges direction and/or magnitude unpredictably with time, although there may be predominant characteristics within the load profile. Major types ofloads with random time profiles are:

1

Wind: Wind velocity causes forces as described in Section 2.5.1 ofthese seminar notes, that is by the decrease of wind momentum as the air strikes the pipe, creating an "equivalent pressure" on the pipe. Wind loadings, even though they may have predominant directions and average velocities over a given time, are subject to gusting, i.e., sudden changes in direction and velocity. As the observed time period lengthens, the observed number of changes increases in an unpredictable manner as weIl, eventuaIly encompassing nearly all directions and a wide range ofvelocities. A typical plot ofwind velocity vs. time is shown in Figure 4-1.

4-2


WIND SPEED (MPH)

TIME (SEC)

Figure 4-1 Note that to date, almost aU wind design (excluding specialized conditions such as vortex shedding) is done using a static equivalent of adynamie loading model, increasing the static load by a "gust factor" to account for potential dynamic efIects. 2

-

Earthquake: Seismic (earthquake) loadings, are caused by the introduction of random motion (accelerations, velocities, and displacements) orthe ground, and corresponding inertialloads (the mass orthe structure times the acceleration), into a structure through the structure-to-ground anchorage. The random ground motion is actually the sum of an infinite number ofindividual harmonie (cyclic) ground motions. Two earthquakes may be similar in terms of predominant direction (along a fault, for example), predominant harmonie frequencies (if certain ofthe underlying cyclic motions tend to dominate), and maximum ground motion, but their exact behavior at any given time may be quite different and unpredictable. A typical plot ofearthquake ground acceleration vs. time is shown in Figure 4-2.

Time, sec

Figure 4-2 (Figure 4-2 from Response Spectrum Method in Seismic Analysis and Desi~ of Structures by Ajaya Kumar Gupta.)

4-3


Note in many cases, for simplification's sake, seismic design is done using a static equivalent of a dynamic loading model, as described in Section 2.5.2 of these seminar notes, however dynamic simulations of seismic loads are usually more accurate. Dynamic analysis of earthquake loads is discussed in detail in Section 5 of these notes.

Harmonie: With this type ofprofIle, the load changes direction and/or magnitude following to a sine profile, ranging from its minimum to its maximum according to a fixed time period. For example, the load may be described by a function of the form: F(t)

= A + B sin(Ct + D)

Where: F( t)

=

A

=

B

variation of maximum and minimum force from mean, lb

C

= =

D

=

phase angle, radians

force magnitude as a function of time, lb mean force, lb

angular frequency, radians/sec

Major types of loads with harmonic time profiles are: 1

Equipment vibration: Ifrotating equipment attached to a pipe is slightly out oftolerance (drive shaft out of round, for example), it may impose a small cyclic displacement onto the pipe at the point of attachment, where the displacement cycle would most likely correspond to the equipment's operating cycle. The displacement at the pipe connection may be so small as to not even be noticeable, but dynamically it could cause significant problems. The loading vs. time, which can be easily predicted once the equipment's operating cycle and variation from tolerance are known, is shown in Figure 4-3.

L

Displocement ot Pump Flonge

time

--

Figure 4-3 Analysis of equipment vibration is discussed in detail in Section 5 of these seminar notes.

4-4


2

-

Acoustic vibration: Iffluid flow characteristics are changed within a pipe (for example ifflow conditions change from laminar to turbulent when the fluid goes through an orifice), slight lateral vibrations may be set up within the pipe. Often these vibrations fit harmonie patterns, with predominant frequencies somewhat predictable based upon the flow conditions. For example, Strouhal's equation predicts that the developed frequency (in cycles/sec, or Hz) of vibration caused by flow through an orifice will be somewhere between 0.2 V!D and 0.3 V!D, where V is the fluid velocity (ft/sec) and D is the diameter of the orifice (ft). Wind flow around a pipe sets up lateral displacements as weil (a phenomenon known as vortex shedding), with an exciting frequency in the area ofO.lB V!D, where V is the wind velocity and D is the outer diameter of the pipe. Vortex shedding is discussed further in Section 5 of these seminar notes.

3

-

Pulsation: During the operation of a reciprocating pump or a compressor, the fluid is compressed by pistons driven by a rotating shaft. This causes a cyclic change (vs. time) in the fluid pressure at any specified location in the system. If the fl uid pressures at opposing elbow pairs or closures is unequal, this creates an unbalanced pressure load in the system. Since the pressure balance changes with the cycle ofthe compressor, the unbalancedforce changes as weIl. (Note that the frequency of the force cycle will most likely be sorne multiple ofthat of the equipment operating cycle, since multiple pistons will cause a corresponding number offoree variations during each shaft rotation.) The pressure variations will continue to move along with the fluid, so in a steady state flow condition, unbalanced forces may be present simultaneously at all elbow pairs in the system. The load magnitudes may vary, and the load cycles may or may not be in phase with each other, depending upon the fluid velocity, the distance of each elbow pair from the compressor, and the length of the piping legs between the elbow pairs. A typicalloading vs. time profile for pulsation is shown in Figure 4-4.

L Unbalanced Force on Elbow Pair

time

--

Figure 4-4 Analysis of pulsation is discussed in detail in Section 5 of these seminar notes. Impulse: With this type of profile, the load magnitude ramps up from zero to sorne value, remains relatively constant for a time, and then ramps down to zero again. For rapid ramping times, this type of profile resembles a rectangle. Major types ofloads with impulse time profiles are:

4-5


Reliefvalve: When system pressure reaches a dangerous level, reliefvalves are set to open in order to vent fluid and reduce the internal pressure. Venting through the valve causes ajet force to act on the piping system; this force ramps up to its full value, from zero, over the opening time ofthe valve. The relief valve remains open (and thejetforce remains relatively constant) until sufficient fluid is vented to relief the overpressure situation. The valve then closes, ramping down the jet force over the closing time of the valve. A typical force-time profile for a relief valve load is shown in Figure 4-5.

1

Force

Time

Relief Valve Opening Time

Duration lof Load Mass to Vent Flowrate

=

Relief Valve Closing Time

Figure 4-5 In some cases, analysis ofreliefvalve loads is done using a static equivalent of adynamie loading model, as described in Section 2.5.3 ofthese seminar notes, however dynamic analyses are usually more accurate. Dynamic analysis ofrelief valve loadings is discussed in detail in Section 5 of these seminar notes. 2

-

Fluid hammer: When the flow offluid through a system is suddenly halted at one point, through valve closure or a pump trip, the fluid in the remainder of the system cannot he stopped instantaneously as well. As fluid continues to flowinto the area of stoppage (upstream of the valve or pump), the fluid compresses, causing ahigh pressure situation at that point. Likewise, on the other side of the restriction, the fluid moves away from the stoppage point, creating a low pressure (vacuum) situation at that location. Fluid at the next elbow or closure along the pipeline is still at the original operating pressure, resulting in an unbalanced pressure force acting on the valve seat or the elbow. The fluid continues to flow, compressing (or decompressing) fluid further away from the point of flow stoppage, thus causing the leading edge of the pressure pulse to move through the line. As the pulse moves past the frrst elbow, the pressure is now equalized at each end of the pipe run, leading to a balanced (i.e., zero) pressure load on the frrst pipe leg. However the unbalanced pressure, by passing the elbow, has now shifted to the second leg. The unbalanced pressure load will continue to rise and fall in sequentiallegs as the pressure pulse travels

4-6


back to the source (or forward to the sink). A typical force-time profile for a fluid hammer load in a single leg is shown in Figure 4-6. The ramp up time (and corresponding profile) roughly coincides with the elapsed time from full flow to low flow, such as the closing time of the valve or trip time of the pum p. Since the leading edge of the pressure pulse is not expected to change as the pulse travels through the system, the ramp down time is the same. The duration ofthe load from initiation through the beginning of the down ramp is equal to the time required for the pressure pulse to travel the length of the pipe leg. Length of Leoding Edge Based on Closing Time

----p r--

Unbalanced Force

Leg Length Speed of Sound

Time

1.. ,-1 _ CD

Valve stops the flow at time t=O, and a compressive pressure wave moves away from the valve

Valve Closing

(2) At some time t=t1 the high pressure wave, moving at the speed of sound in the fluid, exists between two elbow-elbow pairs. There is an unbalanced force acting ta the left equal to P - P times the Area inside the pipe. The unbalanced force acts until the pressure wave passes the second elbow in the pair.

Figure 4-6 Analyses offluid hammer loadings are often done using a static equivalent ofa dynamic loading model, as described in Section 2.5.3 of these seminar notes, however these types of analysis can he extremely inaccurate. Dynamic analysis offluid hammer loadings is discussed in detail in Section 5 ofthese seminar notes. 3

Slug flow: Most piping systems are designed tohandle single-phase fluids (i.e., those which are uniformly liquid or gas). Under certain circumstances, however the fluid may have multiple phases. For example, slurry systems transport solid materials in liquids, and gases may condense, creating pockets of liquid in otherwise gaseous media. Systems carrying multi-phase fluids are susceptible to slug flow.

4-7


Flow

I~

.. 1

Slug Length (2-Phase Flow)

Figure 4-7 In general, when fluid changes direction in a piping system, it is balanced by the net force on the elbow. This force, shown in Figure 4-7, is equal to the change in momentum with respect to time, or: F

= dp

1 dt

= pv

2

A· .J(1-cos8)/2

Where: dp

=

change in momentum, lb-sec

dt

=

change in time, sec

p

=

fluid density, Ibm/in3

v

=

fluid velocity, in/sec

A

=


e

=

angle of the bend, degrees

Normally this force is constant, and is small enough that it can be easily absorbed through tension in the pipe wall, to be passed on to adjacent elbows which may have equal and opposite loads, therefore zeroing the net load on the system. Therefore these type of momentum loads are usually ignored by the stress analyst. However, ifthe fluid velocity or density changes with time, this momentum load will change with time as weIl, leading to a dynamic (changing) load, which may not be cancelled by the load at other elbows. For example, consider a slug ofliquid in a gas system. The steady state momentum load is insignificant, since the fluid density of agas is effectively zero. Suddenly the liquid slug hits the elbow, increasing the momentum load by a factor ofthousands. This load lasts onlyas long as it takes for the slug to traverse the elbow, and then suddenly drops to near zero again. A typical profile is shown in Figure 4-8. The exact profile of the slug load depends upon the shape of the slug - the force can be calculated at any given time from

4-8


"/(1-

F = pv 2 A· cosE» / 2 , where the A used is the cross-sectional area of the slug instantaneously impactingthe elbow. The time duration ofthe load depends upon the length of the slug divided by the velocity of the fluid. Force Profile 8ased on

\IU

(~eading Edge of

9

Profile 8ased on Trailing Edge of Slug

~ Time 1..

Slug Length Velocity of Flow

-1

Figure 4-8 Analysis of slug flow is usually very difficult because the exact shape and length of the slug is rarely known (and rarely constant). Dynamic analysis of slug loads during two-phase flow conditions is discussed in detail in Section 5 ofthese seminar notes.

4.1.2 System Response Time Versus Timing of Load Change Sorne systems may react more or less quickly than others to changing loads. The implication ofthis is illustrated by the following examples. Consider a system where the restraint loads respond fully to (and therefore completely counteract) an imposed load in 25 milliseconds. For the imposed load profile and configuration shown in Figures 4-9a and 4-9b, respectively, the restraint loads would follow a force versus time profile similar to that shown in Figure 4-9c. This is due to the fact that before the restraints can fully react to the applied load, the load has been removed. Therefore, the induced reaction loads (and by extension, the member forces and moments, and stresses) are much lowerthan wouldoccur under a staticloadofthe same magnitude (each restraint under a static load would see a reaction equal to -P/2, for a total of -P).

4-9


p

-

-

Force

5

10

15

20

25

30

35

Time (Milliseconds)

APPLIED LOAD PROFILE (a)

IP

~

li

L.

tR2

tRI SYSTEM CON FIGU RATION (b)

Time (Milliseconds) 5

10

15

20

25

30

35

40

-PIS Total System Reaction

(R 1 +R2)

-P

Expected Static Response SYSTEM RESPONSE (c)

Figure 4-9

4-10


Now consider a system identical in all respects except that its restraint loads respond fully to any imposed load in only 1 millisecond. For the imposed load prome and configuration shown in Figures 4-9a and 4-9b, the restraint loads would follow a force versus time profile similar to that shown in Figure 4-10 - virtually identical to that of the applied load. Therefore, when a system responds rapidly to applied loads, the induced reaction loads, member forces, etc. are approximately the same as those which would occur under the same static load.

Time (Milliseconds) 5

10

15

20

25

30

Total System Reaction (R1+R2) _P

35

Expected ____ Static Response

SYSTEM RESPONSE

Figure 4-10 Thepertinentquestionisthen, whatisafastresponsetime,andwhatisaslowresponsetime. In truth, there is no absol ute answer - what is really important is the relative response time ofthe system as compared to the rate of change ofthe applied load. For example, what if the load applied to the system in Figure 4-9 had a duration of 25 seconds, instead of 10 milliseconds? The restraint loads would have sufficient time to fully respond to the applied load, and the reactions would be the same as for a static load. In fact, a static load is simply a dynamic load with a duration long enough that aU systems have the opportunity to respond fully to it. From the above, it is evident that system response to dynamic loads can produce at least two possible results, based upon the system response time. Slowly responding systems result in response loads lower than the applied loads, while rapidly responding systems result in response loads approximately the same as the applied loads. Whathappens when the system response is somewhere in between? Consider the systems described in Figure 4-9 and 4-10 (i.e., with response times of 25 milliseconds and 1 millisecond, respectively), this time loaded with a harmonic load, cycling between P and -P, with frequency of 1 cycle per 25 milliseconds. The responses of the two systems are shown in Figures 4-11a and 4-11b. The frrst system again lags behind and fails to fully develop response loads. The second system responds almost instantly andjust about fuUy responds to the applied load, as before.

4-11


Devoloped Response (R1 +R2) p

\

Applied Load

( Time

-p

(a)

p

Time

Force

r-~4-----~--r-~4-----~~r--++-----~~--~

-p

(h) Figure 4-11 Now consider a system which has a response time somewhere in the middle - about 12.5 milliseconds. Upon initialloading, the system initially attempts to respond to the load P, with restraint loads each equal to -P/2. Since the system response lags, it does not fully develop these restraint loads, but, after 12.5 milliseconds, will have total system response (restraint loads) of somewhere around -0.7P. Considering the cyclic load, the applied load on the system will be, at 12.5milliseconds, -P,for a net loadon the systemof-l. 7P(seeFigure 4-12).

4-12


Applied Load

p \ \ \

Force I---'\~t--+--I----t--~ Ti m e \ 25ms \ \ \.~ Response

\--

-p

\.

.......

Net System Load (Applied + Developed)

Figure 4-12 The system now attempts to resolve the net load of -1. 7P with two restraint loads of +0.85P. Assuming that at time T=25 milliseconds, these loads have actually reached only +0.6P (due to the response lag), or a total of +1.2P, the externalload will now be +P, so the net system load will be +2.2P, as shown in Figure 4-13.

Net System Load (Applied + Developed)

2P

Developed Response (R1+R2) p

Applied Load

\ \

\

Force

-p

1----'\----'1------+---'--.+--1----:.

\ \ \ ',j

Ti m e

25ms

\"

Figure 4-13 This net load will then be resisted by total restraint loads (system response) of -2.2P, which will have reached approximately -1.5P by T=37.5 milliseconds, at which time the load will have reversed again, creating a net load on the system of -2.5P. Continuing in this way, the net load on the system will be approximately 2.8P at 50 milliseconds, -3.0P at 62.5 milliseconds, 3.1P at 75 milliseconds, and so forth. The total developed load (total restraint loads) is shown as a function oftime in Figure 4-14.

4-13


Devoloped Response (R1 +R2) (

2P

"

p

""

\

1 1

\ \

~

Applied Load

\ Time Foree I---+-_---;'--I-_+----I-_ _---t;.._+---+-..... I --+---I---+-_ 1

175ms 1

-p

\

" _.....

1 1 /

Figure 4-14 This may continue until the developed load spirals out of control, and the structure fails. From this example, it is clear that there is a third possibility for a system response under dynamic loading - the induced load may far exceed the applied load. This is the type of response with which the engineer must normally he concerned.

4.1.3 Lack of System Equilibrium As illustrated above, system response loads (reactions, internal forces and moments, etc.) are not necessarily equal and opposite to the applied dynamic loads, so the net sum ofthe forces and moments acting on the system are something other than zero. This means that the system does not meet the laws of static equilibrium, and cannot be solved using traditional static solution methods.

4.1.4 Movement of Dynamically Loaded System Since the sum of the forces and moments acting on a dynamically loaded system may not be zero, there may be unbalanced loads acting on the system. According to Newton's laws of motion, an unbalanced force on a system results in motion, due to the acceleration expressed by F = MA. This information is useful in identifying problems due to dynamic loads - a simple rule of thumb states that if the piping system is moving, it is heing subjected to a dynamic load.

4.1.5 Relation of Induced System Loads to Applied Loads As shown above, the internal system response loads may he quite different than the dynamically applied externalloads. The response can be classified by calculating the ratio ofthe system response to the applied load (or to the system response expected for a staticload of the same magnitude) - this ratio is called the Dynamic Load Factor, or DLF.

This concept is illustrated in Figure 4-15. In Figure 4-15a, a weight rests on a spring, and the spring compresses 1 inch. In Figure 4-15b, the same weight is suddenly applied to the spring. In this case, the spring oscillates, eventually coming to rest at the static deflection. However, initially the maximum deflection overshoots the static compression by an additional 1 inch.

4-14


i 1

Disploced 1" from Uncompressed Length

~ ~

Ii

Weight ot Rest

Settles ot Stotic Displocement

1. \

,~Time T

2"1

~

/

L

String is Cut, Loading the Spring Instontly

./

~

\....-----~

1".J

~T

;

Displocement Path of the Moss

(a)

(b)

Figure 4-15 The Dynamic Load Factor for this case is the maximum dynamic response (2.0 inches) divided by the static response under the same load (1.0 inch): DLF

=

Max. Dvnamic Displacement Static Displacement

=

.2...Q

1.0

=

2.0

In fact, the DLF for an instantaneously applied constant load (not a harmonic or impact load) is always exactly 2.0. This is easily proven by equating work done by a weight while compressing a spring with the energy stored by the compressed spring: WX

= KX2/2, or:

2W

= KX (twiee statie weight load)

Where: W

=

weight, lb.

x

=

spring compression, in.

K

= spring stiffness, lb/in

Note: The force in the spring is KX., which, equal ta 2W, is exactly twice the static load, for a DLF of2.0. System responses may be classified based upon the value of the resulting DLF. Possible system responses are: 1)

The system response may be much lower than the applied load. In this case, the DLF will be much smaller than 1.0. This type ofresponse is called a flexible response. Example: When one end of a rope lying on the floor is pulled quickly,

4-15


most ofthe rope will not feel the pull and will continue to lie undisturhed in the pile on the floor. 2) The system response may he much higher than the applied load. In this case, the DLF will he much greaterthan 1.0. This type ofresponse is called a resonant response. Example: When one end of a whip is swung over a small distance, the other end of the whip moves in a large arc and cracks with a great force. 3) The system response may be about the same as the applied load. In this case, the DLF will be approximately equal to 1.0. This type ofresponse is called a rigid response. Example: When one end ofa baseball bat is swung, the free endfollows along simultaneously, over roughly the same distance as the batter's hands. In which of these categories a system's response falls is dependent upon the relation hetween the system characteristics (system response speed) and load characteristics (force time profile). A system, in and ofitself, is not flexible, resonant, or rigid. For example, in the first example above, the rope no longer has a flexible response if the duration of the applied load is increased significantly - i.e., if someone pulls indefmitely, eventually the entire rope will get up and follow the pulled end.

4-16


4.2 The Dynamic Equation of Motion As noted earlier, the dynamic response of a system is based upon the dynamic characteristics of both the system and the load.

4.2.1 Evaluation of a Single Degree-of-Freedom System Dynamic response can be studied by examining a simple system - that of a single-de greeof-freedom oscillator, as shown in Figure 4-16.

M

K

Figure 4-16 The single-degree-of-freedom oscillator consists of a mass M attached to ground by a spring with a stiffness K and a dashpot with a damping value of C. The spring pulls on the mass with a force proportional to its extension or contraction (or the displacement of the mass); the dashpot provides a frictional force proportional to the velocity of the mass. Any unbalanced force accelerates the mass. The hehavior of a single degree-of-freedom oscillator can then he descrihed by the dynamic equation of motion: M

x (t) + C x (t) + K x(t) = F(t)

Where: M

=

mass of oscillator, slug

x(t)

= =

acceleration of mass as a function of time, ft/sec 2

= = = =

velocity of mass as a function of time, ft/sec

C

x(t) K

x(t) F(t)

damping of oscillator, slug/sec or lh-sedft

structural stiffness of oscillator, lb/ft displacement of oscillator as a function of time, ft imposed extemal force as a function oftime, lb

This equation cannot he explicitly solved, unless the damping term, C, is zero and the imposed load is harmonic (i.e., of the form F(t) = asinb(t+c». Therefore, the damping value

4-17


is often dropped (since it is usually small) in order to simplify the equation. The equation can he simplified further by taking the simplest external harmonic load - a load of zero. If there is no externalload, and dam ping is approximately zero, the equation descrihes the free vibration of an undamped single-degree-of-freedom oscillator: M x(t) + K x(t) = 0 Its solution, as a function of time t, is:

x(t)

= Xo

x(t)

= . Xo

sin .JK/M t

(KIM) sin .JK/M t

Where: Xo

.JK 1M

=

initial displacement of oscillator, ft

= = natural (angular) frequency, radians/sec (J)

(note: cyclic frequency in Hz = ro /2rc) The system characteristics of a single degree-of-freedom oscillator can he completely descrihed by its natural frequency and its damping value. System response can be determined once the Dynamic Load Factor (as a function of the natural frequency and damping value) for the applied load is known.

4.2.2 The Dynamic Load Factor As mentioned earlier, in most cases the dynamic equation of motion cannot be explicitly solved for an externally loaded system. The system response can often be estimated through the use ofnumeric solution techniques, such as a time history analysis, where acceleration, velocity, displacement, imposed load, unbalanced force, etc. can be calculated through integration at subsequent time steps. Once these values at all time steps have been calculated, the maximum response to the imposed load can be determined by multiplyingthe maximum displacement (occurring at any time throughout the load duration) times the stiffness K. Dividing this maximum response load by the maximum imposed load gives the Dynamic Load Factor.

Iftime history analyses are done like this for a number of different types ofloads, and for systems with various characteristics (expressedin terms ofnatural frequency and damping), DLF functions for various types ofloads can he estimated. Use of pre-determined functions simplifies the analysis, since time history analyses need not be redone every time a dynamic load on a system is evaluated. For example, the Dynamic Lo~d Factor for a harmonïcally applied load, as a function of damping and natural frequency of the single-degree-of-freedom oscillator has heen discovered to be:

4-18


DLF

=

Where: DLF

=

dynamic loading factor (ratio ofinduced to applied dynamic load), dimension less

Cc

=

ratio of system damping to "critical damping", where "critical damping"

=

2 .JK M ,dimensionless (representative values for piping systems typically range from 0.02 to 0.1) CO[

=

forcing frequency of applied harmonic load, rad/sec

COn

=

natural frequency of oscillator, rad/sec

Note the implications ofthis function when comparing the forcing frequency of the load to the natural frequency of the oscillator. Approximating Cc as zero, when: ron

«

CO[,

the DLF approaches 0 (flexible response)

COn

=

CO[,

the DLF approaches infinity (resonant response)

COn

»

CO[,

the DLF approaches 1.0 (rigid response)

As Cc increases, the DLF approaches 1.0 (increasing the flexible response and decreasing the resonant response). Damping prevents the DLF from ever actually reaching infinity.

A set of curves (for various damping values), in terms ofDLF versus COn can be plotted for a specific harmonic load, as shown in Figure 4-17. Knowing the natural frequency and damping of a system, the induced load in the system can be easily determined by reading the DLF off of the curve and multiplying it by the applied load.

r

1 % Critical Damping 2% Damping 3% Damping

DLF

10% Damping

1.0

Figure 4-17

4-19


This demonstrates that potential problems may occur when the natural frequency of the oscillator is very close to the forcing frequency of an imposed load. DLF curves can be estimated for othertypes ofloads as weIl. For example, the instantaneous DLF of a rectangular load is relatively unaffected by damping, and is:

DLF = 1 - cos Oln t, for t <= td DLF = cos Oln (t-td) - cos Oln, for t > td Where: t

=

instantaneous time at which evaluation is made, sec

td

=

duration of rectangular load, sec

The plot of the maximum DLF versus COntcV21t for a rectangular load is shown in Figure 4-18. Note that as the maximum DLF for this type ofload is 2.0, and as the duration td increases, the DLF curve moves to the left, exciting lower frequency oscillators. 2.0

1

1

V

1

1.6

1

1

l

1

1.2

li J:

i4: ....

9

0.8

0.4

o

1 1

/

FIt) Forr td r

/1

J

.,

0.05

0.10

1

1

1

i

1 0.2

0.5

1.0

2.0

5

10

(Figure 4-18 from Structural Dynamics Theor.y & Computation by Mario Paz). An impulse load (such as due to a relief valve, fluid hammer, etc.) is similar to a rectangular load, except that it may have ramp-up and ramp-down times that are notinstantaneous. The effect of an extended, rather than an instantaneous ramping time is to lower the maximum DLF from 2.0. This is illustrated in Figure 4-19, which shows three force-time profiles, each with different ramp up times and durations. The corresponding DLF curves are also shown for these prof:ù.es, demonstrating that as the force durationincreases, the curve shifts left, encompassing more low-order frequencies; and as the ramp-up time approaches zero, the maximum DLF nears 2.0.

4-20


1..35

.

1.135

•••••••

. 875

(

.75

d )

".

.

.

.................

.

'.,

••••••

,,,

.

.

.................

'.,

. ••

ft

••••

"

. •••••••

.

r c e

n 0 r l'II a. 1 i z e

. •••••••

..1J4"Ilt---------"lto--
1.. F 0

'.,

.......

.

'.'

'\"

•••••••

; ....... .

.

. . . . . ...... ." .......... - ...... . ....... . ................ '." ....... , ................. ","

".

· ....... :........ : ................ :. ·~PROfH.E· 'II~' ....... :........ : ....... ~ ....... .

. 635

. ...... :· ....... :. ........ :......... .:....:.P·FiOFÎLE· 'Ii::i .

.5

" . . . . . . . . ..... -,· ................. '., ...... -, ....... "' ....... .

.375 .35

. . . . ...... · ................ '." ...... -, ................ .

.135

. ..... -,· .................

.8

".

15

.8

38

45

68

75

Ti .... e

.

".,

...................... " .

185

98 (l'IIsec)

138

135

158

FORCE VERSUS TIME PROFILE 3. 3.7

. •••••••

"."

3.4 · ......

','

.

. ••••••

"

.

. . . . . . . . . . . . . . . . .

.

"."

••••••

_,

.

. . . ............ . .... ....... ., ................ ' . ' ....... "

~

.

.

. . . . . . . . . . . . . . . . .

,,"

••••••

_,

. ••••

_

.....

ft

••••

••

. ..... . . ................. . .

'.'

"

~

:PROF ILE 113

3.1 1.8 1)

L F

1.5

........:./..... ~ ....... ~ ........ :........ ; ....... ~ ........ :........ ; ....... ; ....... .

:. '1:. :. +:. :. . .:+ . ·.'J'"+.i:.:..•+~:+ PROFILE ·112···· ... :........ : ....... ........ :........ : ....... ....... .

·.. .•. . . . .. . . . . . [. .. . . . . . ;. . . . ': . . . . 'l'~ ~ ~

.6 .3 .8

~

..: .......

~

:

:. . . . ..

~

f ....... ; ....... .

il ···~~~iE·~··'······ .•....... , ...... , ..... ,............. ,

1.3 · .. . .9

4" :

~

:. . . ..

~

:

........ ........ .......

:- ....... : ....... ~ ....... .

t •.• •,• ·• • :• • • • l••••••• ·••••••••• :•••.••• l•••••••••·•••••••:••••••• l•.•• '• •

~

+'~~~~~~~~~~~~~~~~~~,-~~~~~-r~~~~~~

:"8

4

8

13

16

38

34

Ma.tural Frequency

3S

33

(Hz)

DYNAMIC LOAD FACTOR VERSUS NATURAL FREQUENCY Figure 4-19

4-21

36

48


4.3 Evaluation of Multi-Degree-of-Freedom Systems In reality, a piping system is not as simple as a single degree-of-freedom oseillator, but ean he modeled (at its simplest) as a series ofmasses and springs, as shown in Figure 4-20.

~2

~1

Figure 4-20

4.3.1 Modal Analysis Ignoring damping, the equations ofmotion deseribing the response ofthis multi-degree-offreedom oscillator are: [M] {x(t)} + [K] {x(t)}

= {F(t)}, or:

Ml x let) + KI XI(t) - K2 (X2(t) - XI(t» M2 X2(t) + K2 (X2(t) - XI(t»

= FI(t)

= F2(t)

For free vibration, these equations reduee to: [M] {x(t)} + [K] {x(t)}

= fOl, or:

Ml x let) + KI XI(t) - K2 (X2(t) - XI(t»

=0

M2 X2(t) + K2 (X2(t) - XI(t» = 0, or: Sinee free vibration is harmonie in nature, these equations ean be solved by substituting: XI(t) = al sin co t X2(t) = a2 sin co t x let) = -al co2 sin co t x 2(t) = -a2 co2 sin co t 4-22


These equations can now be written in matrix form:

[::]

;

[~]

The solution to this (other than when al and a2 are both zero) is found when the determinant of the matrix is zero, or:

Using the quadratic equation, this equation has two solutions in 002: 0012

=

((K1+K2)M2 + M1K2) - [((K1+K2)M2 + M1K2)2 - 4M1M2K1K2 ]1/2 2M 1M2

0022

=

((K1+K2)M2 + M1K2) + [((K1+K2)M2 + M1K2)2 - 4M1M2K1K2 ]1/2 2M 1M2

Two sets of displacements (an, a21 and a12 , a22) can be solved for relative values by substituting the two values of 002 into the earlier set of equations: a21

=

an [KI + K2 - Ml 001 2] / K2

a22

=

a12 [KI + K2 - Ml 002 2] / K2

Since these displacements can only be solved for relative to each other, rather than exact values, they are often unity normalized, with the maximum displacement in each set being given a value of1.0, and aU other displacements in the set being given relatively lowervalues. Or, more often these sets of dis placements are normalized using a computationally more convenient method, caUed mass normalization. This method normalizes the sets of dis placements such that: []T [M] []

= [1], or:

~] Where: []

=

matrix ofmass-normalized relative displacements, where columns correspond to sets of displacements, and rows correspond to directions and mass points

[]T

=

transpose ofmatrix ofmass-normalized relative displacements

[1]

=

identity matrix

4-23


The implication to this analysis is that the displacements XI(t) and X2(t) at any given time are the sum of two independent cyclic displacement functions: XI(t)

=

Cl an sin (J)l t + C2 al2 sin (J)2 t

X2(t)

=

Cl a21 sin (J)l t + C2 a22 sin (J)2 t

=

constants of integration, based upon initial conditions

Where: Cl, C2

Individual equations of dynamic motion can be written for each of the independent cyclic displacement functions by allocating a proportional share of the system mass and stiffness to each:

=fl(t) m2 Z2(t) + k2 Z2(t) =f2(t)

ml Zl(t) + k l Zl(t)

Where: ml

m2

=

share of system mass allocated to cyclic function 1

=

Mlan 2 + M2a 21 2

=

share of system mass allocated to cyclic function 2

Z let) = acceleration, versus time, through displacement set 1 Z2(t) = acceleration, versus time, through displacement set 2 kl

=

share of system stiffness allocated to cyclic function 1

k2

= = = = = =

share of system stiffness allocated to cyclic function 2

Zl(t) Z2(t) flet)

f2(t)

=

(J)22 m2 relative displacement versus time, through displacement set 1 relative displacement versus time, through displacement set 2 share of system loading versus time allocated to cyclic function 1 anFI(t) + a2IF2(t) share of system loading versus time allocated to cyclic function 2

4-24


ln fact, the free vibration response of any system with N degrees-of-freedom (i.e. mass points times displacement directions) is the sum ofN independent cyclic functions. Each ofthese vibration functions are called "modes of vibration" of the system - each one has its own natural frequency (cOi) and has a single degree-of-freedom, vibrating back and forth about the set ofdis placements
Participation Factors and DLFs

As noted, a system has one mode of vibration for each degree-of-freedom (translational and

rotational) at each mass point in the model. In reality, systems are continuous, and therefore have an infinite number of modes of vibration. For example, for a cantilever, the natural frequency of any ofthese infinite modes of vibration can be calculated as:

Where:

mn n

E 1

m L

= = = = = =

natural angular frequency of mode n, rad/sec mode # modulus of elasticity of cantilever, psi moment ofinertia of cantilever, in4 unit mass per length of cantilever, slug/in length of cantilever, in

4-25


Note that this shows that the natural frequency of the second mode of vibration of the cantilever is four times that of the first mode of vibration, and the natural frequency of the third mode is nine times that of the first. The mode shapes correspondingto the first few modes ofvibration ofthe cantilever are shown in Figure 4-21. As noted earlier, the actual magnitude ofthe mode shape displacements may change, but the ratios of dis placements at various points to each other will remain constant.

1

~

I<==>< f==>c=>< pc:><=><

1stMode

2nd Mode

3,d Mode

4th Mode

Figure 4-21 We note that with ascending frequencies, more points on the cantilever are tied to their original position, so higher order modes are tied more tightly to their original position. This me ans that the higher order modes are less likely to vibrate. Conversely, the lower order modes are more likely to predominate in the total system response (in the absence ofimposed loading) than will the higher order modes. (This explains why the free vibration of a system often seems to resemble the vibration of the first mode, and why people often speak of a system having a natural frequency, rather than a natural frequency ofthe first mode.) The measure of the tendency of the mode to get involved in the vibration is called the mode participation factor, and is calculated (in vector form) for each mode as:

{MPFïl

= -{c,I>ïl T [M]

Where: {MPFi}= {c,I>i}

participation factor of mode i, in vector form

= mass-normalized shape of mode i

The degree to which a mode is excited is also influenced by the mode shape and the load distribution. For example, if the cantilever is loaded as shown in Figure 4-22, the first and third modes will be excited (since the load occurs at high points ofthose modes), while the second mode will not be excited at all. 4-26


1 st Mode

I<=>r< 1

2nd Mode

1 1

P0< 1

3rd Mode

Figure 4-22 This demonstrates the concept of the load participation factor, which is a measure of the coincidence of the high points of the load to the high points of the mode shape. Therefore, the mode participation factor is further multiplied by the mode shape, giving the mass participation factor, r:

Since each mode of vibration behaves as a single-degree-of-freedom oscillator, the modal response is also based upon the DLF for that mode's frequency under the imposed load. The final contributor to the modal response multiplier is of course the imposed load itself. Therefore each of the individual modal responses can he calculated by applying, to the system, static loads equal to:

Where: {Fi}

=

equivalent static load vector induced by mode i

DLFi

=

dynamic load multiplier for oscillator with frequency COi

{cI>ïl

=

mass-normalized shape of mode i

[M]

=

system mass matrix

{F}

=

imposed load vector

The results ofthe individualmodalloads are then calculated and summed to provide the total system response (using one ofthe methods describedlater, in Section 4. 6.2.4 ofthese seminar notes).

4-27


4.4 Eigensolver Aigorithm The EIGENSOLVER is the solution process whereby a piping or structural system's natural frequencies and mode shapes are determined. This algorithm uses an iterative method to solve for successive modes, so itcan take much longerthan would a static solution ofthe same piping system. Furthermore, solution time increases in a non-linear way with the number of modes to be extracted. For example, the time to extract ten modes of a typical piping system using an IBM XT (this would take only a few seconds on a 80486 machine) is shown in Figure 4-23. One static solution for this model took less than 15 seconds, also using an XT. .AtE

la

.. IDa

CAESAR Il EISEILSOlVER (Y2.1C Julv 21. 1986! (tsc) ta talet lut tiaenDair and continu•• FrtQuency 1 = FrtQuency 2 = FreQuency 3 = FrfQufRcy 4 = FrealltllCY S = Freallency 6 :; Freauency 7 = FrlQUlllcY 8 :; Freauency 9 :; FreaUt!llcy Il :;

.97416 Hz. 1.54213 Hz. 2.28722 Hz. 2.78333 Hz. 3.41256 Hz. 3.81449 Hz. 4.35438 Hz. 5.63579 Hz. 6.17498 Hz. 7.38423 Hz.

Elused ElaDsed EliDStd Elu5td Elusld Elund El aDsed Elused ElaDsed ElaDsed

Ti .. Tiu Tiu Tiu Ti.1 Ti I l Ti It Tilt Ti u Tilt

= 1: 1:56 = 1: 1: 9 = 1: 1:44 :; 1: 1:46 :; Il 2: 1 = 1: 2:26 = 1: 2:42 1: 1: 3:23 .. 1: 3:41 .. 1: 4:23

Figure 4-23 As noted earlier, associated with each mode is a shape and a frequency, which together define the system's tendency to vibrate; the mode shape defines the shape the system would like to take when it vibrates, and the natural frequency defines the corresponding rate of vibration. The eigensolver returns a set ofthese for each mode, with the dimensionless mode shape called an eigenvector, and the frequency returned as the square of the angular frequency (0)2), known as the eigenvalue.

4-28


Mode #1

=1' ...,

3.412 Hz

I1r

0.974 Hz

......

.. ..,. MIll

1

1

Rrst system cantilever mode in the Z direction.

~

Opening and closing of the two U·s.

~

:,. ....

Mode #6

III'

1.542 Hz

3.8144 Hz

IIf ...,

IIOIIS

1 R

1 U--

Rrst system cantilever mode in the X direction.

Mode #3

III'

2.287 Hz

Two component "twisting" mode.

Mode#7

... IW

4.35 Hz

MIll

fIJ

..II! MIll

,. cac.

AI.

&1.

Opening and closing of the two U's. Note that mode 5 was more "rocking" while this mode is more "stretching".

X-direction swinging of "U"s opposite ta one another.

Mode#4

...,

2.783 Hz

Model8

lit'

5.636 Hz

1

... Bf

Rrst tarsional mode of this piping system.

Figure 4-24

4-29

"Twisting" and "pivoting" about the inside leg of the U.


Given the eigenvalue, the modal frequency can be expressed in angular frequency (radians per second), cyclic frequency (Hz), or period (seconds per cycle): eigenvalue

= m2 (radians squared per second squared)

angular frequency

=

cyclic frequency

= m/ 2rr (Hz, or cycles per second)

period

=

m (radians per second)

2rr / m (seconds per cycle)

Basic mode shapes of a simple piping system, as determined by CAESAR II's eigensolver are shown in Figure 4-24. The absolute magnitude of a mode shape dis placement computed by an eigensolver is unknown, with only the shape being given (i.e. onlythe ratios ofthe displacements at various degrees offreedom are known for each mode, with these ratios heing constant for each mode). For example, the eigensolution can only predict that the magnitude of A (as shown in Figure 4-25) is twice that ofB, and four times that ofC, etc. RISI!'

tolr

NOlIS

~~ ...................... _~

lm! SPICIY

If' COLOIS OIIGIIL

IUNI

IIIIX JI

Ingr.

Figure 4-25 One eigenpair can potentially be calculated for each degree of freedom in the model that contains sorne nonzero mass (node point) and sorne non-rigid stiffness (i.e., is not fully restrained). CAESAR II, as well as most other pipe stress programs, typically omit rotational degrees offreedom from dynamic models in orderto simplify the calculation - this is usually acceptable since rotational modes ofvibration usually have very high frequencies, and correspondingvery low mode participation factors. Figure 4-26 demonstrates how many modes can he extracted for sorne simple systems.

4-30


10

•

=

2 mass points X 6 DOFsjpoint 12 DOFs total 2 X 3 rotational DOFs/point 6 rotafional restraints 3 restrained translational DOFs at node 3

=

12 -

6 -

3

=3

potential modes of vibration

=

5 mass points X 6 DOFsjpoint 30 DOFs total 5 X 3 rotational DOFs/point 15 rot. restraints 8 restrained translational DOFs at nodes 5,20,and 25

=

30 -

5

15 -

=7

8

potential modes of vibration

10

Figure 4-26

Notes on the CAESAR II Eigensolver/Dynamic Analysis: 1) Natural frequencies for more than one mode of a single piping system can be identical. The two systems shown in Figure 4-27 illustrate this.

}---·---·-·-·-·-·-f---·----+-~

~

~

1

----.

If)

;~

Fi rst Mode in "Y" Direction

1

~

Second Mode - same natural frequency, identical mode shape, but in HZ" direction.

J--x Z

T Identical branches will display (4) equal frequencies in the X. and probably the Z directions.

Figure 4-27

4-31


2) Natural frequencies for a system modes can he zero, but this condition results in the termination ofthe eigensolver. Zero natural frequencies indicate rigid body motion, as shown in Figure 4-28.

.----=n

~-----,

Piping System with Y-only Supports

r Suction

F===tF====~f----1\.. Di scha rge

Pump on Teflon Siide Plates

Pump on Teflor Siide Plates

Figure 4-28 If there is no friction at the Y-supports or at the pump slide plates, then

theoreticaily, the system could slide horizontally an infinite distance without causing loads to develop within the system. This is called a "rigid-body" mode because the whole piping system moves as if it were rigid, i.e. there are no displacement strains to cause the system to appear flexible. 3) In almost all cases, the eigensolver will detect modal frequencies from the lowest frequency to the highest. Sometimes, when there is sorne strong directional dependency in the system, the modes may converge in the wrong order. This couldcause a problem if the eigensolver reaches the cutoffnumberofmodes (i.e., 20), but has not yet found the 20 modes with the lowest frequency (it may have found modes 1 through 18, 20, and 21, and would have found number 19 next). In this case, the analyst would have proceeded, believing aU modes below the 21st frequency had been considered in the analysis. CAESAR II checks for this anomaly using a calculation called a Sturm Sequence Check. This procedure determines the numher of modes that should have been found hetween the highest and lowest frequencies found, and compares that against the actual number of modes extracted. Ifthose numbers are different, the user is given a warning. To correct this problem the user can either: a) Ifthe mode cutoffhalted the eigensolution, increase the mode cutoffto the "negative terms" value reported by the Sturm Sequence Check. b) Ifthe frequency cutoffterminated the eigensolution, increase the cutoff frequency sorne smaU amount. This will usuaily allow the lost modes to fail into the solution frequency range. c) Fix the subspace size at 10 and rerun the job. Increasing the number of approximation vectors improves the possibility that at least one ofthem

4-32


will contain some component of the missing modes, allowing the vector to properly converge. 4) Eigensolutions oflargejobs (ajob can he considered "large" ifit has 300 elements or more, and/or requires that 50 or more frequencies be extracted) may require special treatment. In these cases, the user should: a) Set the optimum bandwidth parameters during the problem input phase using the auxiliary parameter spreadsheet ("K" from the piping input spreadsheet). b) Set the subspace size to 12 using the dynamic control parameter spreadsheet. c) Set "No. to converge before shift allowed" to 4 in the dynamics control parameter spreadsheet. d) If the job requires the calculation of more than 100 natural frequencies the Frequency Space Array number on the control parameter spreadsheet should he increased. In this case a value of 200 or 300 should be used. 5) CAESAR II's eigensolver will not solve an artificially small job, having 3 or fewer degrees of freedom. In this case, the program will display a "NOT ENOUGH DEGREES OF FREEDOM" error message. In this case, the user can insert an extra element or two in order to create additional degrees offreedom. Note that this problem might also occur, for example, if the pipe or structural material density was accidently set to zero. 6) CAESAR II's dynamic analyses are alllinear!!! This means that one-directional restraints will not lift off and reseat, gaps will not open and close, and friction will not act as a constant effort force. Therefore, for dynamic analyses, aIl non-linear effects must be modeled as linear - for exam pIe, a one-directional restraint must be modeled as either seated (active) or lifted off (inactive), and a gap must be either open (inactive) or closed (active).

THE PARTICULAR STATUS OF ALL NONLINEAR RESTRAINTS IN A DYNAMIC RUN 18 DETERMINED FROM A PRIOR STATIC RUN. The user must tell CAESAR II (using an entry in the dynamic control parameter spreadsheet) which ofthe static load cases should be used to set the status ofthe nonlinear restraints in the job. For example, assume that in the operating load case (#1) the restraint at node 15Iifted-off, while during the sustained load case (#2) the restraint at node 15 was active. In order to perform a dynamic analysis on this job, the user must tell CAESAR II which ofthe load cases is to be used to set the status ofthe nonlinear restraints. Ifload case #1 (i.e. the operating case) is used, then the restraint at node 15 will be completely omitted for the dynamic run. Ifload case #2 (i.e. the sustained case) is used, then the restraint at node 15 will be modeled in for the dynamic run. RE8TRAINTS NEVER CHANGE STATUS DURING DYNAMIC ANALYSES.

4-33


A DYNAMICS JOB MUST HAVE APRECEDING STATICS JOB WHENEVER: a) There are spring hangers to be designed in the job. The static runs must be made in order to determine the spring rate to be used in the dynamic model. b) There are one-directional restraints in the job, i.e. any restraint with a type preceded by a (+) or a (-) sign. c) Therearefrictionalrestraintsin the job, i.e. anyrestraints witha nonzero "mu" value. d) There are gap restraints in the job. e) There are "large rotation rod" restraints in the job. f)

There are bilinear restraints in the job.

7) Modeling of friction in dynamic models presents a special case, since friction actually affects the damping component of dynamic response, and (as seen previously) damping cannot be included in the solution of dynamic equations. Friction can be accounted for in CAESAR fi through the use ofthe FRICTION FACTOR on the dynamic control parameter spreadsheet. CAESAR fi approximates the restraining effect of friction on the pipe by including stiffnesses transverse to the direction of the restraint at which friction was specified. The magnitude of the stiffness due to the frictional effect is computed by:

=

Fn x

Kfriction

=

stiffness of frictional restraint inserted by CAESAR fi (lb/in)

Fn

=

the normal force at the restraint taken from the static solution (lb)

II (mu)

=

friction coefficient, dimensionless

Fact

=

FRICTION FACTOR from the control spreadsheet, in- 1 (Typical values range from 0 to 1000)

Kfriction

).l

x Fact

Where:

This factor should he adjusted as necessary in order to make the dynamic model conform to the real system's response. Entering a friction factor greater than zero causes these friction stiffnesses to be inserted into the dynamics job. Increasing this factor correspondingly increases the effect ofthe friction. Entering a friction factor equal to zero ignores any frictional effect in the dynamics job.

4-34


normal force @15 = 795 lb (from static load case) friction coefficient (from system input)

= 0.3 =

friction factor 1.0 (trom control spreadsheet) fricfional stiffness

Figure 4-29

4-35

= 795

= 238.5

X 0.3 X 1.0 lb/in


4.5 Accuracy of The Dynamic Model Previous discussions have indicated that the maximum number of modes that may be extracted from a dynamic model is dependent upon the number of mass points (normally mass points and node points are synonymous). Therefore the number of modes that can be extracted, and for which responses can be determined, is based upon how the system was modeled. For example, the two systems shown in Figure 4-30 will yield exactly the same staticresults, butahighlydifferentnumberofmodes,and thereforehighlydifferentdynamic results.

~---------.

. .

~~----~.---~.~--~.---~.----~. Figure 4-30 This raises the questions ofhow many mass points should be used in the model in order to achieve sufficient accuracy, and how many modes should he extracted and summed. The first of these questions deals with the accuracy of the modal calculations, while the second question is concerned with at what point does a finite summation sufficiently approximate an infinite summation.

4.5.1 Mass Point Spacing The first question is: How many are mass points are enough to get sufficiently accurate estimates of a system's modes of vibration? CAESAR II's models are lumped mass models, which assume that halfofthe mass ofeach element is lumped at the element's end nodes. Consider the calculated natural frequency of a lumped mass model of a cantilever vs. the exact solution of the frequency ofits frrst mode (as discussed in Section 4.3.2 ofthese seminar notes):

~-------e. K=3EI/L3

~A A A 6 YVY A A 6 \T'V A '\{]J 2" \l'Tv 2"

Static Model

l

Figure 4-31

4-36

Lumped Moss Model


For the primitive model:

=

(KIM) 112

=

=

[(3EI/L3)/mU2] 1/2 (6EIImL4)1I2

m

=

angular frequency of system

K

= =

stiffness of cantilever, lb/in 3EI/L3

E

=

modulus of elasticity of cantilever, psi

1

=

moment ofinertia of cantilever, in4

L

=

length of cantilever, in

M

= =

lumped mass at node point, slug mU2

m

=

distributed mass per length, slug/in

m

Where:

As seen in Section 4.3.2, for the continuous cantilever, the angularfrequency ofthe frrst mode of vibration is:

ml = (0.597 rt)2 (EIImL4 )1/2 Therefore the accuracy of the primitive, lumped mass model is:

m / ml = (6EIImL4)1/2 / [(0.597 rt)2 (EIImL4)1/2] =

61/2/(0.597 rt)2 = .696

Therefore this primitive, lumped mass model produces a modal frequency that is only 69.6% ofthe true value. Furthermore, the second mode of vibration cannot even he approximated using this model, since there is no other mass point with a degree offreedom in that direction. Adding a new node point (mass point) in the center (as shown in Figure 4-32) permits estimation of the second mode, and provides additional accuracy to the estimate of the first.

..

..

~---------- ~--------

Stotic Model

Lumped Moss Model

Figure 4-32

4-37


The chart below shows the ratio (to the exact solution) of the frequency ofthe fIrst mode of vibration obtained for lumped mass models with various numher of nodes: Ha.d..e..s.

Ratio (to exact solution)

2

69.6% 88.5% 93.7% 95.7% 97.9%

3 4 5

10

This shows that the accuracy of the lumped mass model can be increased significantly through the addition ofjust a few mass points, while further addition ofmany more mass points, althoughimproving accuracy, provides diminishing accuracy per node. Therefore the ''best'' model provides a balance - with enough mass points for accuracy, but few enough for computational efficiency. Note that the inevitable inaccuracy in the model can usually be taken care of through peak spreading (during spectral analysis) or sine sweeps (during harmonic analysis). Also evident from this chart is the fact that convergence to 100% accuracy oflumped mass models is from low to high (as mass points are added). This demonstrates that it is difficult to accurately design systems to fail within the flexible range of a load function, since the true modal frequencies, being higher than those calculated, may actually fail within the resonant or rigid response ranges. Fortunately, in many cases a large amount ofmass points are not needed, since the modes ofmost interest (i.e., those with low frequencies) are those representingthe gross movements of the system, not the cantilever (beam hending) modes. The effect of the number ofmass points on the solution time and the solution accuracy is illustrated by modifying the system seen earlier (in Figure 4-23) by breaking i t down into 1foot sections, as shown in Figure 4-33. Note that mode 10 is the first significantly different mode. This is the fIrst cantilever bending mode for single elements that existedin the simple mode!. Therefore the addedaccuracymay he worth the added computational time only if the 10th mode or above contributes significantly to the solution. (The contribution ofthe extractedmodes as a percent ofthe total solution can he determined in CAESAR II's active mass report.) A good rule of thumb to keep in mind is that the frrst 15% of a model's available modes are usuaily quite accurate, while the remaining 85% of the modes are poorer mathematical approximations. Therefore sufficient mass points should be used so that the frrst 15% ofthe potential modes (mass points times translational degrees-of-freedom less rigidly restrained translational degrees-of-freedom) contribute a large amount (for example 90% or more) of the total system mass.

4-38


IOI.JI;

Et IISHOII

QUIT

Original Madel Showing Nades

ROtin

Et

"aSI~

QUIT

Refined Mass Madel Showing Nades

Eignesolution for Job with Extra Mass Nades:

CAESAR Il EI6ENSOlYER (V2.1C Julv 21. 1986! (HC) tD take lut eioenD.ir and continue. lst Case Results

FreQuency 1 = Frtatltllcv 2 = Frtauencv 3:: FrtautRcv 4 = FrtauellCv 5:: FrtQutRcv 6 = Frtaul!llCY 7:: FrtQutRcy 8 = FrtaUtncv 9 = FrtautRcy 11::

.97484 1.55619 2.31444 2.81788 3.53281 3.92918 4.43837 5.68892 6.11755 7.94314

Hz. Hz. Hz. Hz.

Hz. Hz.

Hz. Hz.

Hz. Hz.

Elapsed Elaosed Elaosed Elallsed EliIlIsed EliIlIsed EliIlIstd Elaostd EliIlISed

Ti.e = 1: 3:210.97406 Till! = 1: 4: 9 1. 54213 Ti. = 1/ 4:59 2.28722 Ti.e" 1: 7: 9 2.78333 Ti." 1: 8: 4 3.41256

Ti.t = 1: 9: 1 Ti. = 1/11/32 Ti.t" 1/13:21 Till! = 1:13:28 EI.ned Ti.e = 1:16: 3

3.81449 4.35438 5.63579 6.07498 7.38423

4: 23

~ 4 times as long to run as first case

Figure 4-33 4-39


There have been many criteria developed for lumped mass spacing. For example, the following recommendations have been excerpted from the paper:

On Mass-Lumping Technique for Seismic Analysis of Piping JOHN K. LIN and ADOLPH T. MOLIN Advanced Engineering Department United Engineers & Constructors, 1nc. Philadelphia, Pennsylvania

ERIC N. LlAO Stone & Webster Engineering Corp. Cherry Hill, New Jersey

The lumped mass spacing recommendations given in the paper are:

where L .. distance between two consecutive lumped masses, in ft D outside diameter of pipe, in in. t thickness of pipe, in in. ~ W" weight per ~t length of pipe, in Ib/ft

= =

DESCRIPTION OF METHOD OF ANA LYSIS

In arranging the lump-mass model for seismic analysis of a truee .dimensional piping system, the spacing of

lumps can easily be determined from Eq. (4). However. the following ground rules are to be followed: 1) There must be at Jeast onelump·mass between two supports. First. dividing the length between supports of the pipe by the spacing oflumps L, one obtains the number of lumps required for that portion. If it is not an integer. the next higher mteger shall be used. 2) At least one lump is required in between bends. Similar to 1 above. if the length between bends is not an eu ct multiple of the spacing L. the next higher integer shaU be used.

4-40


3) Lump·mass is required wherever there is a coneen. trated or heavy weight such as valve and flange. 4) Win Eq. (4) includes the weight of the pipe, watec, insulation, and others when applicable. S) The distance between the fust lump·mass and a fIXed end shall be (Lf2). Sinùlarly. the masses just next to a simple support are about a distance of (L/2) away from the support. 6) The weight of the fust mass next to a fixed end must .c~nsist of lhat of one half of the fust portion and one half of the second portion. A diffcrent lumping mcthod consider the fust mass as repre~nting the cn. tire f~~ portion in addition to one half of the second portion may aisa give adcquatc resuit. Howevcr, it is found that the former approach yields better resuit in many cases. 7) Similarly, the portion of piping bctween a simple . support and a lump contributes only one half of its weight to the lump with the other half of its weight being not accounted for in piping stress analysis. Aiternatively. the cntire portion may be lumped into thc mass. 8) Thesimplified lumping formula, Eq. (4), is appli. cable to a three-dimensional piping system with several supports. 9) The lumping method can bc extended to a piping system consisting of several portions each of which has a differcnt crosS section: D. t. and W. 10) More lumps may be needed for short piping.

to

The following three pages provide mass lumping tables that were generated using the recommendations ofthis paper. It must be kept in mind that these are only recommendations, and not the absolutely "correct" spacing by any means.

4-41


LUflPED "ASS 5PACIN6 1ft.)

IN5ULATION ----} FLUID 56 ------) NO" 5th Tht 1 48. .133 1 SI. .179 1 168. .251

1.156

a.1 in. 8.556 1.156

1. 8 in.

IUS6

1.556

1.156

8.156

2.8 in. 1.556 1.856

1.856

3.8 in. 8.556 1.856

---------------------------------------------------------------------------------------------6.8 6.1 6.2

5.B 6.1 6.1

5.7 5.9 6••

5.5 5.7 5.B

5.4 5.6 5.S

5.3 5.5 5.B

5.8 5.2 5.4

2 41. .154 2 88. .21S 2 161. .343

8.1 8.1 8.2

7.6 7.8 8.1

7.3 7.6 7.9

7.5 7.7 7.9

7.3 7.5 7.8

7.8 7.3 7.7

3 48. .216 3 81. .388 3 161. .437

9.7 9.7 9.8

9.2 9.4 9.7

S.9 9.2 9.5

9.3 9.5 9.6

S.9 9.2 9.5

4.9 5.4

4.S 5.1 5.4

4.5 4.7 5••

4.4 4.7 5.'

4.4 4.7 5.8

7.8 7.3 7.6

6.8 7.1 7.5

6.6 7.8 7.4

6.5 6.8 7.2

6.4 6.7 7.2

6.3 6.6 7.1

8.6 S.9 9.3

8.9 9.1 9.4

S.6 '8.9 9.2

S.3 S.7 9.1

8.4 8.7 9.1

8.2 S.Ô 8.9

B.I S.4 8.S

11.1 11.4 111.6 18.7

9.6 11.1 11.3 18.5

11.2 11.4 11.6 11.7

9.7 18.1 18.3 18.5

9.4 9.8 11.1 11.4

9.7 18.1 11.3 111.5

9.4 9.8 11.1 18.3

9.1 9.6 9.9 18.1

11.1

11.4 11.7 11.8 12.8

18.8 11.2 11.5 11.8

11.4 18.9 11.3 11.5

11.8 11.3 11.6 11.8

18.5

11.7 11.9

111.6 11.1 11.4 11.7

11.3 11.5

11.1 11.7 11.' 11.3

~.1

4 48. 4 S8. 4 121. 4 168.

.237 .337 .437 .531

11.9 11.8 11.1 11.1

18.3 11.6 18.8 11.9

9.9 11.2 11.5 11.7

11.6 18.7 11.9 11.1

5 48. 5 81. 5 121. 5 161.

.258 .375 .511 .625

12.1 12.2 12.3 12.4

11.4 11.7 11. 9 12.1

11.8 11.3 11.6 11.8

11.8 12.11 12.1 12.2

6 41. .281 6 81. .432 il 121. .562 6 161. .718

13.2 13.3 13.4 13.5

12.3 12.7 13.1 13.2

11.7 12.3 12.6 12.9

12.9 13.1 13.2 13.3

12.1 12.5 12.8 13.8

11.5 12.1 12.4 12.8

12.5 12.8 13.8 13.1

11.8 12.3 12.6 12.9

11.3 11.9 12.3 12.6

12.1 12.5 12.7 12.9

12.1 12.4 12.7

11.1 11.7 12.1 12.4

8 28. S 31. 6 48. 8 68. 8 SI. B 1811. 8 128. 8 141. 8 161.

.251 .277 .322 .486 .51111 .593 .718 .812 .986

15.1 15.1 15.1 15.1 15.2 15.2 15.3 15.3 15.4

13.6 13.7 13.9 14.2 14.4 14.6 14.8 14.9 15.111

12.7 12.8 13.1 13.5 13.8 14.1 14.3 14.5 14.6

14.6 14.7 14.8 14.9 14.9 15.8 15.1 15.2 15.2

13.3 13.5 13.7 14.1 14.2 14.4 14.6 14.8 14.9

12.5 12.7 12.9 13.3 13.7 13.9 14.2 14.4 14.5

14.2 14.3 14.4 14.6 14.7 14.8 14.9 15.0 15.1

13.1 13.2 13.4 13.8 14.8 14.2 14.5 14.6 14.7

12.3 12.5 12.7 13.2 13.5 13.8 14.1 14.3 14.4

13.7 13.9 14.8 14.2 14.4 14.6 14.7 14.8 14.9

12.8 12.9 13.2 13.5 13.8 14.8 14.3 14.4 14.6

12.1 12.2 12.5 13.8 13.3 13.6 13.9 14.1 14.3

Iii 28. 18 38. 18 48. 10 61. 10 se. 10 111. lB 120. 10 148. 18 168.

.25'

16.8 16.8 16.8 16.9 16.9 17.8 17.8 17 .1 17.2

14.8 15.2 15.4 15.8 16.0 16.3 16.4 16.6 16.7

13.7 14.1 14.4 15.8 15.3 15.7 15.9 16.2 16.3

16.3 16.4 16.5 16.6 16.7 16.8 16.9 17.a 17.a

14.6 14.9 15.2 15.6 15.9 16.1 16.3 16.5 16.6

13.5 13.9 14.3 14.9 15.2 15.5 15.8 16.1 16.3

15.9 16.0 16.2 16.4 16.5 16.6 16.7 16.8 16.9

14.3 14.7 15.8 15.5 15.7 15.9 16.1 16.4 16.5

13.3 13.8 14.1 14.7 15.1 15.4 15.7 16.0 16.1

15.4 15.6 15.8 16.1 16.2 16.4 16.5 16.7 16.7

14.' 14.4 14.7 15.2 15.5 15.8 16.1 16.2 16.4

13.1 13.6 13.9 14.6

.31U

.365 .588 .593 .718 .843 1.888 1.125

11.5

4-42

ILl

11.5

14.9

15.3 15.5 15.8 16.8


LU"PED "ASS SPACIN6 (ft.)

INSULATION ---- > a.B in. FLUID 56 ------} a.ilS6 a.5SB 1.856 NO" Seh Thk 12 28. .251 18.3 15.8 14.5 12 31. .33i1 18.3 16.3 15.1 12 41. .486 18.3 16.7 15.6 12 61. .562 18.4 17.2 16.3 12 88. .687 18.4 17.4 16.7 12 111. .843 18.5 17.7 17.1 12 128. 1.... IB.6 17.9 17.3 12 141. 1.125 18.6 18.1 17.5 12 168. 1.312 18.7 18.2 17.8

8.il56

1.iI in. 8.556

1.156

0.156

2.B in. 0.556

1.156

1.156

17.8 17.9 18.11 18.1 18.2 18.3 18.4 18.5 18.6

15.6 16.1 16.5 17.8 17.3 17.6 17.8 17.9 18.1

14.3 15.0 15.4 16.1 16.5 16.9 17.2 17.4 17.7

17.3 17.5 17.7 17.9 18.8 1B.l 18.3 18.3 18.4

15.3 15.9 16.3 16.8 17.1 17.4 17.6 17.8 18.8

14.1 14.8 15.3 16.1 16.4 16.B 17.1 17.3 17.6

16.8 17 .1 17.3 17.6 18.1 18.1 18.2 18.3

17 .8

3.8 in. 8.556 15.1 15.6 16.8 16.6 16.9 17.3 17 .5 17.7 17.9

1.056 lUI

14.6 15.1 15.8. 16.3 16.7 17.11 17.2 17.5

.251 .312 .375 .437 .S93 .751 .937 1.893 1.258 1.416

19.1 19.2 19.2 '19.2 19.3 19.3 19.4 19.4 19.5 19.6

16.4 16.9 17.2 17.5 17.9 IB.3 18.5 19.7 19.9 19.8

14.9 15.5 15.9 16.3 17.1 17.4 17.9 18.1 18.4 18.6

18.6 IB.7 18.8 19.9 19.8 19.1 19.2 19.3 19.4 19.4

16.2 16.6 17.0 17.3 17.8 IB.l 18.4 18.6 18.B 18.9

14.8 15.3 15.8 16.2 16.8 17.3 17 .8 18.1 18.3 18.5

18.1 18.3 18.5 IB.6 18.8 18.9 19.1 19.2 19.2 19.3

IS.9 16.4 16.8 17 .1 17.6 18.1 18.3 18.S 18.7 18.8

14.6 15.2 IS.6 16.0 16.7 17.2 17.7 17.9 18.2 18.4

17 .6 17.9 18.1 IB.2 18.5 18.7 18.9 19.8 19.1 19.2

15.6 16.1 16.5 16.8 17.4 17.B 18.1 18.4 18.6 18.7

17.5 17.8 19.1 18.3

.2S8 .312 .375 .588 .656 .843 1.831 1.218 1.437

17.2 17.7 18.1 IB.7 19.1 19.5 19.8 21.8 20.2 20.3

IS.6 16.2 16.7 17 .4 18.1 18.6 19.iI 19.4 19.7 19.9

19.9 21.1 21.1 21.2 2i1.4 28.5 21.6 21.6 21.7

1. 593

28.4 21.5 21.5 28.5 28.6 28.6 28.7 21.8 28.8 21.9

21.8

17.8 17.5 17.9 18.5 19.1 19.4 19.7 19.9 2i1.1 21.3

15.5 16.1 16.6 17.3 17.9 18.5 18.9 19.3 19.6 19.8

19.4 19.6 19.7 19.9 21.1 28.3 21.4 21.5 28.6 21.7

16.8 17.3 17.7 18.3 18.8 19.2 19.5 19.8 21.8 28.2

15.3 15.9 16.4 17 .2 17.8 18.4 18.8 19.2 19.5 19.7

18.8 19.1 19.3 19.6 19.9 28.1 21.2 21.4 211.S 28.6

16.5 17.11 17.5 18.1 18.6 19.1 19.4 19.7 19.9 211.1

15.1 15.7 16.2 17.11 17.7 18.3 18.7 19.1 19.4 19.6

18. .258 28. .312 38. .437 48. .562 68. .758 88. .937 188. 1.156 lB 128. 1.375 18 148. 1. 562 18 16B. 1.781

21.7 21.7 21.7 21.8 21.8 21.9 22.8 22.8 22.1 22.2

18.1 18.5 19.3 19.8 28.3 21.7 21.0 21.2

16.2 16.9 17.8 18.5 19.2 19.7 2i1.2 211.5 20.8 21. Il

21.1 21.2 21.4 21.5 21.6 21.7 21.8 21.9 22.8 22.1

17.8 18.3 19.1 19.6 21.2 28.5 20.9 21.1 21.3 21.5

16.1 16.7 17.7 18.4 19.1 19.6 21.1 28.5 28.7 21.8

21.5 21.8

17.5 18.1 18.9 19.4 28.8 21.4 28.7 21.0 21.2 21.4

15.9 16.6 17.5 18.2 19.8 19.5 2i1.0 21.4 28.6 21.9

21.8 28.3 21.7 211.9 21.2 21.3 21.5

17.3 17.9 18.7 19.3 19.8 28.3 28.6 23.9 21.1 21.3

15.8 16.4 17 .4 18.1 IB.B 19.4 19.9 2i1.3 2a.5 21.8

28 28 28 28 28 28 28

22.5 22.9 22.9 22.9 23.8 23.1 23. 1 23.2 23.3 23.4

18.7 19.7 28.4 20.8

16.7 IB.8 IB.9 19.3 28.2 28.8

22.2 22.5 22.6 22.7 22.8 22.9 23.0 23.1 23.2 23.3

18.5 19.5 28.2 20.6 21.2 21.6 22.0 22.2 22.5 22.7

16.6 17.9 IB.7 19.2 28.1 21.7 21.2 21.5 21.9 22.1

21. 7 22.0 22.3

18.2 19.3 21.0 21.4 21.1

16.5 17.7 18.6 19.1 21.8 211.6 21.1 21. 4

21.1 21.6 21.9

18.8 19.1 19.8 2a.2 28.9 21.4

16.3 17 .6 18.5 19.11 19.8 20.5 21. Il 21.3 21.7 21. 9

14 11. 14 21. 14 31. 14 48.

14 14 14 14 14 14

68. 88. 111. 128. 141. 168.

16 16 16 16 16 16 16 16 16

18. 21. 38. 48. 61. 88. 181. 128. 141. lb 161.

18 18 18 18 18 18 18

10. 28. 38. 48. 61. 88. 188. 2~ 128. 20 141. 21 168.

.250 .375 .51i1 .593 .812 1.131 1. 281 1,5BI 1. 758 1.968

21. -4

21.6

21. -4

21.B 22.1 22.6 22.7

21.3

21.6 21.9 22.2

4-43

21.11 21.2 21.4 21.S 21.7 21.8 21.9 22.1

22.4

22.6 22.7 22.9 23.0 23.1 23.2

21.5 21. 9

22.1

22.4

21.8

22.6

22.8

21. 6

21.7 21.9

22.1

22.4 22.6 22.7 22.8 23.11 23.1

21.8

22.8 22.3 22.5

14.4 15.1 15.5 15.8 16.6 17 .1


lUftPED ftASS SPACIN6 (ft.)

------------------------INSUlATION ----} B.8 in. 1.1 in. 2.11 in. 3.1 in. FlUID S6 ------) 1.856 8.556 1.156 11.156 Il.556 1.15e 8.856 a.5Se 1.15S a.a5e B.5se use NO" Sth Thk ---------------------------------------------------------------------------------------------22 11. .25' 23.9 19.3 17.3 23.3 19.1 17.1 22.7 18.9 17.1 22.1 18.6 16.B 22 21. .375 24.8 21.5 28.3 18.5 23.1 18.6 23.5 28.1 18.3 22.7 18.2 19.8 22 38 • •511 24.11 21.2 19.5 23.7 21.8 19.4 23.4 21.8 19.2 23.1 19.1 21.6 22 61. .875 24.1 22.4 21.1 23.9 23.7 2B.9 22.2 21.1 22.1 23.5 21.9 28.8 22 81. 1.125 24.2 22.8 21.8 24.8 22.7 21. 7 23.9 22.6 21.6 23.7 22.4 21.5 22 III. 1.375 24.3 23.1 22.2 24.1 22.2 24.1 22.9 22.1 23.9 22.8 23.' 22.' 22 121. 1.625 24.3 23.4 22.6 24.2 23.3 22.5 24.1 23.2 24.8 22.5 23.1 22.4 22 141. 1.875 24.4 23.6 23.5 22.9 24.2 23.5 22.B 24.1 23.4 22.7 22.9 24.3 22 161. 2.125 24.5 24.4 23.8 23.2 23.2 24.3 23.7 23.1 24.2 23.7 23.6 23.1 24 24 24 24 24 24 24 24 24 24

.251 .375 .562 .687 .968 1.218 1.531 1.812 2.1162 2.343

25.1 25.1 25.1 25.1 25.2 25.3 25.4 25.4 25.5 25.6

19.9 21.1 22.2 22.7 23.4 23.8 24.2 24.5 24.7 24.9

19.1 2B.4 21.1 22.1 22.7 23.3 23.7 24.1 24.3

24.3 24.6 24.8 24.9 25.1 25.1 25.2 25.3 25.4 25.5

19.7 21.9 22.1 22.5 23.3 23.7 24.1 24.4 24.6 24.8

17.6 19.8 2'.3 21.8 22.a 22.6 23.2 23.6 23..9 24.2

23.7 24.1 24.5 24.6 24.8 25.1 25.1 25.2 25.3 25.4

19.5 28.7 21.8 22.3 23.1 23.6 24.11 24.3 24.5 24.7

17.5 18.9 21.2 21.9 21. 9 22.5 23.1 23.5 23..8 24.1

23.1 23.7 24.2 24.3 24.6 24.8 25.1 25.1 25.2 25.3

19.3 21.5 21.7 22.2 23.1 23.5 23.9 24.2 24.4 24.7

26 11. .312 26 28. .511

2i""

26.1

21.2 22.6

18.9 21.6

25.5 25.7

21.11 22.4

18.8 2'.5

24.9 25.4

28.8 22.2

18.7 2'.4

24.4 25.'

22.'

18.6 2'.3

28 111. .312 28 28. .511 28 3•• .625

27.11 27.1 27.1

21.7 23.2 23.8

19.4 21.1 21.9

26.4 26.7 26.8

21.5 23.11 23.7

19.3 21.' 21.8

25.9 26.3 26.5

21.3 22.9 23..5

19.2 28.9 21.7

25.4 26.' 26.2

21.1 22.7 23.3

19.8 21.8 21.6

31 18. .312

38 28. .5111 38 38. .625 38 48. .758

28.8 28.8 28.8 28.1

22.3 23.8 24.5 25.1

19.8 21.6 22.4 23..1

27.4 27.6 27.7 27.8

22.1 23.6 24.3 24.8

19.7 21.5 22.3 23.1

26.8 27.2 27.4 27.5

21.9 23.5 24.2 24.7

19.6 21.4 22.2 22.9

26.3 26.9 27.1 27.3

21.7 23.3 24.' 24.5

19.5 21.3 22.1 22.8

32 32 32 32

18. 211. 38. 41.

.312 .5811 .625 .688

28.9 28.9 28.9 29.8

22.8 24.4 25.1 25.4

21.2 22.1 22.9 23.3

28.3 28.5 28.6 28.7

22.6 24.2 24.9 25.2

21.1 22.' 22.8 23.2

27.7 28.1 28.3 28.4

22.4 24.11 24.8 25.1

28.1 21.9 22.7 23.1

27.1 27.8 28.8 28.1

22.2 23.9 24.6 24.9

19.9 21.7 22.6 23.11

34 34 34 34

11. 21. 311. 48.

.312 .511 .625 .688

29.8 29.8 29.8 29.8

23.2 24.9 25.7 26.8

21.6 22.5 23.4 23.8

29.1 29.4 29.5 29.5

23 •• 24.8 25.5 25.8

21.5 22.4 23.7

28.5 29.11 29.2 29.2

22.9 24.6 25.4 25.7

21.4 22.3 23..2 23.6

28 •• 28.6 28.9 28.9

22.7 24.4 25.2 25.5

28.2 22.2 23.1 23.5

36 36 36 36

ut .312 28. .588 311. .625 41. .758

311.6 31.6 38.7 311.7

23.7 25.5 26.2 26.8

28.9 22.9 23.8 24.6

311.11 3.8.2 311.3 38.4

23.5 25.3 26.1 26.7

21.8 22.8 23.7 24.5

29.4 29.8 30.8 31.1

23.3 25.1 25.9 26.5

28.7 22.7 23.6 24.4

28.8 29.4 29.7 29.9

23.1

25.8 25.8 26.4

28.6 22.6 23.5 24.3

42 211. .5111 42 31. .625 42 48. .758

33.1 33.1 33.1

2b.9 27.B 28.4

24.8 25.1 25.9

32.6 32.8 32.8

26.7 27.6 28.3

24 •• 25.0 25.8

32.2 32.4 32.5

26.6 27.5 28.1

23.9 24.9 25.7

31.8 32.1

26.4 27.3 28.8

23.8 24.8 25.6

II.

2•• 31. 48. 68. 81. 188. 121. 141. 1611.

17.7

23.3

4-44

32.2

21.6

17.3

IB.7 21.1 28.7 21.8 22.4 23.8 23.5 23.8 24.1


4.6 Types of Analysis There are a number of ways ofperforming a dynamic analysis. These methods are describe,d in this section.

4.6.1 Time History Analysis Time history analysis is the most accurate means of performing dynamic analysis, if the computing power is available. Load data is assembled for a number oftime slices throughout the duration of the dynamic event. This data is then read in by a computer program which simulates the dynamic response of the system, solving the dynamic equation of the system for each time slice. This is done by saving the dynamic parameters (acceleration, velocity, displacement, etc.) for each node point and each mode of vibration, and integrating these values over the time slice. The analyst can then review the solutions for ail time slices to determine the largest displacements, forces, stresses, etc. occurring at any location in the piping at any time during the course of the load. This type of analysis is accurate for loads with known force-time profiles, such as relief valve discharge or fluid hammer, but not for unpredictable loads such as earthquakes. This type of analysis is also very computer intensive, slow, expensive, and therefore suitable implementations of this method are rare.

4.6.2 Seismic Spectrum Analysis Time history analyses cannot be performed for seismic loads, since earthquakes cause random motion. The solution to this is to perform spectrum analysis (also known as the response spectrum method). A typical earthquake time history might appear as shown in Figure 4-34 .

.-

c

Q)

E Q)

u

a

0. Cf)

o

Avg 15-30 seconds

Figure 4-34 This time history is the plot of the earth's acceleration as experienced by the piping system through each ofits supports. The time history plot can be measured using accelerometers resting on the ground near the earthquake site, as shown in Figure 4-35.

4-45


y

ACCELEROMETERS MOUNTED TO MEASURE (' HORIZONTAL AND VERTICAL COMPONENTS RESPONSE PLOTS X -r---~ _ _ _ _ _

Z

L.. _ _ _ _ _

+

SHOCK WAVE

~

JEARTH<--

-~---------------~

Figure 4-35 To try to analyze this complex motion in a time history would be too burdensome. Additionally, the time-dependent waveform is different for each earthquake, even those occurring at the same site. 4.6.2.1 Generation of the Response Spectrum To simplify the analytical definition of the earthquake, it is necessary to get the random waveform shown above into some simple frequency-content plot. The most predominantly used frequency-content plot is the response spectrum. A response spectrum for an earthquake load can be developed by placing a series of single DOF oscillators on a mechanical shake table (as shown in Figure 4-36) and feeding a "typical" earthquake force time history (typical for a specifie site) through it, measuring the maximum response (displacement, velocity, or acceleration) of each oscillator.

4-46


."pon'"hA

"a,im"m For Each of the Single Degree of Freedom Bodies

}".,A~ "

:>

A Response 4'V,V ïfI~

:> Time

B Response

1" .,.. .A. "'1"

'''''V~V

cp . . ,

Measured Earthquake Movemenl

Response

1

V'

pr ,

:> Time

)

Time

Note:

Maximums occur at diffe,.-ent limes.

Figure 4-36 The expectation is that even though ail earthquakes are different, similar ones should produce the same maximum responses, even though the time at which they occur will differ with each individual occurrence. (Responses will be based on the maximum ground displacement and acceleration, the dynamic load factors determined by the ratios of the predominant harmonie frequencies of the earthquake to the natural frequencies of the oscillators, and system damping.) Response spectra for a number of damping values can be generated by plotting the maximum response for each oscillator. A plot of a set oftypical response spectra is shown in Figure 4-37 .

.8

.7

!

1% Critical

.6

Acceleration

CG)

Damping~\

5% Critical Damping f \

.5

10% Critical Damping /

.4

/ /

.3

1 1

.2

) ,//r

•1

\ \ \ \ \

l '........

---~ .

.1

.2.3

.5

.7

1

Computed Frequency CCPS)

Figure 4-37

4-47


Seismie response speetra resemble harmonie DLF eurves, sinee sei smic loads evidenee strong harmonie tendeneies. For example, as damping value inereases, the system response approaehes the ground motion. Seismie speetra usually also show strong evidence offlexible, resonant, and rigid areas. Spectra may have multiple peaks due to filtering by the building and/or piping system; however multiple peaks are usually enveloped in order to account for uncertainties in the analysis. Seismic response spectra peaks are typieaily spread to aeeount for inaceuracies as weIL The idea behind the generation of the response spectra is that a system's modes of vibration will respond to the load in the exact same manner as will a single degree-of-freedom oscillator. For example, the water tower in Figure 4-38 ean be modeled as a single degreeof-freedom oscillator in the lateral direction. Ifits natural frequency is 6 Hz, it would be expected to displaee a maximum of about 6.3 inches when subjected to the earthquake creating the accompanying response spectrum.

{f;=

6 Hz

271

6.0

U1 w I

u

~

MAX. DISPLACEMENTS THAT OCCURRED FOR EACH OF THE SINGLE DEGREE OF FREEDOM BODIES. (MAX. VALUES COULD OCCUR AT ANY TIME DURING THE EARTHQUAKE.)

5.0 4.0

Vl 1-

z

w w u

:::;: oct

3.0 2.0

....J

0. Vl

c;

1.0 123456789 FREQUENCY (HZ)

Figure 4-38 Maximum responses can be plotted in terms of maximum displacements, veloeities, and/or aceelerations. For example, the response speetra in Figure 4-39 is plottedin terms ofail three simultaneously.Tripartitecurves,suchasthesetakenfromNuclearRegulatoryGuidel.60, are fairly common.

4-48


0.2

0..

1

Z

1

10

FRF NENev, . . (Uë)

•

10

100

Figure 4-39 A fairly subtle point concerning the various response spectra is that the displacement and velocity response values are measured relative to the earth, while the acceleration response value is measuredrelative to anon-moveable coordinate system. Stiffness and velocity forces on the structure due to the earthquake are caused by the relative movement of the earth to the mass, while the inertial force on the structure is caused by the absolute acceleration of the mass. Displacement, velocity, and acceleration terms ofthe spectra are aIl related bythe frequency: d

=

v/ro =a/ ro2

4-49


Where: d

=

dis placement from response spectrum at frequency ro , in

v

=

velocity from response spectrum at frequency ro, in/sec

ro

=

angular frequency at which response spectrum parameters are taken, rad/ sec

a

=

acceleration from response spectrum at frequency ro , in/sec2

For example, at 0.75 Hz (4.712 rad/sec), the response spectrum shown in Figure 4-39 has a displacement value of 30 inches. The velocity at that frequency is 30 x 4.712 = 141 in/sec, and the acceleration is 30 x 4.712 2 = 666 in/sec2 . Note that on the right hand side of the graph the response acceleration becomes constant, while on the left hand side of the graph the response displacement is constant. The flat right hand side ofthe response spectrum diagram indicates that there is no dynamic amplification for oscillators having natural frequencies above some rigid value (in this case around 33 Hz.). Oscillators with rigid natural frequencies move along with the earth, so the maximum acceleration of the oscillator is equal to the maximum acceleration of the earth. The acceleration conforming to the high frequency flat portion of the curve is called the Zero Period Acceleration, or the ZPA. This is the maximum acceleration that the earth experiences at any time during the earthquake. The term "Zero Period" is used because a body with a zero period has an infinitely high frequency, and is certainly in the rigid range of the earthquake. The left hand, constant displacement, side of the response spectrum curve indicates that there is a low frequency cutoffwhere the earth moves too quickly to affect the oscillator (it's in the flexible range). In this case, the absolute acceleration is near zero, and the relative displacement is equal to the maximum displacement of the earth. For all modal natural frequencies below the low frequency (flexible) cutoff: 1) The system forces due to mass acceleration are zero. 2) The relative movement ofthe earth and the mass are the same for all single degree of freedom bodies below the low frequency cutoff. (This relative movement is equal to the maximum displacement of the earth.) For all modal natural frequencies above the high frequency (rigid) cutoff: 1) Ali of the loading on the piping system is due to the acceleration of the masses. 2) There is no relative movement between the masses and their supports due to the modal component associated with the high natural frequency. 3) Acceleration loadings are purely static and are mode independent. Nuclear Regulatory Guide 1.60 (published in December, 1973), included on the following six pages, describes the requirements for constructing design seismic response spectra for nuclear power plants. 4-50


P.t!V'iion 1 Oecember 1973

U.S. ATOP.'lIC ENERGY COMMISSION

REGULATORV GUIDE DIRECTORATE OF REGULATORY STANDARD.

REGULATORY GUIDE 1.60

DESIGN RESPONSE SPECTRA FOR SEISMIC DESIGN OF NUCLEAR POWER PLANTS A. INTRODucnON

Criterion 2. "Deslgn Bases for Protection Against Phenomena." of Appendix A. "General Design Criteria fClr Nuclear Power Plants." to 10 CFR Part 50. ·'U..:enstng of ProduCllon and Utilization Facililies." requlres. in paM. th31 nuc1ear' power pbnl structures. systcms. and cllmponents important to safety be uC~Ip1cd 10 withstancJ the effects of earthquakcs. r~
B. DISCUSSION III order to ~pproxlmate the intensity and thereby estln14te the maximum ground aeceleration l of the exoected strongest ground motion (SSE) for a given site. proposcd Appendlx A 10 10 CFR Part 100 speciftes a number of required investigations. Il does not, however, ;:wc ;) method lor defining the response spectral ,lH,,:sp
TIIe' recllrt.ieù

~r"lIlld

accclerallons and response penvide J basis for Ihe rall(\/lal 'design 01 ~Irlll.:lures 10 resist earthquakes. The DcSI~1 I
(\1

p;.I~1

• SC'\: (}Cfanlllons

~I

c~rlhqllakcs

Ihe end u( lhe lCulde.

extensive study has been described by Newmark and 8lun'le in rcferences 1. 2. and 3. ACter revie"";ng these rcferenced documents, the AEC Regulatory staff has dctcrmined as acceptable lhe following procedure for del"ining the Design Response Spectra representlng the effects of the vibratory motion of the SSE. 1/2 Ihe SSE. Jnd the Operating Basis E2rthquake (OBE) on sites underiain by either rock or soi! deposlIs and covering ail frcquencies of interest. However, fOJ unusually soft sites., modification to trus procedure will be required. ln this procedure. the configurations of the horizontal compement Design Response Spectf3 for ~ach of the two mutually perpendicular horizontal axes are shown in Figure 1 of Ihis guide. These shapes agree with those developed by Newmark, 81ume, and Kapuf in 1 c:ference 1. In Figure 1 the base wOIçam eonsists of threc parts: the bOHom line on the !eft part represcnl~ . the maximum ground di.splacement. the bouom !int on 1he nght part represents 'the maximum acceleration. and the middle pan de pends on the maxImum veloeit)'. The horizontal compone nt Design Rc::ponse Spcctra in Figure 1 of this gUIde correspond 10 a maximum hurizontal ground acc~/e1Qtion of 1.0 g. The maximum ground displ:!cement is taXen proportionaJ 10 the maximum ground acceleration, OInd is sel at 36 illches fur a ground acc:eleration of 1.0 g. The numerir.:al values of design displacements, lIelocities, and .ccelera tions for the horizontal component Des.ign Respons.e Speetra are obtained by muJtiplying the corresponding values of the maximum ground dispiacemc/ll ~nd J!:cele~atlor. by the factors gJ1len in Table 1 of this guide. The dlspla~'cment region lmes of the Design Rc:sponse Spectra are parallel III Ihe maxImum ground displacemenl hile :.lI1d ~re shown on the !eft of Figure !. The velu.: Il Y Teglon Itnes slupe dowl1ward from a frequenc)! ni 0.25 0.:1" (control polOt DI to a frequency of 2.5 c;::: :'::lIrol pOlnl C) and Jre shown at the top. The remalOinJ; IwO sets uf lines ~tween the frequencles of 2.5 cps and 33 <:~ (conlrol ;:Joint A), wlth a break at a frequenr.:y of <; cps (cont roi

4-51


e.arthquJÏte or (2) have physical characleristics th al could signlfacanlly affeel the speclral pallern of Input mOUlin, such as bclng underl:ain hy pnur suil ùcpn"llS. the prnœdure descri'~d "hoye will IInt apply. In Uiesc cases, the DeSign Rcspt>llsc Spectr:1 shuuld he ùcvelopeù mdlvu..lually accnrùing ln the sile characterislics.

;Joint B). .:onstllut.: the :1cœler:llion rcgion (If the hUIiJunloil Design Respnnsc Spe,·u.a. For frcquenclcs

;lIi!her

Ih.1Il

3.1 ..:ps. the

InUIOlUIll

gruund acccleralhlll

;tn.: rl'lll~'SC'I\IS IIIl' l)cslgn R<"p\lnsc Spcctra. ·1 il.: vC'rt ...:.ti

..:urr~'!;pul\dlng

,·omp,menl [)cslgn Rcspunsc Spcl:lra

lhe maximum nuntuntld "uund ur 1.0 g arc ~h()wn in Figure 2 of this guide. The numencal v31ues or design displiCcmcnls, vclocilies. Jnd J.relcr .. tlllns in these spcctra arc obtaincd by Illuillplylnl!: lhe correspondang values of the maximum hcJn:llnla/ gruund motion (aceeleration .. 1.0 g and dlspl;&l,;ernent = 36 in.) by the factors givcn in Table Il of IhlS gUide. The dlsplacement region lines of the Design Response Spectra are par.allel to the maximum ground J.asplal,;emenl hne and are shawn on the left of Figure 2. The velocity region Unes sl!)pe downward from a r~equency of 0.25 c1'5 (control point D) to a frequency of :U cps (control pomt C) and arc show at the top. The remaining two sets of Iines between the frequencics ,If .1 ..5 .:ps and 33 cps (control point A). with 1 break at l!le r"rcquency 0 f 9 cps (c:ontroJ point B), eonstitute the .i.:.:eicratlltn region of the vertical Design Rcspansc S p.:.: 1rOi. Il should be noted that the vCrtical Design Kcspun~ Spectra values are 2/3 thase of the horizontal f)cSII/.II Responsc Spectra (or frequendes leu than 0.25: :\H 'requencies tugher than 3.5, tbey are thi: 5amC, while :11<' ral'" vanes betwcen 2/3 and 1 for frequencies Xlween 0.2S and 3.5. For frequencies higher than 33 .:ps. the DeSI~ Response Speclra foUow the maximum pound al,;cderation line. III

C. REGUlATORY POSITION

un"('IC'ru/lfIn

The horizontal and vertical component Design Response Spectra in Figures 1 and 2, respecûvcly, of tlUs guide conespond to a maximum horizontal ground Jcceierallon of 1.0 g. For sites with difrcrent dcceleration values specified for the daim earthquake, t he Design Response Spectra should be linèarly scaled (rom Figures 1 and 2 in proportion to the specified nwumum horizontal ground ac:cderation. For lites that ( 1) are relati't'ely close 10 the epic:entcr of an expected

4-52

1. The horlZOntal component ground Design Response Spectra. without soU·structure interaction c(reets, of the SSE, 1/2 the SSE, or the 08E on sites underlain by rock or by sail should be linearly scaled from Figure 12 in proportion ta the maximum horizontal ground acceleration specifw=d for the earthquake chosen. (Figure 1 corresponds to a maximum honzontal ground ac~leration of 1.0 g and acc:ompanyingdisplacemcnt of 36 in.) The applicable multiplication factors and control points are given in Table 1. For «Sampin, ratios nOl included in Figure 1 or' Table l, a linear interpolation should be used.

2. The YCrtical companent ground Design Rcsponsc Spectral without saü·structure interaction efiects, of the SSE. 1/2 the SSE, or the ODE on sites underlain by rock or by soil should be linearly scaled from Figure 22 in proportion to the maximum horizontal ground Jcceleration specifacd for the eanhquûe chosen. (Figure 2 is bascd on a maximum horizOllllZ1 ground lZCCele,anOil of 1.0 g and aceompanying displacemenl of 36 in.) The applicable multiplication factors and control points are given in Table Il. For damping ratios not induded in Figure 2 or Table Il, a linear interpolation should be uscd. 'This docs not apply to sites which (1) are relativel)' clOIe to the epicezller of an cxpectcd eanbquake or (2) whic:h hue ph)'sicaJ c:huac:terisIÏCI tbat coukl sipifacantl)' arfect the spcc:tll.l mmbinauon input motioa. The Desipa Rcspoll.le

0'

Spec:tn ror such situ abould be dcYeloped on a c:uo-by-aae

basiL


DEFINITIONS ReapolUr SpccCnlm

a pau, .. l'

mC3RS

'11c

rdaliuMlllp uhtained by In.dyzin[l. eV:llualinll. and

maximum

s.alls1iellly _,"'\Im"'nln, • number

respl"SC (acœlcrallon. vclodty. ,JI dlsplacemcnl) uf • family of ide.lized sin'e-depee-oC.freedom dlmped osciUators I l • function of natural trcquencies (or periods) of the osc:illalon to a spcciflCd vibralory molion input at their supports. WMft obtaiDcd tram. recorded eart"ake ,record. the laponse speçtrum tends ta be inqular. with 1 number of peaks and

nr indlvidual rcspllOlc

spcctr. dcrivcd rrClm the records of sianifaçanl pllt carl hquakes.

Maximwa (pak) Grauad ,Ac:ceMntioD specified ror • given site mans that valu. of the acceleraûon which corresponds to zero pcriod in the desicn respollse' s~çtra for tbat site. At zero period the rcsponse .peetra acceleratioa il tdentical for aD dampin, values and is equal ln the maximum· (peak) l10Uftd aççcleraUoR spedfied fcr that lite.

-sn

YIlleys. lDp Ra~ SpeCUll111 is 1 relatiwly lIftooth

TABLE 1

HORIZONTAL DESIGN RESPONSE SPECTRA RELATIVE VALUES OF SPECTRUM AMPLIFICATION FACTORS FOR CONTROL POINTS

,-" of

C,ltIcIII DPnpi..

Amplification Fec:tcn for Conttol Foin. Acca. . .ion'

A(3I_

I*pl8alnnt' -

1

ac.CIIIl

Cl2.5 epù

Dlo.a ...

005

1,.0

".96

2.0 5.0 1.0 10.0

1.0

3.54 2.61

5.95 4.25 3.13

3.20 2.50

2.27

2.72

1.88

.1.90

2.28

1.70

1.0

J.o

1.0

2.05

·u.xi.... ,_tilt........

lI'Ouad

pIIIId dilpllClllMBt 1& eau. propartiDlla1 10 awdaalllll 36 la. for pounclac:cdlntioa of 1.0 payity.

'ACCIIJaaÜCID lJId 4iapIac:eIacnt unpIific:atioa lac:ton IN takaD froIII r_lllJll-.lldaaI .... iBnf_l.

4-53


TABLE Il

VERTICAL DESIGN RESPQNSE SPECTRA RELATIVE VALUES OF SPECTRUM AMPLlFI~ATION FACTORS F9R CONTROL POINTS

Perelli nt

AmpiifiQtion Fileto" for Control Points

of

Acx:el....'ion'

Critlal

o.mP"9

0.5 2.0 5.0

7.0 10.0

1

A(33cpal

1.0 1.0 1.0 1.0 1.0

[)isp&lcement' . a

1

C(~:5 epcl

B(9 cpsl

Z.6!

5.67'\ 4.05 Z.9R

2.27 1.90

:!.17

4.96

3.54

0(0.25 q:l51

:.59

2.13 1.67 1.37 1.25 1.13

Maximum IJound displaœmcnt is taken proporUon.l to maximllm :lcL-eler.ation and Il 36 in. ((Ir pound :u:,,-eler:ltion (lI' 1.0 plvity.

)trt\u~

~

Acçclcroation Impllf1l::llion l'aCtllrS fnr

Ih~

ycrlil:ili dcsip

te.~J'OnlC

ICspOn.'Ie ~pcctra al a giveft amplificalÎl,n (;II:tun arc 2/3 Chose fur hari-

,.-,'\:lra an: cqual tn lbu., rnr horimnlill desll(n

rrcquellCY, whCJca. cI.cement lnnlal dcsi~ responsc spcc:Ua. 't1Icse ratins bcllll\.'Cl\ lbe amplific:;ztion factors fur Ihe lWU dcs~n rcspunsc IpCCUllre in apcemenl ""ilJllbosc recommmd.ed in rcferenœ 1. .

'l'hese ",.. lues WCIC: c:ha",od 10 maltc Ihis l:able c:nnsixtenl with the disnl' \l("rtial camponeftl$ in Scction R of Ihis ,;uide.

~u5Si"n

REFERENCES

1.

Ncwm:uk, N. M., John A. Siume. and lCanwar K. Kapur. "Design Rcsponse Spectra for Nuclear Power PI:lIIls:· ASCE Structural Engineering Meeting. San Francisco. April 1973.

N. M. Newnlarlc. Consulting Engineering S~rvices. UA Study of Vertic.al and Horizontal· Earthquake

4-54

Spectra," Urbana, lIlinois. USAEC Contract No. AT(49-S)-2b67, WASH·12S5. April 1973. 3.

John A. Slume &. Associates. '4Recommcndations for Shape of Earthquakc Response Spccua.·· San Fr:tncisco. Califomia. USAEC Contract No. AT(49.S)-3011, WASH·1254. February 1973.


0.1

0.2

0.5

1

2

5

10

20

50

100

FRF1UENCY,qJI FIGURE 1. HORIZONTAL DESIGN RESPONSE SPECTRA - SCALED TO 19 HORIZONTAL GROUND ACCELERATION

4-55


100

-.= u

!

50

>" ~

Ü 0

...

.",

>

20

10

FREQUENCY,qII FIGURE 2. VERTICAL DESIGN RESPONSE SPECTRA - SCALED TO 1; HORIZONTAL GROUND ACCELERATION

4-56


4.6.2.2 Application to a Multi-Degree-of Freedom System

Reviewing the dynamic equations for an earthquake problem:

M

J

X(t)

x(t)

=

system displacement (relative to ground), as a function oftime

xg(t)

=

dis placement of ground (introduced through system supports), as a function oftime

Figure 4-40 The structural forces within the system are based upon the displacements relative to the ground: Fstruct

=K

x(t)

The inertial force on the mass is based upon the total acceleration, notjust the acceleration relative to the earth: Fmass

=M [x(t) + xg(t)]

Therefore, ifthe spring is very, very stiff; the mass will move along with the earth and there will be no relative acceleration hetween the earth and the mass, but there will still he the inertial force. The equation for earthquake motion hecomes: M [x(t) + xg(t)] + Kx(t) = 0 Moving the ground acceleration to the right side of the equation: M x(t) + K x(t) = -M xg(t) As it turns out, the mode shapes of the piping system (as extracted by the Eigensolver) have a unique set of characteristics. They are a peculiar set ofvectors which serve to rotate the coordinate axes for each node.

4-57


y

v 1 / 1 /

----y Z

/

[~J ) ~~ ~ = ) ~~ ~ \ (X3) (v3) Matrix which rotates the point CD from the X,Y,Z coordinates into alternate coordinates

Figure 4-41 This rotation will be seen to serve a unique purpose. Proceeding by multiplying both the acceleration and the displacements by the matrix of mode shapes, this gives the same accelerations and displacements in a different, rotated coordinate system: x -> v v(t) = cI> x(t); v(t) = cI> x(t) Multiplying these expressions by the inverse of cI>, i.e. cI> -1 converts the rotated coordinate system back to the original coordinate system: cI> -1 v(t)

= x(t);

cI> -1 v(t)

= x(t)

Insertingthese last two expressions backinto the dynamic equations forearthquake motion: M cI> -1 v(t) + K cI>-1 v(t) = -M xg(t) Multiplying both sides ofthis equation by cI> again: cI> M cI>-1 v(t) + cI> K cI>-1 v(t)

= -cI> M xg(t)

The "sets ofequations" represented by the matrix expression above are displayed graphically in Figure 4-42.

4-58


n

n

cp

M

Fully Populated

Diagonal

K Fully Symmetrically 8inded Populated

v(t) Fully Populated

cp-1

v(t)

Fully Populated

This multiplication results in a "Rotated" stiffness matrix that looks like:

This multiplication results in a "Rotated" mass matrix that looks like:

N

N

'" K

M

Diagonal

Diagonal

Figure 4-42 This is a unique property of the dynamic mode shapes of any arbitrary structural system: <1> M <1>-1 v(t) + <1> K <1>-1 v(t)

M

v(t) +

K

= -<1> M xg(t)

v(t) = -<1> M xgCt)

The "diagonalization" ofthe M and K matrix took what was previously an intricately linked set of equations and found a generalized coordinate system that completely decoupled each equation in the system, so that each equation can now be de aIt with as a single degree-offreedom system. The new set ofequations appears in Figure 4-43:

4-59


n

v(t)

K

v(t)

= -CPMXg(f)

This represents N independent equations of the form: ml VI(t) + k l VI(t) = - [<1> M xg(t)h m2 V2(t) + k 2 V2(t) = - [<1> M xg(t)h mN VN(t) + k N VN(t) = - [<1> M Xg(t)]N

Figure 4-43 Note that each of the equations above represents a different single degree-of-freedom body. Remembering how the earthquake shock spectrum was generated, one knows exactly what the maximum displacement for each of the single degree-of-freedom bodies is going to be under the seismic load. Doing the previous multiplications by _ provided a set of equations that were "decoupled", and so were very easy to solve. These equations were of the form:

-

-

M v(t) + K v(t) = -<1> M xg(t) The response spectrum, by definition, gives us solutions to the equation:

-

-

M x(t) + K x(t) = - M xg(t) The form of these equations is very similar except for the right hand sides. Introducing a constant "C" into the first equation, it becomes:

-

-

-

M v(t) + K v(t) = C[ M xg(t)]

4-60


ln this form, the solution to the following equation can be found from the response spectrum:

1\1

v(t) +

K v(t) = - 1\1 xg(t)

Multiplyingthat solution by "c" converts the dis placement solution to the modal coordinate system. From the above:

C

= -<1> MI M

Since the displacement from the response spectrum is in the rotated coordinate system, it must be "unrotated" back to the global coordinate system, so C must be multiplied by <1>-1:

r = C <1>-1 = -<1> M <1>-1 1M This value "_", by which the solution from the shock spectrum must be multiplie d, is the ~ participation factor. Note that this is the same mass participation factor as defined in Section 4.32 as -cI>T M<1>, since <1>-1

=

cI>T M and M

=

<1> M cI>-1; so:

<1> M <1>-1 1 M = -<1> M <1>T MI <1> M <1>-1 = -<1> M cI>T M 1 <1> M <1>T M = -<1>T M cI> Mil M

= -<1>T M <1> The procedure for doing earthquake dynamic analysis can be summarized as described below: 1) Generate the shock response spectrum either by feeding the earthquake time

history through a shake table, or through analytical means such as those described in Regulatory Guide 1.60. 2) Set up the set of dynamic equations for the earthquake problem: M x(t) + K x(t) = -M xg(t) 3) Solve the Eigenproblem to get the natural frequencies and mode shapes. 4) Using the mode shapes, decouple the complex set of equations: M

x(t) + K

x(t) = -M xg(t)

M

vet) + K

v(t)

= -cI> M xg(t)

4-61


-

-

-

5) Read maximum displacements (zmax) for M v(t) + K v(t) = -M xg(t) from the response spectrum curve. (Remember that because of the decoupling, this expression represents a group of single degree offreedom bodies.) 6) Convert this displacement to the modal displacements that we need, in the global

coordinate system, by multiplying it by the mass participation factor, r: Xmax =

r

Zmax = (cI> M cI>-1 / M) Zmax

These individual modal responses are then combined to get the total system response. 4.6.2.3 Modal Results Summation Methods Considering that the response spectrum yields the maximum response at any time during the course of the load, and considering that each ofthe modes of vibration will probably have different frequencies, it is probable that the peak responses of all modes will not occur simultaneously. Therefore the correct means of summing the modal responses must be considered. Available modal combination methods are: 1) Double Square Root of the Sum of the Squares (DSRSS): This combination

method is the most technically correct. The total system response is calculated as:

R

nn = [ 1:1: •

•

IJ

E

iJ· R.1 R.J ]112

Where: R

= total system response

Eij

=

correlation coefficient between mode i and mode j

= varies from -1.0 (for out-of-phase modal responses) to O. (for independent modal responses) to 1.0 (for in-phase modal responses)

Rï

= response of mode i

Rj

=

response of mode j

This method results in a correctly phased response for dependent modal responses and a statistical response (SRSS) for independent modal responses (since they are unlikely to occur at exactly the same time during the load profile). The drawback to this method is that the inter-modal correlation coefficients are usually not known, other than some empirical estimates based on earthquake data (see the excerpt from the USNRC Regulatory Guide 1.92 in Section 4.6.2.5). Therefore, more usable methods have been developed, which are described below.

4-62


2) Absolute: This method states that the total system response is equal to the absolute value of the individual modal responses. (This is effectively the same as using the DSRSS method with aU correlation coefficients equal to 1.0.)

R

=

n L IR.I

i

1

This method gives the most conservative result, since it assumes that the aIl maximum modal responses occur at exactly the same time during the course of the applied load. This is usually overly-conservative, since modes with different natural frequencies will probably experience their maximum DLF during different parts of the load profile. 3) Square Root of the Sum ofthe Squares (SRSS): This method states that the total system response is equal to the square root of the sum of the squares of the individual modal responses. (This is effectively the same as using the DSRSS method with aIl correlation coefficients equal to 0.0.)

R

n

=

[L R.2]1 / 2

i

1

This method is based upon the statistical assumption that all modal responses are completely independent, with the maxima following a relatively uniform distribution throughout the duration of the applied load. This is usually nonconservative, especially ifthere are any modes with very close frequencies, since those modes will probably experience their maximum DLF during the same part of the load profile. 4) Grouping Method: This method attempts to eliminate the drawbacks of the Absolute and SRSS methods, by assumingthat modes are completely correlated with any modes with similar (closely spaced) frequencies, and are completely uncorrelated with those modes with widely different frequencies. R. 2

R=

1

+

P L

k L

q=1

m=j

1R

lq

R

lq

1]1/2

Where: p

=

number of groups of closely-spaced modes (where modes are consid ered to be closely-spaced iftheir frequencies are within 10% ofthat of the base mode in the group)

J

=

number of first mode in group i

k

=

number of last mode in group i

Rlq

=

response of mode l in group q

Rmq

=

response ofmode m in group q

4-63


Effectively, this method requires that the responses of any modes which have frequencies within 10% of each other first be added together absolutely, with the results of each ofthese groups then combined with the remaining modal results using the SRSS method.

4.6.2.4 Combination of Spatial Components Ifmultiple shocks have been applied to the structure in more than one direction, the results must be combined spatially as weIl (for example, the X-direction results with the Z-direction results). Combining these results vectorially points to an SRSS combination method; however a question arises as to whether the spatial summations should precede or follow the modal summations. The two different combination methods (i.e. spatially first or modally first), are associated with the terms DEPENDENT and INDEPENDENT respectively: Spatial before Modal-> DEPENDENT Modal before Spatial-> INDEPENDENT IEEE 344-1975 (IEEE Recommended Practices for Seismic Qualification of Class lE Equipment for Nuclear Power Generating Stations) states: "Earthquakes produce random groundmotions which are characterized by simultaneous but statistically INDEPENDENT horizontal and vertical components." Dependent and Independent refer to the time relationship between the X, y and Z components ofthe earthquake. With a dependent shock case, the X, Y, and Z components ofthe earthquake have a dependent relationship - a change in the shock along one direction produces a corresponding change in the other directions. For exam pIe, this would be the case when the earthquake acts along a specific directionhaving components in more than one axis - such as when a fault runs at a 300 angle between the X- and Z-axes. In this case, the Zdirection load would be a scaled (sin 300/cos 300), but otherwise identical version of the Xdirection load. An Independent shock is one where the X, Y and Z time histories produce related frequency spectra but have completely unrelated time histories. It is the Independent type of earthquake that is far more common, and thus in most cases the modal components should be combined first. Combination of modal response can be illustrated by considering a simple problem - a cantilever beam built in CAESAR II and for which four modes (natural frequencies and shapes) were extracted:

4-64


y

Mode#2 -

8.5 Hz. Z-direction

J-IIIIIIII!~---x

-

8.5 Hz. Y-direction

Z Y

Mode#4 -

Mode#3 -

15.6 Hz. Z-direction

15.6 Hz. Y-direction

Z

Figure 4-44

Goingto the response spectrum plot for the two differentfrequencies, the maximum response can be obtained in terms of displacement, velocity, or acceleration. This response can then be converted into structural displacements, and in tum into forces, moments, and stresses. Modal response is the product of the mode shape, the direction of the excitation, the magnitude, and the dynamic load factor. (For example, note that the modal displacements along the X-direction of the cantilever are zero, thus there will be no contribution to the stresses or dis placements due to the X-component of the earthquake.) For the cantilever, the stress at the wall represents the important response quantity of interest. For each of the modes, the maximum displacements are converted into forces, moments, and stresses. Once these response quantities have been computed for each individu al mode, and for each direction, they must be combined to get an estimate of the cumulative value. There are four modes and three excitation directions, which means there are 3 x 4, or 12, response quantities to combine. Looking specifically at the stresses at the connection of the cantilever to the wall, the 12 values are shown below. (Note that most of the response quantities for the cantilever are zero, while for true three dimensional piping models this will not he the case.) Mode 1 2 3 4

Frequency

8.5 Hz 8.5 Hz 15.6 Hz 15.6 Hz

Modal Stresses at the Wall X-Direction Y-Direction Z-Direction 0.0 18000.0 0.0 0.0 0.0 18000.0 0.0 4000.0 0.0 0.0 0.0 4000.0

4-65


The response quantities are defmed as spatial and modal. Spatial response quantities are those associated with each individual excitation direction (the columns of the table shown above). Modal response quantities are those associated with each individual mode (the rows of the table shown above). Using a GROUP combination method for the modal results, it is first necessary to determine if the modes are closely spaced - i.e., iftheir frequencies are within 10% of each other. Modes 1 and 2 are closely spaced (difference in frequency is 0%), as are modes 3 and 4 (difference is 0% as weIl), so their responses will be added absolutely when the modal responses are combined. The combinations can be made first spatially and then modally, or first modally and then spatially. The differences are shown schematicaUy in Figure 4-45. MobAt. SIJ .... ~·,:Tlol-lS S(c.o·,.:')l) (Gfou? )

lIIi.DI Rodai Strlssls at th. 1111

y

!!adl Fria. 1 8.5 111 2 8.5 111 3 15.6 111 .. 15.6 111

1gO~Oà

è(

l'l.MQ.)

~o+OQ( ~ooo S

3~OOQ '2.

C '-osn 'r' C;;;'11 (f'}_

4<60002...

,)'/?.

C!.oS,€LY

SPk~ll

f

1IaII. Fr". 1 •• 5 Hz

2 '.5 Hz 3 15.6 Hz .. 15.6 Hz

..

Mo 'iJfrL , I~~~__S ~~1JM. ( 4.fl>.f_P f-~l) f" .si')

\ Figure 4-45

Figure 4·45 A difference in the final results arises whenever different methods are used for the spatial and modal combinations. In this case, modal results are combined using the GROUP method, while the spatial results are combined with the SRSS method. Therefore the stresses ofmodes 1 and 2, and modes 3 and 4 are added absolutely in the first case, because they are aU non-zero at the time when the modal summations are made. Nuclear Regulatory Guide 1.92 (published in February, 1976), included on the following 5 pages, de scribes the requirements for combining modal responses and spatial components when performing seismic response spectra analysis for nuclear power plants.

4-66


....Won, .

U.S. NUCLEAR REGULATORY COMMISSION

. Flibruliry 1878 ;.: .:.....

'.';J

OFFICE OF STANDARDS DEVELOPMENT

REGULATORV GUIDE 1.92

COMBINING MODAL RESPONSES ANO"SPA11AL COMPONENTS IN SEISMIC ~. , - . RESPONSE ANALYSIS

case

A. INTRODUCTION

CriteJÏon 2, "Design Bases for Protection Against Natural Phenomena," of Appendix A, "Ceneral Design Criteria for Nuclear Power Plants," to 10 CR Part SO~ "ücensing of Pr04uction anel UtiUzatlon FadJities." requires. in part,; that nuc:1ear power pJlllt ItlUctures, systems, and ccmponents important· to afety be de· signed to withstand the effects of eartbqualœs without lass of capabüity to perform their satèty funetions. Paragraph (aXI) of Section VI, '''AppUcatioo te Engi· neering Design," of Appendix A, '"Seismlc and Geologie Siting Criteria for Nuclear Power Plants," to 10 CR Part 100, "Reactor Site Criteria,'~ .req1JiJes, in part, that structures, systems, and components ùitportant to safety remain functiorW in the event of a SaCe Shutdown Earthquake (SSE). It specifies the use of a suitable dynamic analysis as one method of ensuring that the structures, systems, and components cao withstand the seismic loads. Similarly, paragraph (aX2) of Section VI of the same appendix requiRs. in part. that the structures, systems, and componeats aecessuy for con. tinued operation without undue to'the bealth and safety of the public remain Nnctional iii an Operating Basis Earthqualœ (OBE). Again. the use of a suitable dynamic analym is specified as one methocl of ensuring that the structures, systems, and components cao withstand the seismic 101uis.

:risk

This guide descrtbes methods acceptable to the NRC staff for:

1. Combining the values of the responseof indMdtÎal modes in a respcmse spectrum modal dynamic analysis te fmd the representatiYe maximum value of a particular rC!Sponse of interest for the design of a pen clement of a nuclear power plant structure, system, or component.

l: Combifting tb,e' maximum' values (in. the of time-lùstory dynamic anal)'lis) or the-:'n:PRsenta~ maximum values (in the case of spectrum dynamic analysis) ~f, the response ofa ,pven .. ële~nt 'of .a structure,: system, or çomponent, when N.èh values are calculated independent}y' foi e&eh of the tliièe Ortbogo.. . nal spatiarcampcmenU'(two horizOntahlid :Cne,.WrtiÇaJ.) .' of an· earthqualœ; ,~ combined v&lue: ·WIll·:. he 1he . representatiw ~um valUe of the ëOmbùiéd'iâponse : : of that element of the strueture, syStem, or' component to iimultaneous action of the three tpatial components.

The AcMsOry CommUtee On' Reactètt 'Safe~ds bas ,been consulted conceming this ginde aiul has'<:aocurred in the regulatory position. .:. . ··B. DISCUSSION

1•.CombiDiDg Modal R~onses To .flDd the· values of.. the response of different e1emenu of a JlUC1ear power plant stru'ctu., system, or cOmpoant to a prescribed .n:sp~ spectrum. it is first· nec:asary to calculate the madt shapes' and frequeacies of the s~ctwe, system, or compoàent~' This ~ clone by solwing the foUowing equation for' the eisenwctom and

eigenvalues:

.

[ [K] -

'.

w~ lM1]{ 'n}=O

wh~re {K] is!f1e. stiffness matrix, w n is the natum. frequency foÏ' the nth mode; [M] is the ~IS:' ma~. and {~) is the eigenwctor for the nth mode.:, .

.~ indicate

substmtivedwlps rrom pzmous isIIIOo

_ _ ,..... -'''.'Y ., .... c............ U.L ....._

_in._ ..... _ --_ .... _, .....ui_ . . ._if._..-.._,.. ... 10._ ,,---........ ..... . . ._....... --_ ,._._. .. _._nr. c-_ -

USNRC REGULATORY GUIDES

R ....I•• .." G ...... .;., iuu_ •• _Me _ .... --.... ,. , ... ...."' ........ MC . . . . . . . t . . . . . . . et 1.... '. . . . . . . . . . . . ,.,.. .. the

......_ie_............,..-....- .....______ .....,....-........ . .. eo...... _ · ........_

c ..... ... _ _ ........ _.,,_ ...H 100 ........ ~

con .............." a __ M ..... _ wi....

......-.. ...

~

....

_8ft4--...~

.-.1 ca....-'iII

....eo...__ ... -.._ ............IiCII_.,,_Ca_ _ _ _..._ .., ..................n.a. ...... __ ''''..... .,............... "".......ioft., .. iU _ _

_.-..~_

iftIp. ._

" "_ _ C.-.I......

w......._.

s-Iee--.

~

~ .. .

-...... ...H

4-67

D.C. _ . _ _

~.

_

. '

lIIe ....... _ _ _ iII_ ..

1.'-___ 2. _ _ T., __ _ J. ...... _

'.T'.._ ........

.. _ ,_ _

LOc ............ HMIIIo·

4. _ _ _. '

.................. _ _.....

ca_..... _ _ , _ , ... ....- _

(1)

1. An'II",.. R . . -

Copiee •• . - - . - ...., MI o M _ . " ...rIt... _

......... _

...... U.L ......

:

..

_a..-_ea....-w............ D.C.

~A~Oireee.'.Oftlee of

4-67

...

at....... D........... .

...., . . . -........ .


COADE Pipe Stress Analysis Course

No~es

1/93

1. CombiJlilll SpâiII Campoaen..

Note that it ma)' not be necessary to solve Equation 1 for all modes. In mlllY cases, determlDadoD of only those modes that are sipiflc:ant should lie auftlclent.

1.1 _Rapo. . 10 Tbree Spatial Compoaenu CaJ&:u,. lated Sepuately

The ncxt step is to œtennine the maxiDmm modal displacement relative to the supports. this is done as follows:

Regulatory Guide 1.60, "Design Respouse Spcctra for Seismic Design of Nuclear Power Plants," indicates that design of Seismic Category 1 structures, systems. or components should be based on three orthogonal corn· panent motions (twc horizontal and one vertical) of a prescribed design earthquake. Chu, Amin, and Singh (Ref. 3) have concluded that the representative maxi· mum value of a particular response of interest for (e.g., stress, strain. moment, shear, or displacement) ofa given element of a structure, system, or component subjected to the simultaneous action of the·· thiee .compOôetlts .of the earthqualc.e can be sa1is(actorily ; obtained by talcing the square root of the S\Un of the squares of carrespœding representative maximUm wlues . of the. spec:trum response, or the maximum response values 1rom tlme-hlstory. dynamic analysis, to each . of. the tiuee componeDts calculated independently.

an

(2) whe re lqn \ max is the maximum displaœment vector for the nth: mtde, rD is the modal participation factor for the nth mode and is expressed by

desisn

San is the value of acceleration in the specifted sesponse spectrum corresponding. to w n and desillft cbmping, and superscript T desipates the transpose. Otber maximum values of the responses per mode such I l 1treII. strain. oment. or wu CID be computed !rom the approprilte q max by using the stifIness properties orthe elements f me structure, system, or componeat. Newmark (Rd. 1) has shown that the representaüve maximum value of a . particular response of interest for desip ($Och as components in given directions of stJesspstrain. moment, shear, or displacement) of a given element CID be obtained from the corresponding maximum values of the response of individual modes as computed abo. by taking the square root of the sum of the squares (SRSS) of the maximum values of the respanse of these individual modes of the structure, system, or corn· ponent. The Newmark study, however. does not address the problem of closely spaced modes. Other studies (sec References 2 and 3) have .mown that SRSS. procedure can sipficantly underestimate the true ~ponse in certain cases in whic:h some of the modal frequencies of a structural system are closely spaced (see resulatop: eosition 1.1 for definition of closely spaced modes). The nuclear ind"ustry bas used many different methods to combine the response when closely spaced modes exist. Sorne of these methods can be found in Reterences 2, 4, and 5. A recent unpublished study bas shown that the resulting combined response of nuclear plant f~cilities using any of the methods delineated in rewlatory 20sition 1.2, which covers a broad range otmèthods currenuy being used by the industry, is in good agreement with time-history response. Therefore, any of the methods given in reguJatory position 1.2 is acceptable for combining the modal responses when closely spaced modes exist.

~

ft should be noted that, if the fIequencies of a system are all widely separated. aU the terms in the second summation sign in Equations 4 and 5 of regulatory position 1.2 wiU vanish, and these equations wül degenerate to the SRSS method (Equation 3).

The SRSS procedure used by Newmark (Ref. 1) and Chu, Amin, and Singh (Ref. 3) for comQining the values of the response to three components of an earthquake is based on the consideration that it is' very unlikely that peak values of a response of a given element would oc~r at the same üme during an earthquake. 1.1 Respoase to Tluee Spatial Compooentl CalcuJated Simultaaeously .

The maximum value of a particular response of interest for desillft of a given element cao be obtaïned througb. a step-by·step method. The time·history res· ponses from- e&ch of the three components : of . the earthquake motions cao be obtained and then combined algebraically at each time step or the· response at each time step cao be calculated directly owing. to the simultaneous action of three components. The ·maxi· . mum respanse is determined by scanning the 'combined time-history solution. When this me~od is used, the earthquake motions specified in the tluee c!ifferent directions should be statistically independent. Fora· discussion of statistical independence, seeReféreuce 6 ..

C. REGULATORY POSITION The following procedures for combining the values. of the respanse of inœvidual modes and the respœse tothe three independent spatial components of an earf:hquake in a seismic dynamic anaIysis of a nuclear power plant structure, system, or companent are acceptable to the NRCstaff:

4-68


1/93

COADE Pipe Stress Analysis Course Notes \1. CombiDatiOll of Modal RapOIIIeI 1

1.1 With No aosely Spaœcl Mollet

ln a re!pense spectrum modal dynamic analysis, if the modes are not c1œely spaced (two consecutive modes are defined as closely spaced if their frequenc:ies differ from each other by 10 percent or leu of the lower frequency), the representative maximum value of a particular response of interest for design (e.g., com· panent! of stress, strain, moment, shear, or displace· ment) of a given element of a nuclear power plant structure, system, or component subjected ta a single independent spatial component (response spectrum) of a three.component earthqua1œ should be obtafned by taking the square root of the sum of the squares (SRSS) of correspanding maximum values of the response of the element attributed to individual significant modes of the structure, system, or component. Mathematically, this . can be expressed as foUows:

R=

[t Rkl*J

component shouJd then be obtained by taking the square root of the sum of the squares of corresponding representative maximum values of the response of the element attributed ta each closely spaced group of modes and the remaining modal responses for the modes that ue not closely spaced. MathematicaUy, this an be expressed as foUows:

where ~q and Rmq are modal responses, Ri and Rut within. the qth group. respectively; i is the number of the mode where a group starts; j is the number of the mode where a group ends; R, R§c, and N are as defined preYiously in regulatory position 1.1 of this guide; and P is the num~r of groups of closely spaced modes, excluding individu al separated modes. 1.2.2 Ten Percent Method

(3)

k= 1

(5)

wherc: R is the representative maximum value of a particular response of a given element ta a given component of aD earthquake, Rte is the peak mue of the response of the clement due to the kth mode, and N il the n ·.unber of significant modes considered in the modal respcnse combination. 1

1.:: With OOlely Spaœd Modes

where R, Rt, and N· ~ as defined previously in regulatory position 1.1 of this guide. The second summation is to be done on ail i and j modes whose frequencies are clœely spaœd to each other. Let CAli and "ï be the frequencies of the ith and jth mode. In order 10 ve~fy which of the modes are c10sely spaced, the foUowmg equation will apply:

·In a response spectrum modal dynamic analysis,.if

1s.ome ~r all of the modes .~ closely spaced, any of the -;:..

(6)

toll':,"mg regulatory positions (i.e., 1.2.1", 1.2.2 or 1 May be used as a method acceptarilë.'to the NRC ~ swf .:l combine the modal responses. :1

h

y-?-

1.2.1 Grouping Method

alsol$i
(7)

1.2.3 Double Sum Method

Oosely spaced modes should be divided inlO grou's that include aU IOOdes haviJ;lg. frequencies lying betw' en the lowest frequcncyin the group and a 1 ~requ :ncy 10. ~r,?en~ . ~gher. L . The re'presentative maxr num value of a particular respanse of interest for the Cesign of a given e1ement of a nuciear power plant suuc ure. system. or component attnbuted to each such grOU! of modes should fIrSt be obtained by taldng the sum )f the absolute values of the corresponding peak value; of the respanse of the e1ement attributed to indiv dual modes in that group. The representative :naxi num value of this particular response attributed to ill tt e signifiant modes of the structure, system, or os should be rormeci starting rrom the lowest frequency and vorking towucis sw:casively hilher frequcnc:ies. No one freq\ ency Is to be ln more than one ~up.

1 Gro\:

4-69

where R, ~, and N are as defined preYiously in regulatory position 1.1 of this ~ide. Ra is the peak value of the response of the clement attributed to sth mode.

Eks= [ 1+ {

(Wk - w;)

'(Pk wlc + fJ;

}J-I

(9)

w s)

in which

(10)


f3k,

tdwk

1 where

(11)

b. "''hen the time·history responses from eac:h of the three components of the earthquake motion are calculated by the step·by·step method and combined algebraicaDy al eac:h lime step, the maximum response cao be obtained from the combined time solution. 2

tJk

are the modal frequenc:y" and the damping ratio in the kth mode, respectively, and td is the duration of the earthquake.

2.

wk and

responses are calculated using the time·history method instead of the spectrum method.

2

=f3k+-

Combination of Effects Due to Thœe Spatial Components of an Eartbquake

1

3. If the applicant has used the methods described in

Depenciing on which basic method is used in the 1 scismic analysis. i.e., respanse spectra or time.rustory 1 method. the following two approaches are considered 'acceptable for the combination of three·dimensional earthquake effects.

this guide, the Preliminary Safety Analysis Report (PSAR) should indicate in each applicable section which of the alternative acceptable methods were used for the structures, systems, or components covered by tbat section.

!

2.1 Respon8e Specua Methocl

D. IMPLEMENTATION

The purpose of this section is to provide information to applicants and liœnsees regarding the NRC stafrs plans for utilizing this œguiatory guide.

When the response spectra method is aclopted for sei smic analysis, the representative maximwn wlues of the structural respcnses to each of the tJuee campoDents of earthquake motion should be c:ombined by taking the square root of the sum of the squares of the maximum. representative values of the codilectiOllal respOD.SeS caused by eath of the three components of earthquake motion at a particu1ar point of the structure or of the mathematical model

Except in those cases in which the applicant proposes an alternative method for complying with specified portions of the Commission's regulations, the methods described herein will he used by the staff in the evaluation of submittals for construction permit applica· tions docketed after the date of issue of this guide.

2.2 Tune-History Analysis Method

If an applicant wishes to use thls regulatory guide in developing. submittals for applications docketed on or before the date of issue of this guide, the pertinent portions of the application will be evaluated on the basis of this guide.

When the time.mstory analysis method is employed 1 for seismic analysis. two types of anal}'Sis are generally 1 perfonned depending on the complexity of the problem: a. When the maximum responses due to cach of the three components of the earthquake motion are ca1culated separately, the method for c:ombining the three-dimensional effects is identic:al to that deicribed in 1 regulatory position 2.1 except that the maximum

1. R. L. Wiegel, editor. Etuthqu.tIke EngiMt:Ttng. Englewood Cifrs, N.J., Prentice·HaU,lnc:.,1970, chapter by N. M. Newmark, p. 403.

2. A. K. Singh, S. L. Chu. and S. SinBh, "Influenœ of Closely Spaced Modes in Response Spec:trum Method of Analysis." Proceedings 01 the SpecÜJJty Conlermce on Structural Design' 01 Nucletlr Plant Facilities. Vol. 2. Chicago. December 1973. (Published by American Sode ty of Civil Engineers, New York, New York.) 3. S. L Chu, M. Amin, and S. Singh, "Spectral Treatment of Actions ofThree Earthquake Components on Structures," Nucletrr Engineering tmd Dnign, 1972, Vol. 21, No. l, pp. 126·136.

method is usecl, the earthquake motions specifiee! in the three difrermt directions should be statistically indepen· dent. For 1 discussion of statistical independence, seo Refer· ence 6.

2When this

4. E. Rosenblueth and 1. Elorduy, "Response· of Unear Systems ta Certain Transient Disturbanœs,"

Proceedings, Fourth World Conference on Eanhqu.ake Engineering, Vol. l, Santiago, Chüe, 1969. 5. N. C. Tsai, A. H. Hadjian et al., "Seismic Analysis of Structures and Equipment for Nuc1ear Power Plants," Bechtel Power Corporation Topical Report 4.A, Revi· sion 3, November 1974. 6. C. Chen, "Definition of Statistically IndepeJ\dent Tune Histories," JounuJl of the St7Uctunzl Division, ASCE, February 1975.

4-70


4.6.2.5 Missing Mass Correction

Since the response of the rigid (highest order) modes are purely static and mode independent, this response can be calculated without actually extracting aIl of the rigid modes, through a method called missing mass correction. This is important, since, although displacements may be obtained with good accuracy using only a few ofthe lowest frequency modes, the force, reaction, and stress results may require extraction offar more modes (usually far into the rigid range) before acceptable accuracy is attained. By performing a missing mass calculation, a correction representing the quasi-static contribution ofthe higher order modes not explicitly extracted for the modal/dynamic response can be included, providing greater accuracy with reduced calculation time. The residual response ofnon-extracted rigid modes is calculated by summingthe active mass (over all of the extracted modes) for each degree-of-freedom at each of the mass points,. The difference between the total mass modeled at each of the points, and the sum of the active mass at each ofthe points, is called the missingmass, and represents the residual mass active in the rigid (non-extracted) modes. This missing mass is then multiplied by the ZPA from the response spectrum and applied to the structure as a static load. The static structural response is then combined with the dynamically amplified modal responses as ifit were a modal response as weIl. (Actually this static response is the algebraic sum of the responses of aIl non-extracted modes - representing in-phase response, as would be expected from rigid modes.)

CAESAR II's implementation of the missing mass correction procedure assumes that the missingmass correction represents the contribution ofrigidmodes, and that the ZPAis based upon the spectral ordinate value at the frequency of the last extracted mode. Therefore, it is recommended that the user extract modes up to, but not far beyond, the rigid frequency cutoff of the response spectrum. Choosing a cutofffrequency to the left of the spectrum's resonant peak will provide a non-conservative result, since resonant responses may be missed. U sing a cutoff frequency to the right ofthe peak, but still in the resonant range, will yield conservative results, since the ZPNrigid DLF will be overestimated. Extracting a large number ofrigid modes for calculation of the dynamic response may be conservative, since aU available modal combination methods (SRSS, GROUP, ABS, etc.) give conservative results versus the algebraic combination method which gives a more realistic representation of the net response of the rigid modes. CAESAR II provides two options for combining the missing mass correction with the modal (dynamic) results - SRSS and Absolute. The Absolute combination method of course provides the more conservative result, and is based upon the assumption that the dynamic amplification is going to occur simultaneously with the maximum ground acceleration or force load. Research suggests that the modal and the rigid portions orthe response to typical dynamic loads are actually statistically independent, so that an SRSS combination method is a more accurate representation ofreality. 4.6.3 Force Spectrum Analysis (for Impulse Loadings)

A similar method can be followed for non-random loads, such as any load profile for which the force vs. time profile is known - for example, the ramp up, constantforce, andramp down of a relief valve firing (or fluid hammer or slug flow), as shown in Figure 4-46.

4-71


~

~

Plot ot mo,lm,m d,nom', dlsplo"m,nt for each of the oscillators being exposed to the forcing function

)

Single DOF Oscillators

Relief Valve Discharge Force Profile

Force

~/ Hydraulic Shakers

1

Time

Relief v a l v : - - - I - - - - H e l i e f Volve Opening Time Totol Duration Closing Time Needed to Vent Overpressure

Figure 4-46 A look at the equation for the earthquake problem explains why the force spectrum solution is very similar to the earthquake solution: M x(t) + K x(t) = -M xg(t) The term on the right hand side is nothing more than a dynamic force acting on the piping system, i.e. F = Ma. So the analogous equation to be solved for the force spectrum problem is: M x(t) + K x(t) = F(t) Where: F = the dynamic load (water hammer or relief valve) 4.6.3.1 Generation of the Response Spectrum Instead of the displacement, velocity, or acceleration spectrum used for the seismic problem, aDynamie LoadFactor spectrum is usedfor a force spectrum problem. ADLF spectrum gives the ratio of the maximum dynamic displacement divided by the maximum static displacement. Whereas the earthquake response spectrum analysis method started with the time history of an earthquake excitation, the force spectrum analysis method is done in exactly the same way - except that the analysis starts with the force vs. time profile. Just as for the earthquake, this time history loading can he applied to a shake table of single degree-of-freedom bodies, with a response spectrum (in this case, DLF vs. natural frequency)

4-72


being generated by dividing the maximum oscillator displacements by the static displacements expected under the same load. Note that for this type ofload, as the ramp-up becomes steeper, the DLF gets closer to 2.0; while as the load duration increases, the curve shifts to the left and more lower order modes will become excited.

2.0

Longer Duration Extends Curve Le ft

..

DLF

tt

Quicker Ramp-Up Shifts Curve Up (Towards 2.0)

1.0

Natural Frequency

Figure 4-47

4.6.3.2 Application to a Multi-Degree-of Freedom System The mathematics ofthis method proceed similarly to those for the seismic response spectrum method. Once the eigensolution is complete, the next step is to decouple the set ofequations: cp M cp-1 v(t) + cp K cp-1 v(t)

= cp nt)

which can be re-expressed as:

-

-

M v(t) + K v(t) = cp F(t) The DLF curve was generated for the equation: M x(t) + K x(t) = F(t) So,just as for the earthquake problem it is necessary to find a constant by which to multiply the solution in order to make the problems compatible. By comparing the two equations, it is obvious that the required constant is CP.

-

-

So the solution to M v(t) + K v(t) = cP F(t) can he found as: v max = C (F / K)(DLF) = cP (F / K)(DLF) Where DLF is the dynamic load factor selected from the response spectrum. As with the seismicproblem, vmaxis in the rotatedcoordinate system, so, inorderto get back

to the global piping coordinate system we must "un-rotate" these displacements, again using the mode shapes: xmax = cp-1 V max 4-73


These are the maximum displacements in the piping system due to the relief valve or water hammer load.

4.6.3.3 Summation of Responses Once the force spectrum modal quantities are computed, they must be combinedjust as for earthquakes. The SRSS method with a 10% closely spaced mode criteria seems to give the most reasonable answers for the force spectrum method. The force spectrum combinations usually are much simpler to apply than the earthquake combinations because normally there are no separate spatial components to combine - i.e., there are not X-, Y-, and Z-shocks acting simultaneously. However, in the event that there is more than one potential force load (such as when there is a bank ofreliefvalves that can fire individually or in combination) the spatial combination method may be used to indicate the independence of the loadings. For example, if two relief valves may or may not fire simultaneously (i.e., they are independent), the two shocks should be defined as being in different directions (for example, X- and Y-), and the combination method selected should be "Modal before Spatial". Ifunder certain circumstances, the two valves will definitelyopen simultaneously (i.e., the loadings are dependent), the combination method should be "Spatial before Modal". (Otherwise, the direction defined for a force spectrum loading has no particular meaning.) Including the missing mass correction is especially effective when performing force spectrum calculations. This is because force spectrum loads due to causes such as fluid hammer and slugflow act axially along the pipe, and therefore excite most highly the extremely rigid axial extension modes. Therefore the rigid response represented by the missing mass correction often provides a very high proportion ofthe total force spectrum response. 1t is recommended that the user extract modes up to, but not far beyond, the peak ofthe force response spectrum and use the SRSS method for combining the missing mass.

4.6.4 Harmonie Analysis Harmonic analysis looks at dynamic problems where the forces or displacements (i.e., pulsation or vibration) acting on the piping system take sinusoidal forms. In these cases, when damping is ignored, the dynamic equation of the system becomes:

M x(t) + K x(t) = Fo sin ro t Where: M

=

mass matrix of system, slug

x(t)

=

acceleration vector with respect to time, ftlsec 2

K

=

stiffness matrix of system, lb/ft

x(t)

=

displacement vector with respect to time, ft

Fa

=

vector of amplitude ofharmonic force, lb

ro

=

forcing frequency of harmonic load, rad/sec

t

=

time, sec 4-74


This equation may be solved directly, giving the system displacements, forces, stresses, and reactions; thus obviating the need for response spectra. This type of differential equation has a solution of the form: x(t)

=A sin

ro t

Where: A

= vector representing maximum harmonic displacements of system, ft

Since acceleration is the second derivative of displacement with respect to time: x(t) = -A ro2 sin rot Inserting these equations for displacement and acceleration back into the basic harmonic equation of motion yields: -MA ro2 sin ro t + K A sin ro t

= F 0 sin

ro t

Dividing both sides of this equation by sin ro t: -M A ro2 + KA = F 0 Reordering this equation: (K - M ro2 ) A = F 0

This is exactly the same form of equation as is solved for alllinear (static) piping problems. The appealing thing about this is that the solution time for each excitation frequency takes only as long as a single static solution, and, when there is no phase relationship to the loading, the results give the maximum dynamic responses directly. Due to the speed of the analysis, and because the solutions are so directly applicable, it is advisable to make as much use of this capability as possible. Two considerations must be kept in mind: 1) An assumption of zero damping in a harmonic problem permits a dynamic load factor ofinfinity. This may have to he overcome by a slight tuning of the forcing frequency up or down to get a more realistic DLF. 2) If multiple harmonic loads occur simultaneously, and they are not in phase, system response is the sum of the responses due to the individualloads:

Where:

Ai

=

dis placement vector of system under load i

Bï

=

phase angle of load i, degrees

In this case an absolute maximum solution cannot be found. Rather, solutions for each load, and the sum ofthese, must be found at various times in the load cycle. These combinations should then be reviewed in order to determine which one causes the worst load case. 4-75


The biggest use by far of the harmonic solver is in analyzing low frequency field vibrations due either to fluid pulsation or out-of-round rotating equipment displacements. The approach typicaIly taken towards solving this type of problem is described briefly below: 1) A potential dynamic problem is first identified in the field - either in terms of large cyclic vibrations or high stresses (fatigue failure) being present in an existing piping system, raising questions ofwhether this represents a dangerous situation. As many symptoms of the problem (quantifiable displacements, overstress points, etc.) are identified as possible, for future use in refining the dynamic model. 2) Amodel of the piping system is built using CAESAR II. This should be done as accurately as possible, since system, as weIl as load, characteristics affect the magnitude of the developed response. 3) The engineer next postulates the cause of the load, and from that, an estimate of the frequency, magnitude, point, and direction ofthe load. This is somewhat difficult because the dynamic loads can come from many sources. Dynamic loads may be due to internal pressure pulses, external vibration, flow shedding at intersections, two phase flow, etc., butin almost all cases, there is sorne frequency content of the excitation that corresponds to (and therefore excites) a system mechanical natural frequency. If the load is caused by equipment, then the forcingfrequency is probably sorne multiple ofthe operating frequency; ifthe load is due to acoustic flow problems, then the forcing frequency can be estimated through the use ofStrouhal's equations (from fluid dynamics). Using the best assumptions available, the user should estimate the magnitudes and points of application of the dynamic load. Note that the point of application is not necessarily a point of high system response. 4) The loading is then modeled using harmonic forces or displacements (normally depending upon whether the cause is assumed to be pulsation or vibration) and several harmonic analyses are done, sweeping the frequencies through a range centered about the target frequency (in order to account for uncertainty). The results of each of the analyses are examined for signs of large displacements, indicating harmonic resonance. If the resonance is present, the results of the analysis are compared to the known symptoms from the field. Ifthey are not the similar (or ifthere is no resonance), this indicates that the dynamic model is not a good one, so it must be improved, either in terms of a more accurate system (static) model, a better estimate of the load, or a finer sweep through the frequency range. Once the model has been refined, this step is repeated until the mathematical model behaves just like the actual piping system in the field. 5) At this time, there is a good model ofthe piping system and a good model ofthe loads (or, more accurately, a good model ofthe relationship of the load characteristics to the system characteristics). The results ofthis run are evaluated in order to determine whether they indicate a problem. Since harmonic stresses are cyclic, they should be evaluated against the endurance limit of the piping material; displacements should be reviewed against interference limits or esthetic guidelines.

4-76


6) If the situation is deemed to he a problem, its cause must be identified, where the cause is normally the excitation of a single mode ofvibration. A modal extraction of the system is done; one (or more) of these modes should have a natural frequency close to the forcing frequency ofthe applied load. The guilty mode can be further identified as that one having a shape very similar to the shape of the total system vibration, since this mode shape has certainly been dynamically magnified far heyond the other modes (and thus predominates in the final vibrated shape). 7) Once the guilty mode has been identifie d, it must be eliminated. This is done most easily by adding a restraint at a high point (and in the direction thereof) of the mode shape. Ifthis cannot be done, the mode may also be altered by changing the mass distribution ofthe system (for example, by adding operators to valves). Ifno modification of the system is possible, it may he possible to alter the forcing frequency of the load. If the dynamic load was postulated to be due to internal acoustics, it is recommended that the pipe not be rerouted at this point, as rerouting the pipe will change the internaI flow conditions (which may resolve or amplify the problem, but in either case will void CAESAR II's "good model" of the system). After modifying the system, the harmonic problem (using the single forcing frequency determined as a "good model") is then re-run, and the stresses, displacements, etc. are re-evaluated. 8) If the dynamic problem has been adequately solved, the system is now reanalyzed statically to determine the effects of any modifications on the static loading cases. (Remember, adding restraint normally increases expansion stresses, while adding mass increases sustained stresses.)

4-77

5

This section discusses earthquake spectru- analys1s where each support point on the piping sys.tell is exposed to the SUIe seiSllic excitat10n. foll~ng sect10n d1scusses independant support Spectrwl excitation wtlereby different groups of support points are exposed to different se1S11ic exc1tat1ons. (For eaa.ple. p1p1ng in a systel supported in the rack .ight be exc1ted by one shock. while all p1ping in the s..e syStei. supported frai the ground would be excited by a d1fferent

The

sIIOCk. ) There are tllree ujor sources of earthquake excitation tllat can be

accessed fl"Clll CAESAR II:.

1. Predef1nec1 El Centro. (AYaUable in the CAESAR clata base.) 2. Predef1ned Nuclear Regulatory Guide 1.60. (AYanable in the CAESAR clata base.) 3. User def1ned spectra.

EL CENTRO (Predef1 ned spectl"lll n..e • EI.CEIITIIO) Spectl'Ull clata is taken fl"Clll Biggs: Dyn.iCS- and 15 baRd on the May earttlquake north-soutll cœponent. 0.33 g' s. 13.7 ipso and B.3 ipss. apply for ellstic s,rsteas having 5 Eu.ple D-4 on page 11lustrates

-IntrocluCtion to Structural 18. 1940 El Centro t.l1forn1l The recorded lIUi_ values were This spectl'Ull 15 intended to to 10 percent crtt1cal ~ping. the use of tIIis predefined spect.....

NAC REGULATORY GUIDE 1.60 The predefined spectl'Ull nlIIIes Ire: 1.60H.5 1.60H2 1.60H5 1.60H7 1.60H10

1.60V.5 1.60V2 1.6OV5 1.60V7 1.60VlO

Horizontal Horizontal Horizontal Horizontal Horizontal

and Ind Ind Ind and

vertical vertical Vertical Vertical Vertical

spectrl spectra spectrl spectra spectra

for for for for for

0.5S

~ping

10.OS

~ping

2.OS ~ping 5.OS ~ping 7.OS claçing

Associated w1t11 the Reg. Guide 1.60 spectra 15 a vllue for the 1IIIX1mum ground aecelerltion (or ZPA - zero pertod acceleration). The default value for the ZPA used 1n CAESAR and found in the control earallf!ter spreadsheet 15 0.5 g's. Frai IEEE Std 344-1975 Sect. 3.1 Earthquake Environaent-:

-An earthquake of _gn1tude 6.0 or higher on the Richter seale -y persist for 15 to 30 seconds and proctuce a lIUillUl horizontal ground aeeeleratfon frai 0.1 to 0.6 9 or higher vitll the lII,jor energy content usually occurring in tIIe first 5 or 10 seconds. The t.Ypical broaclband randoll mtion proctuces clauging effects over a frequency range fl"Clll 1 to 33 Hz. Usually the vertical cœponent of the ground mt10n is ass.... to be between 67 and 100 percent of the horizontal below 3.5 Hz ••nd equal to the horizontal above 3.5 Hz.· Often used in cOnjunction with Reg. Guide 1.60 1s Reg. Guide 1.92 which provfdes essentfally three _tIIods for cœb1ning mâl responses and directs tIIat the SRSS cœbinat10n of the tllree spatial responses be used. Most t.Yp1cally tIIis translates into: 1.

~erfoN

_thod. mul

cœbinations first us1ng the -grouping-

2. Perfonl spatial c0lD1nat10ns second using tIIe -SRSS· _thod.

USER DEFINED SPECTRUM (N..e· up to 24 char.cter user define«! label) User defined spectra -y be entered vitll period or frequenty as the range. and d1spl.c....t. veloc1ty or acceler.t1on as the ordinate. TheM spectra -y be read 1n frai an aseff data file or entered d1rectly 1nto a spectrua table dur1ng dyn.1c input process1~i.

-377-

~ P,'l.1 ..>.. ~AMC,~12.

1)~~~~O

ItJ

---::II....

.::n-/II(i'1~ ~....,

oc..CAS 1014.) 1+ C-

Co-,?..ATE.

Srl4f1C

t..DAL) VrfE

tr4-

Sr/2.E(f(i'>

Afto/ù

7#é.

$'0'- fA. TI OAJ .

'5 TA- .,-, (!,.

L0 A- () S .'

1- w + P ) LISE 0 f - /.A.J+ P+ T + /)+ F 3 - w+P+ -;+/)"f- F (o?[) 4- t-..J+P-I-I= (St.1s) 5'- l/3-l)4 (êXP)

-378-

~"""f

St.-tSrA/;..JE D

!5'pÉ.c..mlA.M

i>y'ItJAVI4IC

EXN4P\.E D-4 EARTHQUAIŒ - UNIFO"" SUPPORT MOTION - EL tENTRQ Klnger destgn Ind nonltnelr restrltnts extst tn thts rune The sutte elSes 11II11Zed were: 1 2 3 4 5

-

V+P V+P+T1+04f V+P+T1+04f V+P4f D3-D4

••• for hlnger destgn ••• for hlnger destgn (OPE) (SUS) (EXP)

Use 1.0 tt~s the El Centro Elrthqulke tn the hortzonul dtrecttons Ind 0.667 tt~s the El tentro elrthqulke tn the Yerttell dtrectton. tOllbtne the sutte susutned Cise (Cise 4) wtUl the eerthqulke result. The operlttng Cise (Cise 3) Should be used tG set the nonltnelr restrltnt sutus for the d1n_te lul,S1s. The tAESAR dynIa1e input sequence for Ulis probl. 1s slKMl below. Prtor to ent.er1ng the d1n_te input the user sllould haye fint run 1 sutte 11II11s1s for the lold elses ltsted lbove.

CAESAR Il (VERSION 2.1C) [PI'IE-RELEASEl

PIPING/STRUCTURAL ANALYSIS

<---_.

CURRENT JOIINNIEI - - ) D-4

1 - INPUT 2 - STATICS - DYNAftICS

<-

~

4 -

OUTPUT

S - FILES 6 - JOBNNIE 7 - UTILlTIES 8 -

RETURN Ta DOS

ENTER OPTION

->

1 - UJfPED MSSES

2 - SNU8JIERS ~

CAESAR

- stCICIC DEFINITIDNS 4 - P\LSE T~ &DEIVtTJCIN S - tUND 8U5T TMLE SENERATIDN

Il

6 - PULSE &f'ECTfUII FORCE SETS D~IC

ANALYSIS INPUT

(

7 - HMl'DtIC LDAD8

•

8 - stCICIC CASES • - 8TATIC/D~IC CDNBINATIDN8 A - caNTRG. PMNEIERS • C D E

-

SAVE D~IC 1""'" CHECIC D~IC 1""'" PEJIIFCRI'I DVNMIC ANALYSIS R.tlU'n ta CAESAR _1n .en",

~r SHOCIC LDAD CASE EDITlNB

DPTIDNSI 1 - Acld anDÜIW' shaell lCNd ca. . _ • • • • • • 2 - O.I.t. an . . 'sUne ca_ 3 - Edit an .. istino ca. . 4 - Finishsd ~•

•••• ?

-379-

TG SELECT

...xx

LDAD CASE

CCNTRlaJTIONS - DlrecUCift c.n ... l, Y _ Z _ CA" ... el ......_ • IIlrecUCift c_ln. _ .IU,.ect1C1ft _ t _ . Th. far_t f_ lIirectiCift c_in_ _ _ t __ i. Cc_,cy,CZ) far •••~l.. CO.707,O,0.7071. Ed1tine kava on t~ ,,~ic Il..,.,... K,.oll tM ln"ut. My U"• •tarU"e .U;" A" •• t_1 ... CI) .Ul ... t.k_ • • • c~t.. Er,._. h __ 11--. Il.t_t ... CIft .ny U_ IIUnUne. Only _t._ "ode. f_ I"lIep_lIant s.ap,,_t üclt.Uon. U . .·., !!!

."0.1\

,,_c.

set. •

<->

Dl,._U_,

Btart

Nod.,

.toP

NoeS.,

l"cr_t

1.0 0 . . .7

1.0

1,,_

Dai

<_>To Exit

1 - LU9'ED M8SEB 2 - SNI"'"

CAESAR D~IC

Il

ANALYSIS INPUT

3 4 5 " 7

- ..x:x DEFINITIONS - ~ TABLE IENERATlaN - -.IND . . , . TQU; SEfERATlCIN - ~ SPECTRUIt FDfICE SETS - IWR»CIC LOAD8

• - 6HDCIC CA&EII

• -

8T"TlCID~IC

Cl»œlNATlCIN& _ _ __

• - BAVE D~IC INPUT C - Q4ECIC D'tIIWfIC INPUT D - PERFONt

D~IC

ANALYSIS

E - Returft to CAESAR

STATIC/D~IC

_i" _

LOAD CASE EDITINS

OPTIONS.

- AllII _oth_ .taUc/dyn_ic load - Dal.t. an .. i.t'ne c . . . Ed,t an •• '.t'''e c . . . - F'n'sh'" h_e

STATIC/D~IC

c._I• • • • • • • •

CClœINATlaN CAlE

1 - p,._ .... lIyn_lC load ca_ ".,..,.• •ith Editl"" kava an the , , _ l e Il..,.,... K,.OU tM input. My U,,_ .tartine .ith _ a.t_iek Cil .111 ... tr••tad • • • c~t. Errars ha_ lleen lIet_t... on env U_ sho.l\ IIU"k1ne.

• -D- and .t.tic 1oac1 c . . . nuaàer• •,th an -S-.

Not.

Ent_ .aeh .t..tlc/dyn_'c load ca. . t.o Il. U"e.

caM,,,"

on a seperat.

Fact_ 84

Dl

1".

DaI

1.0 1.0

~

<_>TOExit.~

-380-

:s CAEIM D~IC

IHDCIC DEFINITla.

4 - I"\LSE TAIlLE 8ENEMTICIN 5 - IUND BUST TAIlLE GEJERATICIN • - I"\UIE IlPECTRUPI f"DRCE ET.

11

ANALY81S INPUT

7 - HMIOIIIC LC1AD8 • - &HOCIt CAIIEII

• - STATIC/D~IC ~INATIDN8 A -CONTJaL

~

D~IC INPUT C - OECIC D~IC JNPUT D -' ~ D~IC ANALY81. E - ftMurt! t.o CAE8M _ln .....

• - lAYE

Il..,.

le..,.....

CCNT'ftCL PARNEta LJST - ttodUy t.he "al_ or P"'~_ Oft e.cI'I U . . ""lCh

p,.ec-.a.

t.h. descrlpt.lan. bSUn. Oft t.he f t _ l e ec,.oU the iftllUt.. Er-r-ar. ha_ b . - lIet_t.1Id Oft ."y U . . bUnlri"9 •

~

( - - st.Uc Loacl c._ far NaI'IUn• .,. A_t,..iftt. .t..tus ( - St.Sffnee. Factar for- F,.ScUan CO.o-Not. uslld) ( - - ..... No. of Ei . . .".luee c.lcul.tlld co-Not. uelld) ( - - F,..quency cutoff cta)

0.1 20 0.0:5 0.5

( - Cl_ly lIpac'" ftIIII. Cr-St._S. ( - E.,.t"~Ir. Dur-atian CFor- D&ft8S _t.hacU Csec.) < - 8t.,.uct._.l Du!pSftQ CX of crSUc.U ( - Z_o .... '011 Accwl . .aUan CFor 1te9. Bu' ... 1.60)

N N

< -
• •04.... :s 0.0

o

Co·.)

ua!Id)

( - - (Not. ue!Id)

l'IODAI.. ( - - Sp.Ual ........1 t:a.bSn.tian fS,.st ( - - Sp.Ual t:a.blnaUan ...thOil CSRSS/ABS) GROUP < - - ......1 eo.blnaUan ...thOil C~/lO"/DSRSS/ABS/SR&ln y ( - - Inclua P~_t.t1c to.panent. CY/N)

SRS8

1 - L.UI'FED MIISES 2 - lNU8IIER8

:s -

8CICIC

DlEF'INIT~GN5·

4 - l"tL8I TAIlLE IlENERATICIN

CAE6M D~IC

IJ

ANALY6I& JNPUT

5 - MIND . . , . TA8LE 8DERATJCIN • - PUUIE 8PEC'nU't FORCE SET8 7 - HMfDiIIC ~ •• -- 8HDCIC CA8E8 6TATIC/D~IC ~JNATJCIN&

• - &MIE ~IC IJIPUT C - CHEQC D~IC JJIPUT D - PEJlFCIRI't DVNMIC ANALYSIS E - Rllturn to CAESM _in - . u

-381~

EXAMPLE 0-5 EARTHQUAKE - UNIFORM SUPPORT MOTION - RE6 6UIOE 1.60 This job 1s ltllNr, wittlout lIanger design. Use Reg Guide 1.60 spectra v1t11 2' clulp1n,. TIIe est1 . .ted ZPA " ...1 _ ground accelerat10nl for the site 1s 0.35 g s. TIIe stat1c cases analyzed for the job vere:

1 - II+P+T+D-+f 2-II+P+F 3-- 01-02 CoID1ne 1.0 t1.s the Reg Gutde values and 0.5 ti_s the Reg Guide values witll the sustlined stlt1c case nUlliber 2. Use an SRSS calbination of the Stlttc and ~1c cases. The abbreviated dyn. .1c input for t111s Job 1s SIlCM! below: First the user IlUSt def1ne the probl. and run the stltic analys1s vitll the 10id cases as descr1bed Above, tIIen fl"Oll the CAESAR ..in

.....

: 3 - DYNAMICS 8 - SHOCK LOADS 1 - AcIcI a

new sIIock load case

Shock na• • Factor, Direction. Stan Node, Stop Node, Increment 1.6OH2 1 X 1.60H2 1 Z 1.6OY2 ~

Y

\.v 1 - Add a nev shock load case

Shock nUle, Factor, Direction. Stan Node, Stop Node, Increment 1.60H2 1*0.5 X 1.60H2 1*0.5 Z 1.6OY2 ~*O.5 Y

~.ù

9 - STATIC/DYNAMIC COMBINATIONS 1 - Add anotller stat1c/dyn. .ic 10id case LOid case, Factor

COMBINATlON(SRSS 1

52 1

02 1

1 - Add anotller static/dyn. .ic lad case Lad case. Factor

COMBINATlOli (SRSS 1 02 1 02 1 A - CONTROL PARMETERS SPECTRUM

0.35 SPATIAL

<---<---<----

Dyn..1c Analysis Type Zero Per10d Acceleration (g'sl Spatial or Modal Combination First

B - SAVE DYNNUt INPUT C - CHECK DYMAMIC INPUT D - PERFORM DYNAMIe A1ALYSIS

-382-

........ ,

U.S. NUCLEAR REGULATORY COMMISSION

.

·'......,117.

.

REGULATORY-GUID·E ..

~

OFFICE OF STANDARDS DEVELOPMENT

REGULATORY GUIDE 1.92

COMBINING MOOAL RESPONSES ANO·SPATIAL COMPONENTS IN SEISMIC . RESPONSE ANAL YSIS 2. Combining the maximum Valua (in. the case of time-history dynamic analyais) or the-:tepresentatM maximum values (in the case of spectrum dynauiic analysil) of the mponse of •. 8iVen.element ofa structure,system, or component, when suêh values lIe calculated independen~y for each of the tIiiee ortho..,., na1 spatial·components·(two horizontal arid:œe \'ertiéal) . of an earthquake.The combinee! mue: the representative maximum value of the càmbinechesponse: .. of that element of the structure, system, orcomponent to sùnultaneous action of the three spatial components.

A. INTRODUCTION

Criterion 2, "Design Bases for Protection Against NaturaI Phenomena," of Appendix A, "General Design Criteria for Nuclear Power Plants," to 10 CFR Part 50, "Licensing of Pr04uction and Utilization Fadlities," requires, in part, tltat nuclear power plant structures, systems, and components important to safety be designed to withstand the effects of earthquakes without loss of capability to perform their safety functions_ Paragraph (aXl) of Section VI, ""Application to mgineering Design," of Appendix A, "Seilmic and Geologic Siting Criteria for Nuclear Power Plants," to 10 CFR Part 100, "Reaetor Site Criteria," requires, in part, that structures, systems, and components important to safety remain functionil in the event of a SaCe Shutdown Earthquake (SSE). It specifies the use of a suitable dynamic analysis as one method of ensuring that the structures, systems, and components cm withstand the seismic loads. Similarly, paragraph (aX2) of Section VI of the same appendix requires, in part, that the structures, systems, and components necessary for continued operation without undue risk to the health and safety of the public remain funetional in an Operating Basis F.arthquake (OBE). Again, the use of a suitable dynamic analysis il specified as one method of ensuring that the structures, systems, and components cm withstand the seismic loads.

wqrbe

The Acmsé;ry Committee on lùictor Safep,ards bas ,been consulted conceming this guide aÎld haSCœcurred in the regulatory position.. .. B. DISCUSSION

1.CombiDing Modal Rt!SJJOIIIM!S To . fmd the values of the response of different elements of a Jluclear power plant structUJe, system, or component to a prescribedœsponse spectrum, it il filSt necessary to calculate the mode shapes and fœquencies of the structure, system, or component. 'This il done by solving the following equation for the eiFDvectolS and eiFDvalues:

[[X] -w~ lMl] ttf>n}=O

This guide describes methods acceptable to the NRC

stafffor: 1. Combining the values of the response of individual modes in a response spectrum modal dynamic analysis to fmd the representative maximum value of a particular response of interest for the design of a given element of _ a nuclear power plant structure, system, or component.

"_lai

ac_......

..in,

......

{tI>n} is the eigenwetor for the nth mode .. -Unes indicate substantive cbaDges from prmous ÏSIae.

~

.r.

i'.""

,

where IK] is t:ile stiffness matrix, w n il the naturaI fiequency for the nth mode, [M] is the ~ass matrix, and

_"0'

ui_

st."

.0 _

eo...._ .

c:o-n_. w_IntIt_. D.C. _ . AtMfttioft: Dock_. _

Commen•• _

USNRC REGULATORY GUIDES

Dry Guida .;. issu....0 dacribe .nd m .........iIabIe .0 .... public m ..hocl. 0 .... HIIC ...fI o, impl.....n.in' specifie ...... o, .he Commission 's qgua-tiona. to clelineate techniques used by the a.aft in ev"u~ apecafic probl. . . or .ahdatad accident•. or to pravicle guid.,.ce 10 appli· cants. Ae,ulatory Guides ftOt subatitU1H for Nl'ulM:"'. and complillnce witt. Ihem i. not rwquared. Methoda and solutions differenl from 1hou aet out in ....._ wiU ... _ ...... _vide. .he .intI........ to the ""anc. or continuance of • penniI or licen.. by.he Commi..ion. Comment. and . . . . . .'ions for impravementa in ...... guides .re encourag_ e' ail '.m... and guides wiH be reviMd. a. appropriet•• to accommodMe COIR· Il mena end to retlec1: Mw mfor. .ion or exp.rience. This guide .a. l'eviMd _ • resuIt of . .batanli.,. comment. received 'rom the public ..ut additional

(1)

... _

Sec....ry of _

u.s. _ _•

Senrico Section. The guida ... iuuedin_tollowin. _ _ d . . . . . . .:

1. _ . 1Ioec:t_ 2. ......rcIl_T... IIeect.... M--.. F.ciIit_ 4. Emrir _ _ Sitin, 5. _ _ _ Plant Pntt_

3.,..... ..._

1.-. 7 . T _....-

,._"*-

.. Occupaâo.... ~

10. .........

m.y _ _ _ _ by - ' -.-..,......... _tolhe U.5. N _ .........INY CO........... W........... D.C.

CopiaI of......- ......

.....AttentiDn:DirecIor.Otficeof~.O." ......."'.

-383-

2. Com_DI S..tiaI Components

Note that it MaY not be nec:essary to solve Equation 1 for all modes. In many cases, determination of only thase modes that are significant should be sufficient.

2.1 Respoaae to Tbree S..tial Componeats Cal&» latecl Separately

The next step is to determine the maximum modal displacement relative to the supports. This is done as follows:

(2) where \qn \ max is the maximum displacement vector for the nUl m~de, rn is the modal participation factor for the nth mode and is expressed by

San is the value of acceleration in the specified response spectrum corresponding to w n and design damping, and superscript T designates the transpose. Other maximum values of the responses per mode such as stress, strain, oment, or shear can be computed from the appropriate q max by using the stiffness properties of the elements structure, system, or component. Newmark (Ref. f 1) has shown that the representative maximum value of a 'particular response of interest for design (such as components in given directions of stress, strain, moment, shear, or displacement) of a given element can be obtained from the corresponding maximum values of the response of individual modes as computed above by taking the square root of the sum of the squares (SRSS) of the maximum values of the response of these individual modes of the structure, system, or component. The Newmark study, however, does not address the problem of closely spaced modes. Other studies (see References 2 and 3) have ·shown that SRSS procedure can significantly underestimate the true response in certain cases in which some of the modal frequencies of a structural system are closely spaced (see regulatorv eosition 1.1 for definition of closely spaœd modes). The nuclear industry has used many different methods to combine the response when closely spaced modes exist. Some of these methods can be found in References 2, 4, and S. A recent unpublished study has shown that the resulting combined response of nuclear plant facilities using any of the methods delineated in regulatory position 1.2, which covers a broad range of methods currenuy Seing used by the industry, is in good agreement with time-history response. Therefore, any of the methods given in regulatory position 1.2 is acceptable for combining the modal responses when closely spaced modes exist.

~

~e

It should be noted that, if the frequencies of a system are all widely separated, all the terms in the second summation sign in Equations 4 and 5 of regulatory position 1.2 win vanish, and these equations will degenerate to the SRSS method (Equation 3).

Regulatory Guide 1.60, "Design Response Spectrafor Seismic Design of Nuclear Power Plants," incticates t,hat design of all Seismic Category 1 structures, systems, or compOllents should be based on three orthogonal component motions (two horizontal and one vertical) of a prescribed design earthquake. Chu, Amin, and Singh (Ref. 3) have concluded that the representative maximum value of a particular response of interest for design (e.g., stress, strain, moment, shear, or displacement) of a given element of a structure, system, or component subjected to thesimultaneous action of the three components of the earthquake can be satisfactorily obtained by taking the square root of the sum of the squares of corresponding representative maximum values of the spectrum response, or the maximum response values from time-history dynamic analysis, to each of the tluee components calculated independently. The SRSS procedure used by Newmark (Ref. 1) and Chu, Amin, and Singh (Ref. 3) for combining the values (jf the response to three components of an earthquake is based on the consideration that it is very unlikely that peak values of a response of a given element would occur at the same time during an earthquake. 2.2 Response to Tbree Spatial Components Calculated Simultaneously The maximum value of a particular response of interest for design of a given element can be obtaïned through a step-by-step method. The time-history responses from each of the three components· of the earthquake motions cao be obtained and then combined algebraically at each time step or the respoose at each time step cao be calculated directly owing to the simultaneous action of three components. The maximum response is determined by scanning the combined time-history solution. When this method is used, the earthquake motions specified in the three different directions should be statistically independent. For a discussion of statistical independence, see Reference 6 ..

c.

REGULATORY POSITION

The following procedures for combining the values of the response of individual modes and the respoose to the three independent spatial components of an earthquake in a seismic dynamic analysis of a nuclear power plant structure, system, or component are acceptable to the NRC staff:

1.92-2

-384-

r 1. CombiDation of Modal Respon.s 1.1 With No a_y Spacecl Modes In a response spectrum modal dynamic analysis, Ü the modes are not closely spaced (two consecutive modes are defined as closely spaced if their frequencies differ from eachother by 10 percent or less of the lower frequency), the representative maximum value of a particular response of interest for design (e.g., components of stress, strain, moment, shear, or displace· ment) of a given element of a nuclear power plant structure, system, or component subjected to a single independent spatial component (response spectrum) of a three-component earthqualœ should be obtained by taking the square root of the sum of the squares (SRSS) of corresponding maximum values of the response of the element attributed to individual significant modes of the structure, system, or component. MathematicaIly, this can be expressed as foDows:

R=[t Rkl*J

component should then be obtained by taking the square root of the sum of the squares of corresponding representative rœximum values of the response of the element attributed to each closely spaced group of modes and the remaining modal responses for the modes that are not closely spaced. MathematicaDy, tbis can he expressed as follows:

Rm

~e~ ~q and Rrnq are modal responses, ~ and Wlthin the qth group. respectively; i is the number of the mode where a group starts; j is the number of the mode

where a group ends; R, ~, and N are as defined previously in regulatory position 1.1 ofthis guide; and P is the numbP.r of groups of closely spaced modes, excluding individual separated modes. l'latF.

Roi 'l. .. ~~t. ~ '-~~

1.2.2 Ten Percent Method / -

A

(3)

k= 1

(5)

where R is the representative maximum value of a particular response of a given element to a given component of an earthquake, RJc is the peak value of the response of the element due to the kth mode, and N is the numher of significant modes considered in the modal response combination. 1.2 With Close1y Spaced Modes

where R, Rt, and i-{ are as defined previously in regulatory position 1.1 of this guide. The second surnmation is to he done on aIl i and j modes whose frequencies are closely spaced to each other. Let Wj and Wj be the frequencies of the ith and jth mode. In order to verify which of the modes are closely spaced, the following equation will apply:

-In a response spectrum modal dynamic analysis,ü sorne or aD of the modes are closely spaced, any of the following regulatory positions (i.e., 1.2.1, 1.2.2. or !J,J) May he used as a method accepiablëto the NRC Staff to combine the modal responses.

w·-w·

(6)

< 0.1

_J_ _l W·1 -

(7)

aIso 1 $" i
1.2.1 Grouping Method

1.2.3 Double Sum Method

Oosely spaced modes should he divided into _groups that include aD modes having frequencies lying between the lowest frequency in the group and a 1freq~ncy 10 _percen~ higher. 1 The representative maxunum value of a particular response of interest for the design of a given element of a nuclear power plant structure, system, or component attnbuted to each such group of modes should fust he obtaïned by taking the sum of the absolute values of the corresponding peak values of the response of the element attributed to indi~dual modes in that group. The representative maxunum value of tbis particular response attnbuted to aD the significant modes of the structure, system, or

where R, ~, and N are as defined previously in regulatory position 1.1 of this guide. Rs is the peak value of the response of the element attributed to sth mode.

~ould be fonned st~g f~om the lowest frequency and working towards successJVely higher frequencies. No one frequency is to be in more than one group.

(10)

Eks= [ 1+ {

@f wk + (3;

lJ-

1

(9)

ws)

in which

1Groups \

(Wk - w;)

1.92-3

-385-

:.

(R;t~

responses are calculated using the time-history method instead of the spectrum method.

(11) where WJc and (jk are the modal frequency and the damping ratio in the kth mode, respectively, and td is the duration of the earthquake. 2.

Combioation of Effects Due to Tluee Spatial Co.. pooents of an Earthquake

Depending on which basic method is used in the seismic analysis, i.e., response spectra or time-history method, the folloWÎDg two approaches are considered acceptable for the combination of three-dimensional earthquake effects.

b. When the time-history responses from each of the three components of the earthquake motion are calculated by the step-by-step method and combined algebraically at each time step, the maximum response can be obtained from the combined time solution. 2

3. If the applicant bas used the methods described in this guide, the Preliminary Safety Analysis Report (PSAR) should indicate in each applicable section which of the alternative acceptable methods were used for the structures, systems, or components covered by that section.

2.1 Response Spec:tra Method

D. IMPLEMENTATION

When "the response spectra method is adopted for seismic analysis, the representative maximum values of the structural responses to each of the three components of earthquake motion should be combined by taking the square root of the sum of the squares of the maximum representative values of the codirectional responses caused by each of the three components of earthquake motion at a particular point of the structure or of the mathematical model.

The pUlpose of this section is to provide infonuation to applicants and licensees regarding the NRC staffs plans for utilizing this regulatory guide. Except in those cases in which the applicant proposes an altemative method for complying with specified portions of the Commission's regulations, the methods described herein will be used by the staff in the evaluation of submittals for construction permit applications docketed after the date of issue of this guide.

2.2 ïune-Bistory Analysis Methocl When the time-history analysis method is employed for seismic analysis, two types of analysis are generally performed depending on the complexity of the problem: a. When the maximum responses due to each of the three components of the earthquake motion are calculated separately, the method for combining the three-dimensional effects is identical to that described in regulatory position 2.1 except that the maximum

If an applicant wishes to use this regulatory guide in developing submittals for applications docketed on or before the date of issue of this guide, the pertinent portions of the application will he evaluated on the basis of this guide. 2When this method is used, the earthquake motions specified in the three different dilections should be statistically independent. For a discussion of statistical independence. see Reference 6.

1.92-4

-386-

REFERENCES 1. R. L. Wiegel, editor, Earthquake Engineering, Englewood Cliffs, N.J., Prentice-Hall, Inc., 1970, chapter by N. M. Newrnark, p. 403.

2. A. K. Singh, S. L. Chu, and S. Singh, "Influence of Closely Spaced Modes in Response Spectrum Method of Analysis," Proceedings of the Specialty Conference on Structural Design of Nuclear Plant Facilities, Vol. 2, Chicago, December 1973. (Published by American Society of Civil Engineers, New York, New York.) 3. S. L. Chu, M. Amin, and S. Singh, "Spectral Treatment of Actions of Three Earthquake Components on Structures," Nuclear Engineering and Design, 1972, Vol. 21, No. l,pp. 126-136.

4. E. Rosenblueth and J. Elorduy, "Response of Unear Systems to Certain Transient Disturbances," Proceedings, Fourth World Conference on Earthquake Engineering, Vol. 1, Santiago, Chile, 1969. 5. N. C. Tsai, A. H. Hadjian et al., "Seismic Analysis of Structures and Equipment for Nuclear Power Plants," Bech tel Power Corporation Topical Report 4-A, Revision 3, November 1974. 6. C. Chen, "Definition of Statistically Indepefldent Time Histories," Jounud of the Structural Division, ASCE, February 1975.

1.92-5

-387-

RevIsion 1 Oecember 1913 U.S. ATOMIC ENERGY COMMISSION

REGULATORV GUIDE DIRECTORATE OF REGULATORY STANDARDS

REGUlATORY GUIDE 1.60

DESIGN RESPONSE SPECTRA FOR SEISMIC DESIGN OF NUCLEAR POWER PLANTS A. INTRODUCTION Criterion 2, "Design Bases for Protection Againsl Natural Phenomena," of Append1x A, "General Design Critcria for NucJear Power Plants," to 10 CFR Part 50, "Licensing of Production and Utilization Faeilities," requires, in part, that nuclear power plant structures, systcms, and components important to safety be designed 10 withstand Ihe effects of earthquakes. Proposed Appendix A, "Seismic and Geologie Siting Criteria," to la CFR Part 100, "Reactor Site Criteria," wlmld require, in part, that the Safe Shutdown Earl hquake (SSE) be defined by response spectra COI rcsponding to the expccted maximum ground .lc·l·cler.lIiuns. This guide describes a procedurc acceplablc tu the AEC Regulatory staff for defining lespOllse spectra for the seismic design of nudear power planls. The Adviory Committee on Reactor Safcguards has becn consulted conceming tbis guide and has concurred in Ihe regulatory position. B. DISCUSSION

ln order to approxlmate the intensity and thcreby estimate the maximum ground acceleration l of the expecled strongest ground motion (SSE) for a given site, proposed Appendix A to 10 CFR Part 100 specifies a number of required investigations. ft does not, however, ~ve a method for defining the response spectral l'llTlCSpllnding to the expected maximum ground ;lcceler
.,.dft

..,tU be ecœpt8b" if th.y pro. . . . . .11 for , . . ",.,.,.. ntQIIHtI" IN . .uanc» 0' mnhn'*'GI of • perm.t or "'*'- by tM Con'WNIlIOft

-388-

'ubllllhecl ..,.... wiU bI

corn""""

~

""'1ed

Su~ndlrdl. CO..w'IW!"tlJ .nef .uggnllQrtt for encour...,a .ftCI shovid bit Mn' fO the Sec:retlry

Aflenuon: Dtrw:tor 0' R..,.ory ~".,~nwml

'"

t'" .,..... ..

01 tho Com.........". uS. Atomoc E-ev C _ _ . W... ,ngton. O.e. A"...,_: Cil .... 'ubloc " ' -... S .. II. The tui1iltl ..... uacI

ln

the fofl"""l"I len

br~

d,YIII(.In1.

to

D....od'QUY. as aporopt .... , 10 accOf"N'tlOde ...

ta r.'''' , . . Infor"WltQn 01 •• per-.cw.

In this prOCedure, the configurations of the horizontal comJXment Design Response Spectra for each of the two mutually perpendicular horizontal axes are shown in Figure 1 of this guide. These shapes agree with Ihose developed by Newmark, Blume, and Kapuf in 1 eference 1. In Figure 1 the base wagram consists of Ihree parts: the bottom line on the left part represents . the maximum ground displacement, the bottom linr on the nghl part represents the maximum acceler:llion. and the middle part depends on the maximum velocity. The horizontal component Design Response Spectra in Figure 1 of this guide corresporid 10 a maximum hurizontal ground acceJerrztion of 1.0 g. The maximum ground displacement is taker. proportional to the maximum ground acceleration, and is set at 36 inches for a ground acceleration of 1.0 g. The numerical values of design clisplacements, velocities, and accelerations for the horizontal component Design Response Spectra are obtained by multiplying the corresponding values of the maximum ground clisplacemenl :md i!!:ce!era!lOT! by the factors glven in Table 1 of this guide. The dlsplat'ement region Imes of the Design Response Spectr~ are parallel III the maximum ground displacemenl boe Jnd are shown on the left of Figure 1. The velu..:lty reglon lmes slope downward from a frequency of 0.25 cr~ (conlrol point DI to a frequency of 2.5 ct::; ;.:ntrol pmnl C) and are shown at the top. The remainlng two sets uf lines belween the frequencies of 2.5 cps and 33 cp!> (control point A), wlth a break at a frequency of 9 cps (cont roi Coc"" Of pubh.hed VU .... nwy . . obta.ned Dv reQo,lQ.t ;,-.;;i •..JIi,,.'V Ih~ d.vtt.o", _ .."" ft> tho U.s. Ato""" E _ Com......"'n. _""on. O.e 70!!0<15.

G........ t.uec:I 10 dIncr'tae and ...... 8Ve.iebl. to .hI publiC nwthodl ecc:ap. . . . 10 'hP AEC R.gul.tcwy •• et, of Irl"lCMetnenlll'lg IC)IClftC.-n1 of 'M Co""","IOft', ,..,iM.ona. 10 dtlineM' t.d\nlqUR UMd bv .,... "8ff .ft "'uettnt . .ohe problllml or pol"...... tICC'Ôlfttl. or to pro." ",.-..c. 10 IIIIIP'lCenft. R.....'Otv GuidM .r. not lUtai.. ,,,... far "".'0'" .nd C'OfIIPI..nc:e ...." them • not r-aUl,ad. Mltnoct. end lOfutto", d.ff. .nt 'rom lhole . , OUt N"I AtlguaalC)ry

ft•

extensive study has been described by Newmarit and 91ume in references l, 2, and 3. After reviewing the!e rcferenced documents, the AEC Regulatory staff has determined as acceptable the following procedure for deflning the Design Response Spectra representing the effects of the vibratory motion of the SSE, 1/2 the SSE, and the Operating Basis Earthquake (OBE) on sites underlain by either rock or soi! deposits and covering ail frcquencies of interest. However, for unusually soft sites, modification ta trus procedure will be required.

R.ectOl'l Il. . .,.,, _ T ..t IIIMCtOfI Fuels _ M........ Fecolll_ EnYWOlVneft'.l et'ld SitU''U _ .... _l'lent

1. PO.....

1. J. ... !i.

'ro,*,,_

6. 'roGo.",' 7. Trerac:oo;tlUon

8. Oc:cupeuo,.' H••'th i. Antitrust A.v..... 10. Ge ....

a'

~5.

poan! B), ,onstilut.: the ~œler3tion regi()n of the homon!JI Desi8l1 Respnnse SpedrJ. For frcquenl:ics hl!!her Ih.1Il .U .:ps, the maJumum gmund ac:c:elerallull hnc a'prl'scnls 1111' Design R"'PUIISC Speclra. '1 Il.: vella.: .. 1 ,'omplllJent DeSIgn Responsc Spe':lra .:orr'lospondmg tLl the maximum horizontlll gr()Und (/('('('lcrutlCln 1.0 gare !Jl()wn in Figure 2 ()f this guide. The nurnerieaJ values of design displ4lCements, velocities, and J,celcrations in these spcetra are obtained by lllultiplYlOg Ihe correspondmg values of the maximum hoJn':ontal gruund motion (acceleration 1.0 g and dlspliiccment = 36 in.) by the factors given in Table Il of Ihis gUide. The displacement region lines of the Design Response Spectra are parallel to the maximum ground displacement line and are shown on the left of Figure 2. The ~Ioeity region ünes siope downward from a f:equency of 0.25 cps (control point D) to a frequency of J.S cps (,onlrol point C) and are shawn at the 10p. The remaining two sels of lines betw.n the frequencies of J.S .:ps and 33 cps (control point A), wilh a break at t~ frcquency of 9 cps (control point B), constitute the a.:ceieratilln region of lhe vertical Design Response Spc':lra. Il should be noted that the vertical Design I{espon~ Spectra values are 2/3 those of the horizontal OeSll(rl J{esponse Speetra for frequencies less than 0.25; !or frequencies tugher than 3.5, they are thé same, while [Ix- rail" vanes between 2/3 and 1 for frequencies tx-Iween 0.2S and 3.5. For frequencies higher than 33 cps. the Design Response Spectra foUow the maximum gJOund al:celeration bne.

earthquJite or (2) have physical characterishcs that could significanlly affect lhe SpeClll1 pattern of input mOllulI, sUl:h as helng IInderl3111 hy puur snil dcpc~ltS. the procedure descriJw.d ahnve will IInt apply. III tli~ cases. the DeSIgn RespUl15e Spectra shmald he developed indiVldually aCl:nrding tn the site characterislics.

ur

=

The horizontal and vertical companent Design Response Spectra in Figures 1 and 2, respectively, of tbis guide correspond to a maximum horizontal ground acceleralion of 1.0 g. For sites with different acceleration values specified for the design earthquake, [he Design Response Spectra should he linearly scaled from Figures 1 and 2 in proportion to the specified maximum horizontal ground acœleration. For sites that (1) are relatively close to the epicenter of an expected

C. REGUlATORY POSITION 1. The horizontal component ground DeSIgn Response Spectra, wilhout saü-structure interaction effects, of the SSE, 1/2 the SSE, or the OSE on sites underlain by rock or by sail should he linearly scaled from Figure 12 in proportion ta the maximum horizontal ground acceleration specifaed for the earthquake chosen. (Figure 1 corresponds to a maximum honzontal ground ac~eleration of 1.0 g and accompanyingdisplacement of 36 in.) The applicable multiplication factors and control points are given in Table I. For damping ratios not induded in Figure 1 or' Table l, a linar interpolation should be used. 2. The vertical component ground Design Response Spectra, without saü-structure interaction effects, of the SSE, 1/2 the SSE, or the OSE on sites underlain by rock or by sail should be linearly scaled from Figure 2 2 in proportion to the maximum horizontal ground acceleration specifaed for the earthquaite chosen. (Figure 2 is based on a maximum horizonllli ground flCCeluanon of 1.0 g and accompanying displacement of 36 in.) The applicable multiplication factors and control points are given in Table Il. For damping ratios not inciuded in Figure 2 or Table Il, a linear interpolation should be used. 'This does not apply to sites whic:h (1) ale relatiYely cloR to the epiœnter of an expec:ted eanhquate or (2) whic:h haft physical characteristics dlat couJd sipUf"acantly affect the spectral combinauon input motion. 1be DcsiIn RespoDle Spectra for such IÏta sbould be cleYdoped on 1 ca»by-cue

0'

basis.

1.60-2

-389-

DEFINlnONS rdahun5hip uhtained by analyzing. ev:lIualina. and statistically ,,·..,mhininl a numher nf indMclual response

RapGll8r Spectnam mcaRS a plul .. 1 1he: Rlaximum

raponse (ac:c:cleration. velocity. ur displacen1Cnt) uf a family of idealized sintie-depee-of-frceduRl damped osciUators IS a func:tion of naturaJ frequencies (or

spectra derived (mm the records of SÎpifant eart hquakes.

plst

periods) of the oscillaton to a specifled vibratory

Mlximum (pak) GrGUDd Ac:aIeIatioa specifled for a

motion input at their supports. When obtaincd from 1 recorded eartWake record, the response speçtrum tends to be irrepIlar. with 1 number of peaks and

...uers.

liven site means that value of the acceleration which corresponds to .ro period in the design response s~etra for tbat lite. At zero period the design response _peetra acceleration is identica1 for aD dampUlg nlues and is equal to the maximum (pnk) Found ac:ceJeration

Daip . ., . . . SpectnalD is a relatMly smooth

specified for tbat lite.

TABLE 1

HORIZONTAL DESIGN RESPONSE SPECTRA RELATIVE VALUES OF SPECTRUM AMPLIFICATION FACTORS FOR CONTROL POINTS

PwcInt

Amplification Fectan for Control Poinu

of

CrhicIII Dlmping

0.5 2.0 5.0 7.0 10.0

Aca_etion' AI33_

1.0 1.0 1.0 1.0 1.0

set

Dilpa..nwnt'

t

c:pI'

4.96 3.54 2.61 2.27 1.90

CI2.5_

5.95 4.25 3.13 2.72 2.28

1

DC0.25_

3.20 2.50 2.os 1.88 1.70

'lIaximuaa pUlld dbpIaœmeDt is tlken pmportiollal to maxiIIlam poUlld ICCIIleIation, and js 36 iL for pound acœIIIration of 1.0

srmty.

1 AèœJaatioD and cliaplacement amplif"xation factors are tateD froaI recollUftllldations p.en in ref.-cnce 1.

1.60-3

-390-

TABLE Il

VERTICAL DESIGN RESPONSE SPECTRA RELATIVE VALUES OF SPECTRUM AMPLIFICATION FACTORS FOR CONTROL POINTS

Percent of Critical Dlmping

0.5 2.0 5.0 7.0 10.0

Amplification

FilClon

Acceleration' A(33cpsl

S(Scps)

1.0 1.0 1.0 1.0 1.0

3.54 2.61 2.27 1.90

for Control Points DtsptKement'

3

C(~:S

cps)

5.67.\ 4.05 2.98 2.59

4.96

~.17

S

010.25 cpsl

2.13 1.67 1.37 1.25 1.13

'Maximum (!I"ound displaœment is talten proportional to maximum Itrnund acc:eleration and Ï5 )6 in. for pound ac~. clcrati(ln (lf 1.0 tuavity. S Acceleration amplif"lCation facll."s for the vertical design rClIronle lJIL"Ctra an: cqual tn l''ose fClr horizontal desilln rcsponsc spectra at a given frequency. whereas diqtbcemenl amplifïcatilln fadors arc 2/3 tbose for horizuntal clcsilm responsc: 5pectra. Thesc ratins œlW\.ocn the amplification factors fur Ihe Iwo desi&n respunse spcctra are in apeemenl with tbose recommendecl in rcferenc:e 1.

'These valua were changed to malte Ihis table oonsi5lenl Wilh the dis· cUSlliun uf vertical componenl'; in Seclion B of this Kuidc.

REFERENCES 1.

...

Ncwmark. N. M.• John A. Diurne. and Kanwar K. Kapur, "Design Response Spectra for Nuclear Power l'I:mls," ASCE Structural Engineering Meeting, San Francisco. April 1973.

Spectra," Urbana, Illinois. USAEC Contraet No. AT(49-5)-2667, WASH-12S5, April 1973.

3.

N. M. Newmark Consulting Engineering Services, UA Study of Vertical and Horizontal Earthquake

1.60-4

-391-

John A. Diurne & Associates, ··Recommendations for Shape of Earthquake Response Spectra," San Fnncisœ, Califomia, USAEC Contr.ct No. AT(49-5}3011. WASH-1254, February 1973.

FRflUENCY,c:pa FIGURE 1. HORIZONTAL DESIGN RESPONSE SPECTRA - SCALED TO 19 HORIZONTAL GROUND ACCELERATION

-392-

fiOO

u

!

.i

~.

~

III

>

20

0.2

0.&

1

2

5

10

20

10

100

FAEOUENCY,cpI FIGURE 2. VERTICAL DESIGN RESPONSE SPECTRA - SCALED TO 19 HORIZONTAL GROUND ACCELERATION

-393-

EXAMPLE 0-6 EARTHQUAKE - UNIFORM SUPPORT MOTION - USER oEFINEo SPECTRA There are non11near restraints 1n the job and the stat1c load cases run were: 1 - T+P+w+O+f 2 - P+W+F

3 - 01-02 Use operat1ng case 1 to f1x the nonlinear restraints for the dynamic job and use 3.75 as the friction stiffness mult1plier. The follow1ng shack spectra has been defined: User def1ned shack na.e • TABLE1 Per1 od (sec) 0.001698 0.0286

Accelerlt1Qn (g's) 0.3756 0.984

2.0078 2.0078 1.140 3.078 3.078 1.813 2.256

0.058

0.071 0.091 0.1140 0.1410 0.172 0.20 0.25 0.323

2.256

1.03623

User def1ned snack name • TABLE2. Table 2 1s defined by 3 straight 11nes on the frequency displacement spectrum:

uaax • 10.375 1n. u.ax· 3.832/f 1n. IIIIU· 7. n /f/f in. User def1ned shock na.e • TABLE3 Frequency 0.01 0.5 1.9 3.75 6.39 10.4 15.56 20.01 20.3

(Hz)

Accelerltion (1pss) 58

lOS 210 315 509 789 655 312 119

Table 1 is to be used for mast of the lines at this particular s1te. and so enter 1t into an ascii data f11e te be read in so that it only has te be typed once. Note that TABLE1's values are def1ned 1n tenlS of g's. CAESAR expects accelerat10n in units of length per second squared. For English units th1s _Ins that the values extracted fl"Olll the entered table (1f 1t is entered as shawn above) sllould be scaled by 386. Note how this 1s done in the shock load case editing below. (The same -*386. - could have been used when the data wu entered in the ascii file.) The shack load cases should be set up as follows: SHOCK LOAl> CASE Il:

1.0 * TABLEl 0.6667 * TABLE1 1.0 * TABLE1 Use the spatial

SHOCK LOAl> CASE 12:

1.0 * TABLE1 1n the X direction 0.6667 * TABLE1 1n the Y direction

1n the X direction 1n the Y direction in the Z direction combination method f1rst.

1.0 * TABLE1 in the Z direction Use the .aGal cœbination _thod fint. SHOCK LOAl> CASE 13:

1.0 * 1.0 * 0.667 * Use the

TABLE3 1n the X direction TABLE3 in the Z direction TABLE2 in the Y direction .aGal cœbinltion _thod first.

Once the shock cases Ire run cœbine each dyn_ic load cise with the sustained stltic case 12. Use the ABS _thod of cœbinltion.

-394-

To use shoct TABLE2 1ft! need to generlte the points on 1ts cuneo The intersection of the constant veloc111. constant d1spllc~t l1ne cln be found: 10.375 • 3.832 / f;

f •

3.832/10.375 • 0.36935 cps

The intersection of the constant veloc1ty. constant Iccelerat10n 11ne can be found: 3.832 / f • 7.77 / f/f; f· 7.77 / 3.832 • 2.0277 cps; and th1s corresponds to a displlcl!lleflt of 3.832 / 2.0277 • 1.8898 in. Select a point arb1trlr11y further out on the constant accelerat10n 11ne: ut33HZ· 7.77 1 33 / 33· 0.007135 in. 50 the d1splacement spectrl for TABLE2 will appear: Frequency (Hz) 0.001 0.36935 2.0277 33.0

Displacelent (in.) 10.375 10.375 1.8898

0.007135

ascif file that contains the data points for TABLE1 can be entered using a full screen word processor or EDLIN.COM. Any line in the file that starts w1tll an asterfslt (*) will be treated 1fke a COIIIIIent tfne.

The

TIIe abbreviated dynamfc input for thfs job appears below: Ffrst the user _ust deffne the problem and run the static analys1s w1th the load cases as descrfbed abave. Then frOll the CAESAR 1III1n

menu:

3 - DYNAMICS 2 - SHOCK DEFINITIONS 1 - SpectnIID N_ and Type NIIIIe. Rlnge Type, Ord1nate Type, Rlnge Interpol, Ord1nlte Interpol 'TABLEl PERIOD ACCEL LOG LOG TABLE2 FREQ DISPL LOG LOG TABLE3 F ACCEL LOG LOG Note the 1 s1gn preceedfng TABLEl. TII1s tells CAESAR to reld the TABLE1 input from an Isci1 data file. 3 - Spectrua Table ••• bufld spectru. table 2. (in. ) ORDINATE

(Hz)

RANGE

0.001 10.375 0.36935 10.375 2.0277 1.8898 33.0 0.007135 ••• bufld spectru. table 3. (Hz)

(1n./sec/sec) ORDINATE

RANGE

0.01 0.5 1.9 3.75 6.39 10.4 15.56 20.01

58

108 210 315 509

789 655 312 119

30.3

-395-

8 - SHOCIC CASES

1 - Md new cISe Shock case 1 shock contributions: 5hOCk Maille, Factor, Direction, 5tart Mode, 5top Node, lncrl!llent

TABLE1 TABLE 1 TABLEl

1.0*386 X 0.667*386 Y 1.0*386 Z

5PATlAL(SRSS), MOOAL(GROUP)

1 - Add new case Shock case

2

shock contributions:

Shock Nalle, Factor, Direction, Stan Node, Stop Node, Increllent

TABLEl 1.0*386 X TABLE 1 0.667*386 Y TABLE 1 1.0*386 Z MOOAL(GROUP I,SPATlAL(SRSS)

1 - Add new case Shock case 3 shock contributions: 5hoct Neme, Factor, Direction, Stan Node, Stop Node, lncrl!llent

TABLE3 TABLE3 TABLE2

1.0 X 1.0 Z 0.667 Y

SPATI AL (SRSS l, MODAL (GROUP)

9 - STATIC/DYNAMIC COMBINATIONS 1 - Add new case Load Case, Factor

52

1.0

Dl 1.0 COMBINATION(ABS)

1 - Add new case Load Case, Factor

52 D2

1.0 1.0

COMBINATION(ABS) 1 - Add 1 new cise Lad Cise. Flctor

52 03

1.0 1.0

COMB(AIS)

-396-

A - CONTROL PARAMETERS SPECTRUM l 3.75

<---<---<----

Oyna.ic Analys1s Type Stat1c Load case for nonl1near restre1nts Stiffness factor for friction

B - SAVE DYNAMIC INPUT C - CHECK DYNAMIC INPUT o - PERFORM DYNAMIC ANALYSIS

-397-

INDEPENDANT SUPPORT MOTION SPECTRUM ANALYSIS Earthquake ground motions are caused by the passing of acoustic shock waves thru the earth's soil. These waves are usually hundreds of feet long. If supports having foundations in the sail are grouped together within a several hundred foot radius of each other they will typically see exactly the same excitation from the earthquake. If all of the supports for a particular piping system are attached directly to ground type supports, each support will be excited by an essentially identical time waveform. This type of excitation is known as uniform support excitation. Often pipe is supported from rack, building or vessel structures as well as from ground type supports. These intermediate structures serve to in sorne cases fil ter and in sorne case accentuate the effect of the earthquake. In this situation, the supports attached to the intermediate structure are not exposed to the same excitation as those that are attached directly to ground foundations. Ta accurately model these systems different shocks must be applied to different parts of the piping system. This type of excitation is known as Independant support motion excitation. (ISM's). While the different support groups are exposed to different shocks, there is also a relative movement between support groups that doesn't exist for uniform support exciation. The movement of one support group relative to another is termed pseudostatic displacement. For uniform support excitation there are spatial and modal response components available for combination. A separate response property is computed for each mode for each direction of excitation. Only after all individual- responses are.computed are they summed together in accordance with the user's directions. For independant support excitation there are spatial and modal response components available for each different support group. In addition there is the pseudostatic component of the earthquake that can be added into the dynamic response as well. Components that act along the same spatial direction are termed /ldirectional/l components. Fortunately the NRC has made recommendations for combining each of these different response properties: 1 - Directional Inertial components should be combined using the AaS method. 2 - Modal and spatial responses should be combined using the SRSS method without consideration for closely spaced modes. 3 - Directional pseudostatic components should be combined using the AaS method. 4 - Spatial pseudostatic components should be combined using the SRSS method. 5 - The total response should be found by combining the pseudostatic an inertial components using the SRSS method.

-398-

For running ISM load cases in CAESAR this translates into:

* The modal combination method should be SRSS. * The spatial combination method should be SRSS. * The directional combination method should be ABS. * The pseudostatic combination method should be SRSS.

* It makes no difference whether modal or spatial components are combined first.

The maj or di fference' runni ng ISM type earthquake loads comes in building the shock load cases. Whereas in the uniform excitation case the shock acts implicitly over all of the supports in the system, in the ISM case different shocks act on different groups of supports. The shock 10ad case input form appears: Shock Name,

Factor,

Direction,

Start Node,

Stop Node,

Increment

"Name", "Factor" and "Direction" are al1 that is input for uniform support excitations. For ISM type shocks the group of nodes the shock acts over must also be specified. Users shou1d define restraint support node points with this in mind. For the simple mode1 below, define the ISM cases such that one times the El Centro earthquake acts over the ground supports and one-ha1f times the El Centro earthquake acts over the building supports. This input shock load case would appear: Shock Name,

Factor,

Direction,

Start Node,

Stop Node,

Increment

-------------------------------------------------------------------ELCENTRO ELCENTRO ELCENTRO

1 X 5 30 1 y 5 30 1 Z 5 30

ELCENTRO .5 X 40 60 ELCENTRO .5 y 40 60 ELCENTRO .5 Z 40 60

so '1ç'

.s~

. rD". 6"

"ID

Ir

.3S 20

ID

-399-

2$

30

CiIOOIJQ

The control parameter spreadsheet items to use the NRC recommendations appear below: SRSS SRSS y SRSS ABS

<---<---<---<---<----

Spatial Combination method Modal combination Method Include Pseudostatic Components Pseudostatic combination method Directional Combination Method

For additional discussion, the user is refered to the dynamics example problem C.

-400-

CAESAR II

VERSION 2.1C

DYNAMIC ANALYSIS OF INDEPENDANT SUPPORT EARTHQUAKE EXCITATION

PROBLEM: The cryogenie plplng system shown on the follow;ng page ;s to be designed in aceordance with 831.3 using the ground, building, and uniform speetra shown. Two analysis are to be run: 1) Assume the pipe (structural steel) supports

are rigid.

2) Inelude the flexibility of the structural steel supports by including the steel frames in the analysis.

Finally, compare the results from the two analysis. Design parameters are: Ambient Temperature: Operating Temperature: Pipe: Insulation: Fluid: Columns: Beams:

100 deg F -59 deg F 8 in Sch lOS 4 in 22.3 lb/cu ft 0.232 SG W14x82 WIOx12

CONTENTS: Geometry and Shock Spectrum Definitions ••••• Structural Steel Input •••••••••••••••••••••• Dynamic Input Echo •••••••••••••••••••••••••• Results Comparison &Discussion •••••••••••••

-401-

1.C 4.C 5.C 8.C

CRYOGENIC PIPING DYNAMICS EXAMPLE The ;sometr;c of the complete model ;s shawn in the figure below. This drawing shows the piping, pipe supports, and the structural steel frames.

P,PE,:6"

~

~

1~,,".H.4'T"IOU~" 1.~.3 lII.f_ ..(,.;r.. 1I....f!MGooI'"i T~~~: \00 OpeIL..'T",u.r,

i~IUo1UQ ...~

"1=

- ....,,'F-

,.,»: le

loSa

~.

104'$

,~.

10441

c:::.... " ..... ~ "", •••~ ~_:. , •••"OIl''.2.

'.'S'3

l.e -402-

(TY~.4)

The excitation spectra to be applied to this model are: GROUND RESPONSE SPECTRA BUILDING RESPONSE SPECTRA ENVELOPE RESPONSE SPECTRA

----------------------- ------------------------- -------------------------

Name="GROUND-RESPONSE" T, sec

°

0.05 0.2 0.5 1.0 2.0 3.5 5.0 10 • 0

V, i n/sec

0.787 7.874 21.653 39.37 18.89 43.7 11.8

5.9

Name="BUILDING-RESPONSE" T, sec 0.05 0.2 0.5 1.0 2.0 3.5 5.0 10.0

V, i ni sec 0.787 1.3 3.4 27.3 30.4 21.12 21.3

5.359

Name="ENVELOPE" T, sec 0.05 0.2 0.5 1.0 2.0 3.5 5.0 10.0

V, i ni sec 0.787 7.874 21.653 39.37 30.4 43.7 21.3 5.9

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

ENVELOPE SPEcrRUM

Acts on aIl support points in the model

UNIFORM SUPPORT EXCITATION 2.C

-403-

y

SPECTRDr-I A~ts on aIl points supported d~rectly from the building.

GROUND SPECTRUN

Acts on ail points supported directly from the ground.

INDEPENDANT SUPPORT EXCITATION

For the analysis with steel supports, the structural steel must be included as part of the piping model. This can be accomplished by using the "INCLUDE STRUCTURAL INPUT FILES" option fram the "KAUX" feature of the CAESAR II spreadsheets.

3.C -404-

The structural steel model for this problem can be generated by invoking the C2S program. The input listing from the C2S session is shown below:

SECIO=l, W14X82 ;COLUMN CROSS SECTION SECIO=2, W10X12 ;BEAM CROSS SECTION MATID=l, YM=29E6 POIS=0.3 G=11E6 DENS=0.283 DEFAULT SECID=l ANGLE=90 EOIM 1038 1039 DY=15-0 ;DEFINE ALL COLUMNS EOIM 1043 1044 DY=15-0 EDIM 1048 1049 DY=15-0 EDIM 1053 1054 DY=15-0 DEFAULT SECIO=2 ANGLE=O EDIM 1039 1040 DZ=-2-0 ;DEFINE ALL BEAMS EDIM 1044 1045 DZ=-2-0 EDIM 1049 1050 DZ=-2-0 EDIM 1054 1055 DZ=-2-0 FIX 1038 ALL FIX 1043 ALL FIX 1048 ALL FIX 1053 ALL

4.C -405-

The dynamics input for this problem is summarized in the figure below. Details of the dynamics input are contained on the following pages.

Define 3 shock response spectra: 1) "GROUND-RESPONSE" for supports 5, 1038, 1043, 1048, 1053

1 • UII'O MSSES

2 - SIUIEIS

2) "BUILDING-RESPONSE" for supports

~ - IIIIX IEFIIITIIIIS 4 - PlUE TUlL IEIEIIn.

5-

65, 70, 80, 115

lia ut TIU ....n.

, - RIEF LDIII ImIIIIl 7 - N.S[ &PEt1IIII FGIŒ sm 1 - IlllllaIC LWS

3) "ENVEL OPE

Il

for a11 supports

, - IIIIX WB ___ A- STATlt/DYIIIIIIIC CIIIIIITIIIIS • - CIIITIIII. PMMETEIS

C• EF-

SAVE .YIIIIIIIC 1!nT DEa ''''''''IC 1ft POfIIIIII ITIMIC AIIUSIS a.t.nt t. anAl ail _

2 shock load cases: 1) Independant Support Excitation case (Esc)

2) Uniform (Enve10pe) Support Excitation case

~e1l

CAESAR what type of dynamic analysis to rune

5.C -406-

CAESAR II DYNAPlICS INPUT DATA CHECKIN6 JOBNAPIE = CRYISPI_7 UNITS: Length (in.) Force !lb.) ftas5 IIbl) SU ff
(HARftONIC/SPECTRU"I"ODES/RANGE~

33

Stauc: load Case for Nonlinear Restraint StatU5 {----- stiffness Factor for Friction (O.O-Hot u5ed) {----- "ax. No. of Eigenvalues c:alc:ulatl!d (a-Not u5ed) {----Frequenc:ycutoff !HZ)

0.1 20 0.03 0.5

{----{----{---{-----

N N

(----
?

0.0

o

PlOOAl {----.. (----5RSS (---y {----SRSS {---SRSS (---..,...... .., ... ...... fII. ...

Control Parameters Input Data Echo Tell CAESAR to do a Spectrum Analysi s. Only the four entries shown are required for thi s jOb All other val ues are the original defaultso

{-----

0

Closely Spac:ed Plode Criteria Earthquake Duration (For DSRSS .ethod) !sec.) Struc:tural Dalping (% of c:ritical) Zero Period Acceleration (For Reg. Guide 1.601 (g'5)

Spatial or l'iodaI COlbination first ...._ _ _ _ _ _ __ Co!:binatior: !leine' (SRSS/AeSl • Parameters modi fied to correspond wi th !'IodaI COlcination P\ethoc [6ROU?!10i.!DSR55?ABSiSRS5~ the NRC 1 S recolllllended practice for Inc:lude Pseudostatic Co.ponents (Y/NI ~ combi ni ng ISM shock components Pseudostatic COlbination Plethod iSRSS/ABSl Directional Colbination Plethod ISRSS/ABSI S~atial

0

N

{-----
o 2

o o N N 100

;: ----- 5ubspace süe iO-Not Used 1 {----- No. to Converge before Shift Allolled (O-Not Used) (----- No. of Iterations per shift (O-Pgl co.puted) (----- % of Iterations per shift before orthogonalization {---- Force orthogonalization after convergence (Y/NI {---- Use out-of-core eigensolver (Y/N) {----- Frequency Array Spaces

60C -407-

DYNA~IC

LOAD CASE SHOCK CONTRIBUTIONS

Nau, Factor, Ilireeti on, Start, End. Ine

LOAD CASE 1 1 6ROUND-RES?ONSE 1.0 SROUNIl-RESPONSE 1.0 6RDUND-RESPONSE 1.0

.. liAlue

y

1005 1055 1005 1055 1005 1055

BUILDINS-RESPONSE 1.0 1 65 115 BUILDINS-RESPDNSE 1.0 y 65 115 BUILDINS-RESPONSE 1.0 Z 65 115

..,---------'ISM Load Case and support group definitions

DYNA"IC LOAD CASE 1 2 "ODALISROUP) ,SPATIAL (SRSS) ..- _ _ _ _ _ _ _ _ _ _ __

.Combi nati o'n Methods Redefi ned for uniform support 10ad case

ENVELOPE 1 1 ENYELOPE 1 y ENVELOPE 1 Z

S?ECTRU" DEFINITIONS Hile, Range, Ordinate, Range InterpDlation, Ordinate InterpDlatiDn 6ROUND-RESPONSE PERIOD YELOCITY LOE LOS .. , - - - - - - -...."GROUND-RESPONSE" Spectrum Definition 0.05 0.787 0.2 7.974 0.5 :1.653

i.O 39.37 :. i)

18.29

3.5 43.7 5.0 11.8

l i

entered "PERIOD vs VELOCITY" J'--------------.. userSpectrum table data

10.0 5.9 BUILDIN5-RESPONSE PERlOD YELOCITY LOE LOS 0.05 0.787 0.2 1.3 O.S 3.4 1.0 27.3 2.0 30.4 .,."; • ..1 ~

-

1Full Definition for the shocK "BUILDING-RESPONSE"

21.12

5.0 21.3 1(;.0 5.359 ENVELOPE PERIOD YELOCITY LOE LOS 0.05 0.787 0.2 7.874 0.5 21.653 1.0 39.37

2.0 30.4 3.5 43.7

.. ---------------1-Full Definition for the shocK "ENVELOPE"

5.0 21.3 10.0 5.9

7.C -408-

In order to keep the documentation for this example brief, the only results presented are those for the "uniform support excitation" case. Using this load case, the model with and without structural steel supports will be compared. The results from these two models are shown in the tables below: DISPLACEMENTS X

y

Z

RX

RY

RZ

35 with . without

0.4253 0.0049

0.0336 0.0076

1.5831 1.0334

0.4298 0.2902

0.5932 0.3832

0.0622 0.0033

45 with without

0.4240 0.0036

0.0379 0.0

3.7952 1.9555

0.2311 0.1635

0.5550 0.2576

0.0412 0.0007

50 with without

0.4219 0.0020

0.0447 0.0

3.7435 1.4764

0.1911 0.0817

0.5695 0.4083

0.1220 0.0002

60 with without

0.3799 0.0366

1.4247 0.5838

0.5930 0.0635

0.3613 0.0292

0.3534 0.0425

0.2322 0.0236

75 with without

0.8484 0.6447

1.3529 0.5631

1.3033 1.1291

0.5127 0.4482

0.4247 0.3346

0.4924 0.2114

90 with without

0.5927 0.4689

0.4228 0.3414

0.2087 0.1815

0.3816 0.3425

0.5229 0.4236

0.4461 0.2465

8.C -409-

RESTRAINT LOAQS FX

FY

FZ

MX

MY

MZ

241 207

319 353

523 353

4761 3114

981 647

1133 1001

40 with without

146 18

1118 597

45 with without

229 4

50 with without

CYJ 1536 434

3848 8100

1116~~ 1

2531 2286

1568 1339

4025 2701

5

with without

55 with without

2029 976

65 with without

956 580

1101 560

70 with without

538 500

895 743

80 with without 115 with . without

1939 1408

1154 596

236 110 743 504

253 200

429 359 STRESSES

AXIAL

-410-

BENDING

TORSION

MAX OCT

CODE

20F with without

80 88

20614 13344

1742 1151

9834 6363

20639 13350

35F with without

22 17

13454 8558

571 280

6366 4041

13468 8559

40 with without

164 122

7179 4779

571 280

3431 2265

7211 4782

45 with without

297 193

11001 7963

571 280

5246 3762

11081 7966

55 with without

429 232

16435 11664

571 280

7832 5504

16582 11667

55 with without

140 86

15886 17125

1009 148

7600 8114

16024 17210

60F with without

340 357

20784 12164

696 414

9920 5911

21114 12520

75F with without

69 59

11489 6208

375 281

5448 2963

11539 6267

9.C

DISCUSSION OF RESULTS The comparison table on the previous page illustrates the differences that can exist when the structural steel models are not included in the analysis. ln some cases 9 the results with the structural steel included are many times higher than the results computed without the structural steel. The steel models add flexibility ta the piping system. More flexibility means lower natural frequencies and more modes to be excited by the shock. A comparison of the natural frequencies of the two models is given below:

NATURAL FREQUENCIES No. 1)

2) 3)

4) 5) 6) 7)

8) 9) 10) 11) 12 ) 13 )

14) 15 ) 16 ) 17)

18 ) 19 ) 20) 21 ) 22) 23) 24 ) 25 ) 26 ) 27 ) 28)

With Structure

--------------

Without Structure

----------------1.706 2.533 3.371 3.936 4.384 5.294 5.929 8.957 11.849 16.367 16.564 20.588 22.954 23.474 25.582 29.685 35.083

1.307 2.244 2.520 3.149 3.443 4.206 4.404 5.250 5.675 5.761 5.988 6.594 7.992 11.855 14.086 14.086 14.086 16.504 15.554 20.333 20.589

I~::I 23.052 23.475 25.582 38.085

10.C -411-

ln the above table, there are on1y 5 extra mode shapes for the system which includes the structure. The 6 extra shapes shown in the two boxes belong strictly to the structural steel frame. (Exciting these modes will obviously cause stresses in the piping, a possibility which could go unnoticed in the simpler "piping on1y" model.) The restrain moment at node 55 in the Z direction is much 1arger without the steel mode l than it is with the steel mode1. Even though the piping is tied to the steel, the steel frame will not support much moment in the Z direction. The steel frame bends slightly about the Z axis, and the moment is carried throught from the pipe. ln the "piping only" model, the rigid anchor at node 55 will not rotate about the Z axis (or any other axis) and so ends up carrying al1 of the moment load.

11.C -412-

RELIEF VALVE LOAD ANALYSIS: Relief device loadings are due to the sudden exhausting of a liquid or gas. By far the most difficult design problems involve the relieving of gas. There are several reasons for this: 1 - The potential energy stored in high pressure gas can cause considerable damage when converted into kinetic energy. 2 - There are many installations of gas relief valves. i.e. in every steam plant. 3 - Relief valves tend to "pop" open. and "close" quickly. resulting in sudden loads on the piping system. Liquid safety valves tend to open partially. or very slowly. deadening the effect of the fluid momentum change. For these reasons the GAS relief problem will recieve the majority of the attention here. CAESAR has a liquid relief synthesis model that functions in an identical manner to that for gas. There are two types of destructive dynamic forces associated with gas relief devices: 1 - Thrust at the valvejatmosphere interface 2 - Accoustic shock due to the sudden change in fluid momentum and the associated traveling pressure waves. A rarefaction wave is initiated at the valve orifice on opening. and a compression wave is initiated at the valve orifice on c10sing. These waves are usually ignored. but may cause damage. The magnitudes of the rarefaction compression waves are estimated in the relief load synthesizer in CAESAR, and are dealt with just 1ike any other (waterhammer) type of traveling pressure wave. The figure below illustrates these two types of loading:

-413-

. P.a..~Swl~ THAT

. l·· - _ . _ .

WAVt:.S ...

..

~'\tn. a~

Qo,",,~ .. Pl?IJ.JC,

,

\

&>p~~~_.

1

:."-,

,

.

..

~~

:..

.

TITE

$'l"~~

~ ~ .~ --,.~ ~i.... ..~ - ... ..

-.-

(~~Qr..) )

PI2.L'~~rU.) o~

LoQ

~

WA."Ff"72,.O~ T

Vt+c.."EOf>~t.AJ,.

·CI..o5.d) (oi!.F~ HA~I"''i

....- .. bœ.t.J

o~.)

1---1 ~-~~lo...Jt( C~ HI"""

P~~SS"'\l.t:) 01>.) VALVE'

The first step in any relief load analysis is to compute the magnitudes of the relieving forces. For open-type and similar vent systems the CAESAR Relief Load Synthesizer can be used.

-414-

1".0,) """

~~'ï

~,~~.

CAESAR assumes that a successful vent stack/relief system design maintains the following gas properties:

Supersonic conditions as the gas flow expands after leaving the Relief Valve Orifice

~sonic

gas flow conditions

Steck

Gas expanding from subsonic to sonic conditions at the vent stack exit Clel'l'\.s~c:urlng V~nt Steck

PIpe

'"

~SUbsoniC

gas flow conditions

' " Obli que shock sy stem mode 11 ed as norma 1. shock Soni C (ChOked\ gas fl ow

conditions at relief valve orifice

The assumptions made in the relief load synthesizer are as follows: 1) The gas can be dealt with using the ideal gas equation of state. The compressibility factor (Z) is a measure of the gases deviation from ideal. In the case of steam. over a wide range of pressures and temperatures the ideal gas assumption is a very good one. Sorne typical cornpressibility factors for steam are shown below: At

10 psi and 400 deg. F. 10 1600 10 3200 100 400 100 1600 100 3200 800 800 800 1600 800 3200 800 1500 1600 1500 1500 3200 2000 800 2000 1600 3200 2000 4000 800 1600 4000 4000 3200

Z = 0.9965

0.9999 1.0053 0.9469 0.9990 1.0022 0.9336 0.9925 1.0029 0.8695 0.9859 1.0040 0.8188 0.9813 1.0049 0.5608 0.9647 1.0093

-415-

2) Gas properties follow an isotherm from the relief valve orifice to the vent stack shock wave. (Temperature of the gas doesn't change in this short distance.) 3) Gas expansion from the shock wave thru the vent stack is adiabatic. (No heat into or out-of the vent stack.) 4) Sonic conditions exist at both the relief valve orifice and at the vent stack exhaust into the atmosphere. (Both can provide the li.iting criteria for flow.)

5) The oblique shock system in the vent stack which may exist over a large distance of the vent stack piping~ can be modelled as a normal shock wave of essentially zero length. The input for the relief load synthesizer is shown below:

<---<---<---<---<---<---<---<---<----

Line Temperature Cdeg. F) Line Pressure (psia) ID of Relief Valve Orifice (in.) ID of Relief Valve Piping (in.) Length of Vent Stack (in.) Ratio of Gas Specifie Heats (k) Gas Constant (R) (ft.lb./lbm./deg. R) Ooes the Vent Pipe have an Umprella fitting (Y/N) Should CAESAR size the Vent Stack (Y/N)

-416-

LINE TEMPERATURE (CEG. F) Enter the stagnation condition temperature of the gas to be relived. (Usually just the gas temperature upstream of the relief valve.) LINE PRESSURE (PSIA) Enter the stagnation pressure of the gas to be relieved. (Usually just the gas pressure upstream of the relief valve.) Note that stagnation properties can vary considerably from line properties if the gas flow velocity in the line is high. ID OF RELIEF VALVE ORIFICE (IN.) Enter the flow passage inside diameter for the smallest diameter in the relief valve throat. (This information is usually provided by the relief valve manufacturer.) ID OF RELIEF VALVE PIPING (IN.) Enter the inside diameter of the piping attached directly to the exhaust of the relief valve. ID OF VENT STACK PIPING (IN.) If CAESAR is to size the vent stack then leave this ID blank. If the vent stack piping is the same size as the relief valve piping, i.e. it is one-in-the-same, then this field May be left blank. Otherwise enter the inside diameter of the vent stack piping. LENGTH OF THE VENT STACK (IN.) Enter the length of the vent stack. This is a required entry. double the lengths of fittings and elbows. RATIO OF GAS SPECIFIC HEATS (k) Some typical values for this constant are: Superheated Steam. Saturated Steam ••• Nitrogen •••••••••• Carbon Oioxide •••• Acetylene ••••••••• Ammonia ••••••••••• n-Butane •••••••••• Ethane •••••••••••• Ethylene •••••••••• Methane ••••••••••• Propane ••••••••••• t..l4-r\..o\i2..AL. ~~

Il ••

1.300 1.100 1.399 1.288 1.232 1.304 1.093 1.187 1.240 1.226 1.127 l.27

GAS CONSTANT (R) (FT.LBF./LBM./OEG. R) Some typical values for this constant are: Nitrogen •••••••••• Carbon Dioxide •••• Acetylene •.••••••• Ammonia ••••••••••• n-Butane •••••••••• Ethane •••••••••••• Ethylene •••••••••• Methane ••••••••••• Propane ••••••••••• "'l4.,.. "'" a. A-~

55.16 35.11 59.35 90.73 26.59 51.39 55.09 96.33 35.05

a It-S ., •• Î'1. 1

-417-

Add

OOES THE VENT PIPE HAVE AN UMBRELLA FITTING (YIN) . Enter a Y or a N. See the figures below to determine if the connection of the vent staCK to the vent piping is via an umbrella fitting.

V~nt St~ck

Cl~l'Ip S~CU"'1n9 V~nt St~ck

V~lv~

Pipe

An umbrella fitting exists whenever the vent stack pipe is not hard piped to the relief valve pipe.

UMBRELLA FITTING EXAMPLE

V~nt

Steck

ReUef Va.lv~ Pipe

NOT An UMbrella. Fitting

SHOULO CAESAR SIZE THE VENT STACK (YIN) Enter a Y if CAESAR should size the vent stack. The sizing algorithm searches through a table of available inside pipe diameters starting at the smallest diameter until a vent stacK id is found that satisfies the thermodynamic criteria shown in the figure above. The computed ID is automatically inserted into the input.

-418-

Example input and output fram the relief load synthesizer is shown and discussed below: 1000 2800 2.141 6.065 50*12 1.3 85.8 Y Y

<---<---<---<---<---<---<---<---<---<----

Line Temperature (deg. F) Line Pressure (psia) ID of Relief Valve Orifice (in.) ID of Relief Valve Piping (in.) ID of Vent Stack Piping (in.) Length of Vent Stack (in.) Ratio of Gas Specifie Heats (k) Gas Constant (R) (ft.lbf./lbm./deg. R) does the Vent Pipie have an Umbrella fitting (Y/N) Should CAESAR size the Vent Stack (Y/N)

Computed Mass Flowrate (Vent gas) ••••• (lbm./hour) Thrust at Valve Pipe/Vent Pipe Interface ••• (lbf.) Thrust at Vent Pipe Exit ••••••••••••••••••• (lbf.) Transient Pressure Rise on Valve Opening (psi.) Transient Pressure Rise on Valve closing (psi.) Thermodynamic Entropy Limit (Should be > 1 ) •••••• Subsonic Vent Exit Limit (Should be > 1 ) •••••• (psia) Valve Orifice Gas Conditions ••• (P,vel,T) 152S.0 46.1 Vent Pipe Exit Gas Conditions •• (P,Vel,T) 64.3 Subsonic Velocity Gas Conditions (P,Vel,T)

388229.300 17658.S50 1051S.7S0 347.091 247.542 2.904 1.041 ( fps ) ( deg.F ) 2135.4 S09.6 1848.S S09.6 140S.S889.4

COMPUTED MASS FLOWRATE (VENT GAS) This is CAESAR's computed gas mass flowrate based on choked conditions at the relief orifice. If greater mass flowrates are expected then the error in either the approach used by CAESAR or in the expected mass flowrate should be investigated. THRUST AT VALVE PIPE/VENT PIPE INTERFACE If there is an umbrella fitting between the vent stack and the relief valve piping then this is the thrust load that acts back on the relief valve piping. (See the figure below). If the vent stack is ha rd piped to the relief valve piping then this intermediate thrust will be balanced by tensile loads in the pipe and can be ignored.

dJ-Vftt 1

1:1

s-

UMBRELLA F"ITTING

UMBRELLA rlTTING

~oth thrust loads act simultaneously during

a valve firing.

Only the valve pipe/vent stack interface thrust acts in this configuration.

-419-

THRUST AT THE VENT PIPE EXIT When there is an elbow in the vent staCK plplng, this is the thrust load that acts on the elbow just before the pipe opening into atmosphere. (See the figures below for clarification.)

UMBRELLA rITTING

NOT

NOT

An Ul'lbrello. tittlng

An Ul'IbreUo. tlttlng

TRANSIENT PRESSURE RISE ON VALVE OPENING This is the estimated magnitude of the negative pressure wave that will be superimposed on the line pressure when the relief valve first opens. This negative pressure wave will move bacK thru the relief system piping similar to the pressure wave in the downstream piping of a waterhammer type system. The magnitude of this wave is estimated as (Po-Pa)*Ap, where Po is the stagnation pressure at the source, Pa is atmospheric pressure, and Ap is the area of the header piping.

-420-

TRANSIENT PRESSURE RISE ON VAlVE CLOSING This is the estimated magnitude of the positive pressure wave that will be superimposed on the line pressure when the relief valve slams shut. This positive pressure wave will mave bacK thru the relief system piping similar to the pressure wave in the supply side piping of a waterhammer type system. The magnitude of this wave is estimated fram: r*c*dv where r is the gas density, c is the speed of sound in the gas and dv is the change in the velocity of the gas. THERMODYNAMIC ENTROPY LIMIT SUBSONIC VENT EXIT LIMIT This. values should always be greater than 1. If either of these computed limits fall below 1.0 th en thermodynamic assumptions made regarding the gas properties are incorrect and the computed thrust values should be disregarded. VAlVE ORIFICE GAS CONDITIONS VENT PIPE EXIT GAS CONDITIONS SUBSONIC VELOCITY GAS CONDITIONS These are the thermodynamic properties of the gas at three critical points in the relief system. These three points are shown in the figure. The entire formulation for the thrust gas properties is based on an ideal gas equation of state. If the pressures and temperatures displayed above for the gas being vented are outside of the range where the ideal gas laws apply then some alternate source should be sought for the computation of the system's thrust 10ads.

-421-

The gas load on the piping system has two components: 1) The pressure times area. or P*A component. and 2) The mornentum change component. When the gas goes around a 90 deg. corner, the momentum of the gas is changed frorn the horizontal direction into the vertical direction. Sorne force must be exerted on the gas to accomplish this change in the direction of the momentum. The magnitude of these forces is found in any first year dynamics book as: Force

= (dm/dt)

V

where: (dm/dt) - is the mass flowrate = (rAV). V - is the gas velocity, r - is the gas density, and A - is the inside area of the pipe.

*18-6. Steady-Fluid Streams Knowledge of the forces developed by steady moving fiuid streams is of importance in the design and analysis of turbines, pumps, blades. and fans. The principle of impulse and momentum may he used to determine these forces. Consider, for example, the diversion of a steady stream of fiuid (liquid or gas) by a fixed pipe, Fig. 18-16a. The fiuid enters the

fnoooA ' ~6.Ir.lE.E.il..l"'~ M~'-~4toJ\GS; s-r1l),'T'\c.~ ~.c.

(a)

+ Timer

,

,,

,

,, ,>

"

pipe with a velocity VA and exits with a velocity vB • The momentum- and Impulse-vector diagrams for the fiuid stream are shown in Fig. 18-16b. The force !F. shown acting on the impulse-vector diagram. represents th~ res~ltant ~orce of ail the external forces acting on the fiuid stream. It .lS. thls loadmg which gives the fiuid stream an impulse whereby the o~gtn.al m~mentum of the fiuid is changed in both its magnitude and dIrectIon. Smce the fiow is steady this force will be constant during the time interval dt. As shown in the figure. the fluid stream is in m;tion. and as a result a small amount of fluid. having a mass dm. enters the pipe with a velocity l'A at time l, Considering this element of mass and -422-

A.lJt>

D'<' ",Ar",," -cS

~L'ôQ~t~

the mass of fiuid in the pipe as a closed system. at time t + dt. a corresponding element of mass dm must leave the pipe with a velocity YB' The average velocïzv of the fiuid stream. having mass m. within the pipe section is constant during the time interval dt. In Fig. 18-16b, its velocity is shown to be v. Applying the principle of impulse and momentum to the fiuid stream. we have dm VA

+ mv + :::Fdt = dmv B + mv

Solving for the resultant force yields (18-15) The term dm/dt is called the massfiow and indicates the constant amount of fiuid which fiows either into or out of the pipe per unit of time. (vB - v.-l) represents the vector difJerence between the input and output velocitv of the fiuid stream. Provided motion of the fiuid can be represented'in the xy plane. it is usually convenient to express Eq. 18-15 in the form of two scalar equations: ~~ ~FI!

dm = dï(VBz -

=

VAz)

(18-16)

dm dï(vBI/ - VAl!)

The force summation in Eq. 18-15 or Eqs. 18-16 may easily be accounted for by accompanying the problem solution with a free-body diagram. Only the entrance and exit velocities of the fiuid are required. These velocities represent the relative velocities of the fiuid with respect to the system. For sorne problems. a kinematic diagram for the velocities will help in determining their values (refer to Example 18-9).

No~e ~hat ~he

relief load

syn~hesizer

also computes the gas mass

flowra~e. This flowra~e should no~ be greater ~han ~he manufac~urers ra~ed flow, and if much less, ~hen ~he process engineer ~ha~ àesigned ~he relief sys~em should be no~ified, as ~he ven~ s~ack choked flow

may be ~he llmi~ing factor on mass flowra~e, i.e. ~he relief system might not be able to sufficiently vent the excess steam !!!

condi~ion

-423-

From the relief 10ad synthesizer the fo110wing values are obtained: 1) The magnitude of the dynamic 10ad that acts on the relief tai1pipe and valve due to the pressure, and change in momentum of the re1ieving gas.

2) The magnitudes of the opening and closing pressure waves that propagate back into the f1uid from the relief valve. Once the magnitude of the relief 10ad is known, sorne estimate of its time waveform must be made. This can be computed from the maximum pounds of gas to be vented at any one firing. The duration of the relief valve blast can be found fram: Duration

=

Mass / Flowrate

EXAMPLE: Assume that on over-pressurizing a system, 3000 lbm of steam must be vented. From the relief load synthesizer, we can see that 388229.3 1bm of steam per hour will get thru the orifice/tailpipe configuration. Once the valve opens i t wi 11 take: (3000)lbm / (388229.3)lbm/hour

=

0.00773 hours seconds

= 27.3

to vent 3000 lbm of steam.

The longer the duration the more conservative the dynamic calculation. Longer durations result in quicker rise times in the frequency response spectrum diagram. This is illustrated below:

~~------~----------~--D~F~~.o

~"~~----

~o~.Ç'~ ....~vf:-

--~~~'-5~._

.--_ . .

_~_-:_-

.l~I...,t; . TI ioOo\ ~ I-k. S' n;, 12.-'"

___. . _-- .- -

L~4. ~w.n..""='TlO~ J...~;~ ~ TH:E __ fu~~~I:::~.ç"l

___ C\JQ.v~ . _2tSé._.Y~_aC4.1.c.t-"~{ _._

-424-

To-- "2._0. __

---_-_} -q ~$;"'~T_ n~~O.L~"::'~_-=-::: --

-

lSA-'"(

-- -

--- -- --

-- --- -

-~

< So

~) - - -

-~

-

-

-----

.

-=- -=-

_.-

~ -=-- ---_.,._-

-

_ ._ $th:>Ik.T___ D~f1...ATIOJ•.LLo~:- T"t_6"_. .E:~~Q.JC'{_~~'.PC~rE ___ ___ G.1l2rt(ti: -i2Js~- _$'4:c:;>~,,:i __

Th__Z ...O_~

____. ____. --_.. -

--- ---

An accurate estimate of the rise and fall time of the valve is also

of use. The shorter the estimated rise and fall times the more conservative the dynamic solution.

Just like for the earthquake analysis, the time waveform is converted into a response spectrum by exposing a table of single degree of freedom bodies to the time waveform and saving the maximum displacement of each.

---

-- -

-

--~-

'. '.""'"F. • i

~ : _. __ . ~~_ --:~_._ _ _

•-; ~ -- _. --- ------- --- - -- - _.__. _~ ___. _ c.~ n..Y.!iL~

n.f::::--'2fl.$:>bi~-fJ5... _.

_ _ _ _ _ _ _ _ _ .~_.!!!M~a......J'-I~~~~L.

, -.-- ---------'F="'T---=cé=

-~....;...---

-t--+rt---'+---~~---.,.....---.---.-

.--------

----- .--:--t--rt-._",-7--~---r-------------_. -+-i~~~~~-------...:......~----_ --~:-:-'-'-----------' -'----'-----'--~....;.-----_ ..

------------,.-\----'------------------ ~--'---'--~-'-~

----,'---,- n""'1 e d

-425-

_---

The input procedure for the time waveform and the resulting response spectrum from CAESAR are shown below: 1 - INPUT 2 - STATICS 3 - DYNAMICS <------- Go to DYNAMICS 4 - OUTPUT 1 - LUMPED MASSES 2 - SNUBBERS 3 - SHOCK DEFINITIONS 4 - PULSE TABLE GENERATION <------ Pulse table generation 5 - WIND GUST TABLE GENERATION 6 - RELIEF LOAD SYNTHESIS

FORCE SPECTRUM GENERATION Enter the force spectrum name -----> Enter the maximum table frequency (Hz.) --->

<40>

Enter the des1rea number of points in the table ----> 20 TI ME mill i sec 0.0 5.0 15.0 20.0

FORCE lb. 0.0 1.0 1.0 0.0

The following Force Response Spectrum Table 1s generatea:

FREQ (HZ)

~lTIPlIER

.1112511 .1141111 .8282511 .1641111 .1562511 .3239999 .6112498 1.8241'18 1.6412491 2.4999998 3.6612491 5.1839991 7.1412491 9.6141111 12.6562511 16.3841111 21.8812688 26.2441111 32.58127'1 41.1111211

.1134219 .1183528 .1118866 .1161185 .8147271 .1315411 .1565656 .1964677 .1544218 .2351151 .3431735 .4832111 .6589428 .8712571 1.1159718 1.3799111 1.6356938 1.8364651 1.9341221 1.9354891

\.

--" lPo,_rs

\~

10

-:â

2D

TI""",e"""~~lseCcwos

LOIJ

F-cu::~\oot E"IJC ,lOS' ~H.4. J.J.or

Ê.'l.G/TCi~.

/4./

-426-

8E "S'i.A. i!>S,."-ITI /fIru.'(

Next ~he poin~ of must be entered.

applica~ion

of

~he

load, its direction and magnitude

For a single relief valve, or for multiple relief valves firing simultaneously this results in a single force set. For an interplay between more than one relief valve there may be multiple force sets. The examples below should help illustrate: EXAMPlE:

1 A Single relief valve fires. There is a force on the tailpipe elbow (P*A + mY*Y), calculated from the relief load synthesizer equal ~o 4532 lb. Because of the umbrella configuration there is also a load on the relief piping equal to the value from the relief load synthesizer of 6789 lb. The following force spectrum loads would be en~ered:

'-'.NI f,/2.El.L.A FtT/IAlt:,

j ,-.

-r\._

··'l·..

ro

J/~?J.."' __ ./!.l;E.S~;~/2.T

Mu.t7 4 E" /)t;ft~"'EJ.) _Tp-u.._..Îl!E. .. J2nY/'TIAll; _ .7~. LoLt4s._.

_ .

...

~

. ...... ..._-_ ..__ ...._ -_ __ __ ...

...

..

.......

_ _ _ _--_ _ _ _.....

..

...

..

......

...

-..

",

.,.

_. _.

4532 lb. in the X direction, force set #1. -4532 lb. in the y direction, force set #1. There will be a single dynamic load case that contains force set #1.

-427-

_

. . . _.-

--

...

-

EXAMPLE:

2 A bank of relief valves are assumed to open and close simultaneously. They are at nodes 210, 310 and 410. The thrust force from the relief load synthesizer is equal to 1331 lb. The following force spectrum loads would be entered:

AioTri

rH*T"

TH67J.E

X ~'pf)"'I;';wTS 'IN A

T i-fE

1(;

A-n.E- No if.ft;;-SE

L(}IH)~I

~~WI/U:.7)_ 11"/lJ~ t:oA1FJg"un()~ . J( .P112.~"·~.:"'1 0"'; fc"~é! ~

7;..~E1J t(P

As

THE

.ID

7f,f~

/"fS

Vé.'2'f"

6".5#1 ~ FI Ll..$.T

J or 71n\J.S/fMJ IN 'THe !>tyG BE'TWG:1J.. THF;O/~, F/CE. A-IJ j) TH-E Ftl1,sr . .Ê2..&>..kJ. rffEn.E /$ A hlZC..E'. /~è-+/..b.J.CE_ r#~.,. É)tJT..s /~n/'Z.F~U,€ (c~6.dt..J

SHrJlZr

MOr.lr:$_

7ï.t14T.

10$

J7z-0M

1~~'~7J

7~

,$ tCA"ISC

PII'IIJ.A77DI-J.

-1311 lb. in the Y direction at 210, force set #1. -1311 lb. in the y direction at 310, force set #1. -1311 lb. in the y direction at 410, force set #1. There will be a single dynamic load case that contains force set #1.

-428-

OtZ.Ir:le~

oç

EXAMPLE:

3 There are two relief valves in a bank. They may open individually or simultaneously. The individual firings may coincide, i.e. the first valve may open, fire and begin to close when the second begins to open. The first may open, close, then open again before the system has come to rest, or while the second is open. The steady state thrust load for each of the relief valves is 1679 lb. The relief valves are at nodes 150 and 160.

~i:)y

STA-TE. 1"J+t2MST

r/2.ol"-1

TH6-. -ên/t;r ~OI'I:.o

... -:-,/JVTH€S/~:'

/fp 7~

i ' ( ? \ëÂ~ . -Rii..1 Er:

LoA-O

1..8. ..

Ti ....... E

". ~~ë"·-F\-;2..I~ .~s.mT2,"(

ThiS type of relief installation is particularly difficult to design for. There are two complexities that act together in the same problem: 1 - The independant interaction of more than one thrust load, and 2 - The repeated application of the same thrust load. unfortunately, this is often the situation that exists at multiple valve relief stations. The force sets are defined as shown below: *Set #1 is the independant firing of the valve at 150. -1679 lb. in the Y direction at 150, force set #1. *Set #2 is the independant firing of the valve at 160. -1679 lb. in the y direction at 160, force set #2. *Set #3 is the simultaneous firing of both valves. -1679 lb. in the Y direction at 150, force set #3. -1679 lb. in the Y direction at 160, force set #3. -429-

Note that the LOAO CASE COMBINATIONS are used to affect the variation in the loading history, (NOT the force sets!!!). The load cases would be defined as follows: *LOAD CASE 1 -- SIMULTANEOUS FIRING Contains 1 times force set #3. *LOAD CASE * * Contains Contains Contains

2 -- VALVE 1 FIRING 3 TIMES, VALVE 2 FIRING 3 TIMES, ALL COMPONENTS ADDED USING THE SRSS METHOD, WITH 10% MODAL GROUPING. 1 times force set #1. 1 times force set #1. 1 times force set #1.

Contains 1 times force set #2. Contains 1 times force set #2. Contains 1 times force set #2. DIRECTIONAL(SRSS),SPATIAL(SRSS),MODAL(lO%) The exact CAESAR input describing this combination loading case is shown below:

FORCE SPECTIUI EDITINS - Ùirectlon lIily !le 1, 'f or l, cosues, or a Olrec:tlon veetor. Eac:n force set aeflnes a \I1IIQU! OynalllC load c:onflguratlon. !-orees glYen nere lIUSt !le USI!CI in COrIJunCtlon ..ltn a force SlEctl"lll. taIt • on tlle rlUllerlC keypad serell tne Input. Hny llne startl" .itn an asterlSk (*J ..U! !le taken as a ~. Errors nave lJeen Cletectea on arry Une SIlOII'I bunnng. (lb.)

Force, Direction, NoCle, Force set 1

1 *FORCE SET Il IS ltE

I~T

1

FIilMi CF IH:: VR..~ AT 15()

. -1679 Y 150 1

*FOII:E SET 12 15 ltE lNDEJ:lEJGNT FIilNS t1= ltE WLIJE AT IbO -1679 Y 160 2

*FORCE SET .3 IS TtE SlfllTArEilB fIRINS tJ= IIOTH VR..1Jai -1679 Y 150 3 )o-------------~------., -1679 Y 160 3

Ins Del

(esc) 10 ExIt

-430-

St«K t:XMTRlflUTlONS - blrE!Ctlon can De x, y or L or can De glYel'l as a dIrectIon cosll'e or dIrectIon vector. fne fonaat tor dIrectIon COSllE or vectors lS (CX, cy,CZJ

for examp!e: iO. 707,0, O.707J. Ealtlng keys on the nuErlC keypad serou the Input. Hny bne startlng wlth an asterlsK (If) will Ile taken as a ccaent. &rors have been CletectE!U on any llne stIown bhnkIng. unly enter noaes for Independant Support ExcItatlons tlSlll' SJ " 1

Force set

f

(or)

ShOCk Nille,

,l**

DIrectIon, S'art NOCle, Stop Nocle, IncrelEl'lt

Factor,

1 LDAD CASE l REPRfSENTS THE SHu..TIlEWS ::;IMU t-IRlMi RELIEF WLIJES AT 150 IWJ IbO. TEST 1 Y

(f

IAiTH

34---------------of.----..I

1 1 1 1 1 bIS Del

L(A) CASE

(escHo ExIt

2 SHOCK t:XMTRlflUTlONS - DIrectIon can œ l, y or ï. or can œ glVl!rl as a dIrectIon cosne 01" dIrectIon vector. fhe forllilt for directIon COSllleS or vectors lS (cx,cy,CZ} for eXilple: (0.707,0,0.707). Ealtlftg keyS on the nuErlC keypad seroU the Input. Hny IUle stal"tIng .utn an asterlSk (If) .il! Ile tilken as a COllent. trrors have _ lIeen Getectet:lmLaftY,jœ. snown Dlln~er_

nocies for Independant Support Excitations USM' 5) Force set 11

(or) Shock Hile, Factor, 1

DIrectIon,

Start NOCle, Stop Nocle,

!lICIl!IIent

* LDAD CASE 2 f/EPRfSENTS TIt: HEPETITlYE t=IHlMi Œ T~ Vli.vt HT

1 If :rT~~/JnEs, AND TIE WLVE HT IbO lliRtE UtES. TEST 1 Y1 :3

FII'1.Ih.l~S.

@ \... 0

TEST 1 Y 1

TESTIY2} TEST 1 Y 2

3

fll4l1.iI:;5 ~ tlQO

TEST 1 Y2 DIRECTIIM. (SRSS) , lOlQ.. (10)

Ins Del

-431-

i 11

5E:C""I4SE

T !oK

i~

'PI as:.TIOfjl'r\..

j)"',..,,.._If!

IF THE 7uEW

($.()V~

lt".DlIJ~

$'P.a.TI.. '

7H-E

Me A,,",- "'Y'5" )

])';ltC.T"IOIW/U.

7~

(,;,..,TfZ.lfS

C 01404 &/4J AT"lOIJ METI+e>D

IHfi- MrrH-oo

tAcVI:!'7f..JoJS

Cp

·'t>I~C."lo..l·'

t:>!=- ,4i)i)/710AJ L),c 71+E

~-PDND.Jrs.

ÊN7'fl./ES WfdU.

c.oli>'fI3I/fJATIOAJ

OF- rirE

Âl>DrnOAJ

AL~ 1),FFftTl.P.JT

~G"7'Ht>J)

.J

IVOVL/)

h .JY'AoJ"'''''/1!

Lo,lr()""" 'i tJJ,.., PDA..I ~7t,,)TS •

JJOiE:

IH"IVi

ZEFt:>Il..E

-PlfZEi;:nOAIAl.

ANy

Mp~

Co,..,a/AJAT/OA.J$ ~ MA/)E

/);}..

SPtI4-rlR-c.

CD,..,BIIJ,4TJ()AJ~.

A schematic of how this combination works should help:

-432-

._-_.

-- "--

--

..

__.

--

--- ... -

_J)YNAMJt. L.I:ut/) _CIJ!'E ._t.Ç{:>AtTA:.I/VS._

Jg

.

.

-IJDJ(JIJl-f1t-J,

~O~~9.Jrs:

. /~ 1/4J"",_~.J?1;: _~~_ . .-€:LCI) __ _

--2.~ rl/2.I.IJ.~._-L); Y/!J.:""-v~_2~. ..;r~_~ll:t(!J'.~_~k~ __ €'->Q_

.

_'_-hRJA/I, ~Pl!.. Jl9Lj/I._.~.u,tl . ~~.~ . .8. ';'IM~ _~e.. i4~._.J€Jk,D_ -.~ ~.- r1./1..Ju(_-1k- .. YAL14--@jbl) -.

. .'.-.o~ 2-

~

il'

&~I

)Î----t J-

ifii,pir ~r-

-- -

li~-~-- -

-"'1"114

S •...,r, LU:

~

Dt.s

---- - - . -----:~~:~~~:~;:l-~::S-çt.ç_

(---1~-:. --.--~._-. • • • ~i)r

1

e~&(,

~~

2

UE~-

A1cPt J.i4-!T ~€~

n il ·. H

»

•

1" ~ ~/2C1Ai>1/tJ4 &:o"'1~/~A-TTI)N ~F- /U()I)A,

l2.i~/'TS

-433-

-lok~

~ES .--

-

High frequency modes can often cause confusing results. example illustrates:

The following

L~.fl~! 'Lè~"'.~''-'~'''~ .. A.Lo1>a. • ...,~ VAr-c.\!e ~

..... - .......

_·~~ôv(,

... D~ . I~Lo~ .F1l..,,~j;JUJ~Y ,1U.cM_. .P./ILJ_

. /AJ'tIL>J. VE-

-r.H-E.

~---- ...... ~ --- --,...,~

!~~tt

IH'Ë

~~/f"'F

M.o.~E:S

OF T~

J>JPI!-J{

Ft-~~ 8L~kfl:a!tj)t72.

f'lylf:: .

"

A-«~BLY

IS

So

.i2.1t7/.t>

Co~~

70 THE ,2c.fr OF TItE ?/~/AJ{ Ir t.,)I/...~ kT .L, "'E A/2I~ II) k"Y, /.é, No /lu)t)1IL j)~?t.AC~~74.rd eAIAS/~&; lull.(..

Iv

nt-

'ZC

Sr~I'r/AJ_ lJo

A-I'P~

-

w/'Tlof'OUï"

.J..014.é)S:

:T.-E.

MOJ>A--t,. 7HE"

mz.: -434-

])/!",P~"Cl~E-..J7! 71h:72.C

L.0tI'I-0

~/~i1ZI'I.4711J1J

.......

,---.A-u..~.

-.MO ...... U-,)~

A.k>N. ~(a2.c.. . Ml:' l'H ~,,-,b).

.

·~iU.IS-

-l.._~t-S- ..

pa. J,.

-

IZéoc..u T/OAJ; {(,AJ.H./;-rlé.

-- -

--

"

- - --

Jkf::j..S .f';Z:OBU"" . ~ X/SIS:)

I~J2uA!. __..ST4JJ..e__.Aj~P('J(J.~T/QJ./ LJE .. T#E.t?rAuU~-t/e_ .4.(;)Â;1.)/NC; . .2 - 12/..(~ ~YWA-1lA ~e. A?j!LIC .I+flC>N (FoIU~ ~~~mUM) $- j)~:S/~N .U,O;y~ walZST CM~.

Let's take sorne exarnple situations and work thru thern verbally:

Ar.;~IAM~

Â'rIUO$PH€7UtZ.

CcIJ I)lnoNlS !-h-7ZEJ

10

U ~ ~ /1-fE. ~12. Z~l"ln= L OAO 5r'/tI/I1EtlèYi.

-435_

---1 ,,:~ -~~~~~. ~~~~y~~ ~- r~ ~ rHE UAJ8~

rca.r.,..1'f'JtI'r

~/rrs

---

8rrrNtt'f!#;t~c--nWlrr~WS-CVt--

_.

~'t:NS

__

AW
--

-~..LlA!./Â/-/Z JA.lU7L~. __

rl/zSr

'TJ.I~

l'ASSes

--_ _ _ _ _ _ ._

#Z. ijoW.

_ -i.= .L/V ___ ; __L._--_L~'r~_ ...l~~~V~Ve- _fO.#J n,UT

y' ~Y.tY:Jc.j ry'

W

1-ré.ï.J

15

AC'TS

A

A Vé.-vr

~~/"

!/Of2,I'Çc>/.JïJllt.

A-T

ïl+~

~QVlIVA-t.h..Iï

1~

... ()F _ 4..Jt$.

B~~'ZEO

c.o_?t>/.J E~;

~/l-t)w

AS t> F

ï 1-1 ~

VL?JT

70;

-436-

~HtJw"",

A-çOYé)

T!f't2rl4 {1" $r~t<

1

T/-rl;7U~

" p A-C

ANi)

1$

' /HAï

. (

~,~~r~ .~-..?.o~-"-'T ' \~/'1.N\.Sr

~.G.€ _JÀ.U~

.DF. bt.Nt:.1..I."" ~. _

1'"0

r-

7Hi;. U?1.1F7IN~

'FOa.C.C.

PIAt"

la

.Ff;,rcr/oAi ~f\/ 1'c

f..S""TIM*'n-O

LlsllJt;~J.THE11-- oF7He lINO A?rI20A-êHES .&La".,;;

F-=- A,,,.,QG

WJ.f~J2E:

CM vi "'-v;z ) + &~ -P, ]

f,

.

AND

IvH.f.7Zt:. l'fA

.~ AIlt 6;~ ~ff/€S

76

COI'WP&t1'W

7ff~ Î~ "'>,,,nl1-r'v/l.~ .

1~ l'~

VA~4)"

rfJ.{J M

.M-Kr 'F(,.()I,.#J /l,~ j~_7~ A\/,n"~ t,4tOC!.IT'(

-437-

=

v/+vz

z.

f

!2T

~

=:

'2.-=

v; :; J3/8

l~. 7 --frr:-

H

S

-

-

-~ ; .1 :_{Z~,.f)I14JJ,__ ~ 1?1, ~l.. (B~_$)I".JJIF

_:=

---;- o~-PP""4ç/rlc!f,164

f(

(lol'j,3J.~ (n)";""

._ _

~3

__

-

.

~l::-

7i_:: sn3 '7=

_J.t)J 7/ 3 '12\1"2. = 1ft! ~"-~. fJ>s _ --Îl. =-S.of",2-Î=:'1~_), l.. oR

Z,"L4 ?~

(J'i,7)

($5:~8)( 7".s"' 1.)(}1.)

AW5fJ)E:

'f 6.rs /)

~

)D, 03

~

2.._

J

F ~ (Ço.03)~ z.. [(~.ooOOZ~JÇb) lb,..,. (t038432) ~

(~.J.4 12.+ D,oqoo7.f/flo 07,7) -2,,4 ___ .. -,~~.~ t{.$·7 -,,~J .I~. l~_~c:é: __k _ +--( -i.lio}"J; r)Jb-? - J

__

_

,,-"3

c.'

........ $~C?l..~1b_~.(".)__

-=

So. 0;'

'(.'3 1

t" l.

-

__ J.

40/3.5)

~1 f:::.

~"..

/.-.

(31.,l..)

~ tf (n) ~. .

'-_(C, "000 7...!}SIc) IK ~+-~~

_ (f',7)t3l.2)(oJ - _ ( 0 . 0 0 00/1 i 2...4-)

F =l0032,f32)~ ~t<.....'l,....

-1

_

-

3.tJl-{/()3) /.;.;

!

l_ilo,. Ù &.f .__

.=-

@2..,l-)(n. )

fr- ~ [7..(,,4) I~ ~ 3

__ .,:...."l.-

h

= 4..~S~ 1 ÛO!) ~

Sc.<. ~

eJ

-

3, '14 (/0 CO) ,.:.... 2.. /~

.t:t.!

2..-

_(:!,oS[/){{v6J~

_(7/rf7 h

1~

~ 1- j

/44

$.url.

k

'Z..

~

Yw. (?l.Doo)~ (13/B+IIA~f&;)1-t :; {~~tjo~l ~ )((0.01$3"2..) ~") Ib~ ~ ~\ 3~ï.8 /1zf _ J.d- 1..f!f /('32...., 2..) ~ t*

-438-

1

)

. -tXAf'V/P("G

Fil.

--~_ ..

-. AN])

/2.~~lt7?: __t1N~/6UI2ATIOJ./

POLAJI$ c2F _LfU,.Ù_._A?pi..j~AïJ()I.J.

A good excercise here is to walk thru the example in the manual and talk about each of its features •••

-439-

CAESAR II

PROBLEM:

VERSION 2.lC

OYNAMIC ANALYSIS OF RELIEF LOADS

Analyze the two relief valve system shown on page 1.A subject to t.he the simultaneous firing of both valves. Process steam conditions: 450 psi, @ 650 deg. F Relief Valve Orifice: JOHNSON #34A-06 2.141 in. ID. Valve Opening Time: 8.0 milliseconds Valve Closing Time: 8.0 milliseconds Relief Ouration: 1.0 sec.

CONTENTS: Geometry and CAESAR gas thrust calculations .•••• Problem Setup and discussion ••••••••••••.••.•••• Dynamic Input Echo .....•.•••••....•..........••• Output Discussion ..•.•••.••••••..••.•.•...•.•.•. CAESAR Output Reports •••••••••••••••••••••••••••

-440-

1.A 2.A 3.A S.A 6.A

RELIEF

VALVE

LOADS

CAESAR GAS THRUST LOAD CALCULATIONS

.

. . . )15 .... al L141 S-7'1 MSI lM 1.3

Input

a.'

•

IELlEF ..... I1IIIESJS

'It...

III' ..... ""'_,

c...tH III' Fi_lit ....t ....t ot '11ft h ../ItIIt h .. Joli........ II~.I "'"' Il .... h . bat ••••••••••••••••••• II~.I 1.llIi..t , , _ . Il.... VII .......10 ••• III' 1 1r.....t " .._ . Ii .. 1ft ,.1 .. tllll1" ••• 1ft. 1 ~. btr....... t ISIIIaU • ) Il ...... s..-a. IIIt b.t .....t ISIIoII' • ) Il ••••••

Output

'Il..

"."I.Tl. "."I.Tl.

Ir.het 51, toM.ta ...... .... haI Es.t 51, toMit...... s..... "1..,,, SIl toMit,... "."I.TI.

~.

bit

( - U. T_.tIra ..... FI <- U. " - ' Iftial I. .t . ) 31.'1 ( - Il al Iohof '"he Dr.he. "'.1 ( - ID al Iohof '11ft '.1... hl.1 ( - Il of _ 5tocl 1&1.1 ( - L__ of 1111 Stocl "'.1 ( - Iota. al Sa, Hah. IINt,." Il.,. 1.41 ( - SIl '-tilt '1' 'ft.I~./I".I_. l' II., • 53l ( - ... t. 1111 h .. _ .. - ' U I fitta .. Iflll (-'SIIoII' CMSII III' tif 1111 k l "III

Iftul

2es.6 1'.7 26.4

71$57.921

ma.12'

1211.121 ".121

44.m

1• •

..

1.311

" , lite FI 1'1....1I.'3 "'"'.2 .2

ml.'

"'.3

~------~~------è

2'1..4,'.~~~ Define the type of shock Convert time history data into shock data

2-_

1 - LaftJ lISSES

Scratchpad to compute steam thrust loads (See above)

IIIIZ IEFlml. MIE 11&1 . . . .TIII IID IIIST ,IlLE . . . ., . . E.IEF ~ l'IIIIESJS P\LS[ SPKD Fa sm 1 - _le LM

J4, • 7-

----f---__ Define where the thrust

loads act

Define the dynamic shock load cases C - SM: HMlIC JRf

• - am: HMIIt JRf

E - PEIFlIII IYMlUC ...ms F - lRn li tIESM .ia _

Combine shock stresses with static sustained stresses to check B3l.l code allowables. Tell CAESAR what type of analysis to run

l.A

-441-

RELIEF VALVE EXAMPLE PROBLEM SETUP

REQUIRED:

Cœpuu the support l~ds, forces and stresses in the vent p1p1ng sys~ sho.n on page 1.A .nen the relief valves ftre s1~ltaneously.

GIVEJI:

Vent1ng ste. stagnation propert1es are ghen on pagel.A. The CAESAR "RElIEF LOA!) SYNTHESIS" optton 1s run to eOllPUte the 1101_ thrust load IIIgnttude at the vent pipe extt. This clynaic: la.d will &Ct cIoWIhMrd at the vent elbow !IOdes 6S and 100. Yenttng will lut for approx1l11tely one second, and the open1ng and c:los1ng ti_ for the relief nlft (as provided by the lllnufaeturer) 1s 8.0 .1111seconds. A sUt1e l~d c:ase is run ftrst to perlo... spring hanger sizing at node 22. TIIe stattc load case '3 is the operat1ng case, and w111 be used te set the nonl1near restratnts for tIIe ayn..ic aRalysts.

SOLUTlON:

TIIe sllocl: table !\aile ts arbttrartly selected as "RElIEF", and ts·deftned as llavtng a FREQUENCY range and a FORCE ordtnau. (A , s1gn preceeds tIIe niUle tn the spectr. deftntt10n because the slloclil table ts te be read f~ an asc11 f11e on the hard dtst.) TIIe aetual shoct deftnttion appears: ,RELlEF FREQ

FORCE

FORCE (lbS.)

0.0 1281.0 1281.0 0.0

The thrust loads aet It points 65 and 100.

Tllese 10lds are deftned 15: PULSE SPECTRUM FORCE SETS, and are entered 15 sllown below: (Only a stngle ·Force Set '" is needed for si~ltaneous relief f1r1ng. Multiple Force Sets are .ast c~nly used with water-lla..er aRalyses.) -1281 -1281

Y 65 1 Y 100 1

Tllere 15 only a single slloc:t contribution, and it 1s defined: RELIEF

1.0 Y l

Only two t~s need te be set on the CONTROL PARAMETER spreadslleet. They def1ne tIIe dyn_te analysts type (SPECTRUM) and the static: load case te be used for sett1ng the nonlinear restratnts, (3). Conservattvely, the.,aal c:càtRation _thod cu be set to ABS tnstead of SRSS te produc:e unquesttoRably c:onservattve results. The c:c.plete dynlllte input echo starts on page 3.A.

starts on page S.A.

-442-

,.'

LOG LOG.

TIIe PULSE TABLE GENERATOR bul1ds the asc:H f11e "RELIEF·, that eontatns the relief valve shoct table. Input to the pulse table generator is the ftlenalle, IlUtm. table frequency, number of points, and the ti8e-history waveform. TIIe ..xilDUII frequeney 33 Hz, and tIIere are 20 points in the generaud table. TIIe points 1n the tt_ htstory wave10nlt are enUred as shown below. These points represent the vahe's opening, 1t's one second vent ttme, and it's closing.

0.0 8.0 1000.0 1016.0

\I~

2.A

Output

I!t'

CAESAR II DYMAIIICS lllPUT DATA CHEIXI.' JOBNAKE

c

RELIEF _7

UNITS: ltn9tb (in.) Farn Ub.) IIIss (lb.) SUU nb./ill.) CONTROL PARAftETERS

SPECTRUII ( - Dvnuic AIlalvsis Type (HARIIOJIIC/SPECTRIII/IIODES/RM&E) ""3 1.' 1 33

{(((--

1.1

(-----

28

(-----

'.13 (----'.5 (----Il Il

Static LNd Case for Nanlinear Restnillt Status ~ CONTROL PARANETERS Input data StiffneS5 Factor for Frictian ".I-IIat usedl echo. Only two entries are lin. No. af Eiqenvalues calculated (I-Not usedl required for this job, the Fr!1luency cutoH (Hl) remaining values shown are the original defaults. tlosely SDaced lIode Criteria EarthQuaie Duration (For DSRSS lethodl (sec.) Structural Da.Dinq (1 of critical) Zero Period Acceleration (For Req. Suide 1.611 (g's)

( - - {Not used} ( - - (Not used}

RDDAl {-- Sutial or lIodal Co.billatiDII Hrst SRSS ( - - SDatial Cotbination lIethod (SRSS/ABS) SRSS ( - lIodal Colbination llethod (6ROUP/lII/DSRSS/ABS/SRSS) ............... y ( - - Include Pseudostitic CDlllanents (YINI ~ The most conservative approach AIS ( - - Pseudostatic CDlbinatian lIetbod tSRSS/ABS) would use the ABS summation AIS ( - - Dir!Ctional Cotàinatian lletbod ISRSS/AIS) method for modal combinations. used)

•

{ - (lot

y

{ - Stur. Seauence check on cDllluted eiqenvalues (Y/NI ( - (Ilot usee!)

Il

6 ( - Esti.ated no. of siqnificant fiqures in eillenvalues lE-12 {-- ~acllbi SII!!II Tolerance lEI' { - DecolD05ition Sinllularity Tolerance 1 2 1 1 N Il li.

{-(-(-{-(-(-(--

Subsaac! siz! (Hot Used) No. to Converqe before Silift Alllllled (Hot Used) No. of iterations Der shift (1-P9. cO.Dutedl l of itentions Der shHt before orthoqonalization Force orthD9analization after CDnverlJenCe (Y /lU Use Dut-of-core eiqensolver (Y/NI Fr!1luency Array Soaces

3.A -443-

FORCE SPECTRUft EDITI16

Forct, Dir,dian, Start, bd, Jnc

y 65 J Y III J

-1281 -1281

Define the point of application ~-------------------------------- the and the maximum magnitude of relief load.

~

DYUIIIC LOAD CASE SHOa allTRIlUTlDIS Stlrt. EH. lne

lut, Flctcr. li rtttilll ,

DYNAftIC LOAD CASE 1..~I....._ _ _ _ _ _ _- - - - - - - - - - - Define which shod table to RELIEF 1 Y 1 .use. (There is only one in

this job.) The "y" in the input here is used for output labeling only.

STATIe/IYMARIt CDKBJIATIDI CASES Lad

CiRo

Flctar

CoftBINATION ..E~.~_ _- - - - - - - - - - - - - - S4 JCAS,._

Dl 1

Static Case #4 is the Sustained case, and is to be combined with the Dynamic case Il.

SPECTRU" DEFINITIONS Nale, Range, Ordinate, Range InterpDlatian, Ordinite InttrPolatioD

-

t

IRRIEF FRED FORCE LD6 lO6 FORCE SPECTRUft DATA FILE - - - - - - - - - - - - - - - - - - This line defines the type

t

Jobnale = RaIEF

of spectrum table to be read in. (Remember, the # sign means read the actual table values in from an ascii fil e of the same name.)

fUSER ENTERED TIftE HISTORY f TIllE lailli seconds! FORCE lIb. ) f .UII'

8.IIUI lIR.HII' IIJS.RIII

t f t

FRED (HZ!

t

.UI25 .11411 .12125 •16411 .15625

.:mu .61125 1.12411 1.64125 2.5"11 3.66125 5.18411 7.14825 9.61411 12.65625 16.38411 21.88126 26.2"11 32.58127 48.11112 tft

-444-

NOTE

tft

••

1281.n 1281.1' .n "Il.TIPLIER .11124 .12512 .12765 .48181 .94582 J.71568 1.99996 1.99989 1.99972 1.m:n 1. 99BS9

~-----------------

1.99717

1.99464 1.99132 1.98322 1.97198

1.95473 1.92915 1.89194 1.83987 EDF on soectrua file

4.A

This is aIl input from the file created by the PULSE TABLE GENERATOR. Lines starting wi th asterisks are comment lines .

RELIEF VALVE LOADING -- OUTPUT DISCUSSION There are four Key reports for a relief valve analysis: MASS PARTICIPATION REPORT - This report illustrates how sensitive each of the piping system's modes are to the relief valve firing. High modal participation factors indicate that the mode is easily excited by the applied dynamic forces. If subsequent displacement, restraint or stress reports indicate excessive dynamic responses then the modes having high participation must be damped or eliminated. Once a particular mode is targeted as being a problem it may be viewed tabularly via the mode shape report, or graphically via the animated mode shape plots. (4-0UTPUT from the main menu.) DISPLACEMENT REPORT - This report gives the maximum possible positive or negative displacement that may occur at sorne time during the relief valve's firing. Values in this report are always positive. RESTRAINT REPORT - This report gives the maximum dynamic load the support should be designed for. STRESS REPORT - This report gives 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. This combination is made by specifying STATIC/DYNAMIC COMBINATION CASES during the dynamic input session. (See the Dynamic input data listing for this problem for an example.l Portions of the output for this job follow:

S.A -445-

CIEW Il MSS HlTlCINn.

flLE.IEl,ID' bU.. YÙft 1.... .t ..... liTE••, 21.1917 65_Ia. SIC PIIY1CINn. SI:II.l --<111_F.a. F.a. cr FIIEI Cf Cl IESCIml.

!IllE

DnMlt SIC LOU CASE 1 2 3

4 S 6 7

,• 1. 11 12 Il 14 15

-lLm41 -2.11m 1.54114 -2.19465

.3 .7 1.4 I.S 1.6 2.. 2.. 3.4 5.. LI U.7 12.2 14.4 16.1

.~u,

.mn

•12246

-.16458

.... ....

.•136M .,.

•12141 -.'1767

16.'

l'

17 Il

•11154 •11261

21.6 22.6 41.'

-••7..

1 1 1 1 1 1 1 1 1 1 1 IELJEF 1 1EL1f1 1 IELIEFI 1ELID'1 1EL1EF1 IELlEF 1 E.JEF 1

IELJEF IEUEF IEl,JEF IEUEF IEUEF 1ELID' E.ID' E.ID' IEUEF

•

I.U" Yllft 1.... .t .....

CIEW Il 'ISP\JCŒJIT IUIIIT flLEIE.lf1 liTE,., 21.1917 .nMIC IEUf1 SIC .1

..

S 1.

15 21

Z2 25 31 ~

41 45

•

.. 55

65 71 75

•

. as

ts

...

•••

dM6 1.1227 1.1227 1.1227 1.1221 1.1221 1.1216 1.1218 1."" 1.1. 1.1992 1.2144 1• • 1.1294 1.&127 1.1751

I.am

las

1.2216 1.2311 1• •3

115

.Z5a .&127 •Im .1Ut

.DW

••

..

•• ....

... .1644

.DIt

1• • 3.GM

•4171

~.,m

.817 .SM4 .5151

••

.a72

.•n .am .~15 •ms .ms .~" ...1 ."1 ••14 .~U5

.3US .3U5 .3115

Il'111

31

•

su

œ

m ft6 1 Till 11'111 tm m

.5642 .5644 .65U

..... ..."

.....121t

.ml

.m7

.n .ma .ma .:ml .:ml .:ml

"4 • •4 • •4 • •4

.71Z2 .ml

.n

.ms .ms

.SS74 .5375 ."1' .61U

•ml .ml .ml .ml

.6162 .6225

••114

.Im

••1"

.1266 .1267

•me

.au

... .l1li

.1112

•• •• • •• .ml •

.11~

.U"

w..tllft.I'.I-

..

Il If Il Ill111H IfIII.- IZllII*

4m 1 Till

.13

m.

l 'III

42ft 3ft2 l'UI

liti' ....

•

• •

al lEU.. lit"

•

• • •

al kti.. titi,

•

• • •

l "11 45

•

1562 Il..

U'UI 22

-446-

26 15 1 fIll

Maximum displacements that occur due to the relief loads. These displacements may actually be positive or negative. Their true sign is indeterminate and always shown positive in the displacement report .

..Ii.. yalft 1.... al ..... 65 _ III.

FILlsII&.ID' IllEI.' 21.1917

Ft FI .... Ml.FlIIIIH Ftllllllt FlIIIIH

m

.ml .ml .SHI

'.6136

Ftrrtl(~.I-

.2I7J

.Im

4.4154 4.5641 4.622' 4.7121 4.ma 4.1144 5.1414 5.1"1 7.440 3.t6ZI 4.1421 4.1421 4.1252 4.lm 4.4"7

mlLS. ...Fl m

• .1S25

~.~

••71

CIESIR Il UPaIT ISCTI. InMIC BIEF SIC Il

S

Illltati_t....I - If U

• •• .... ••, •• .n •• .... .ml •• ••

•IUI

.1Z26

U

Dl

D'

1. U'

65_Ia.

- - T n.. sllti.SIiD.I--

.1

The Participation Factor report shows which modes tend to be exciteLi by the applied dynamic load.

lIlJIF •••••• •••• •••• •••• •••• lIlJIF •••• •••• •••• ••

1.' •••••• 1.'1.' •••• 1.'1.' •••• 1.'1.' . •••• 1.'1.' • 1.' •••• 1.'1.' 1.' •••••• 1.'1.' •• 1.' •••• 1.'1.'

1•• 1•• 1•• 1•• 1•• 1•• 1•• 1•• 1•• 1•• 1•• 1•• ·1•• 1•• 1•• 1•• 1•• 1••

•1P.I53

1

"_l1li

ftIt

Support Reactions due to the dynamic relief loaÙS. The top value is the maximum support reaction. The second value is the largest support reaction due to any one mode. The last number on the left tells which mode.

6.A

tAE511 Il mESS IIEPIIIT HIIIIIlt IElIEf SIIIO: Il

fIUllEl.JEf IITtIIll' 21.1"7

. .5 - - 5 t ' f t . I I •• I I I . I I . I - - m"S ... AlIIi. IElllIIIi '!IISla l1li atT IllIDE IlAI.Al/IIDO. Tallloft atTlIIoII.

_1IIoft

26 21 1 fUI

l 'III

ml

l 'III 414

21 Il "JI

374

412

1 JII)

1 nu

m

SIFl

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STlESSIIIoIIt

1••

1••

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1

J&

1

'"

l 'III 1111 1"1 1 YU)

%.21

414

m

"

412

115

Il "JI 26

1 TIl) l441

15

ZI6:5

, 'Ill

l 'III

l 'III 1174 lm 1 fil)

JlIt) 1151 1475 1 JII)

Il 12 2 fil)

1614 1217 J'UI 4257 36" ) TIl)

1174 lm l 'III

1161 1121 JlIt)

1••

314

2134

2.21

224J 1'" 1 TIl) mJ 1lS2 1 TIl)

lI4

21 1 TIl) 21

12

22

21 J fil) l2

21 J TIl)

ne

1 YIJ)

l

314

l 'III 1361

1"1 2 YU)

1 TIl) "7 464 4 'III

17 Il Il ml li l' Il II Till

lm 763 2ml

"7 464 4 YIJ)

J7

13'3 1ft IYUI J419 174 11 nll

Il fil)

21 11 2 'Ill

25

li

Il Il '111 J7 35 II Il'111

11

3S

2612

lm

1ft l

,(1)

15

1611 1ft

Il '111 41 II

11 "m "

"

Il 'III

541 Il 'UI

'"

"7 464 4 "JI

557

41

1,.

5451

41

li Jl TIll J62 65 J24 13 TIll

l

ms

4 '111

4713 l1I2 4 'UI 4111

16n 4 TIll 7442 ~

4 TIll

1221

1."

1••

l.n

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858

1 YUI 164 481 11 YUI

1ft 421 Il JII)

557 C26

36 3S1

C '111

4 YU)

2lII

2.21

1.15

lm l 'III 2647 1112 2 "JI

1."

1••

l24l

1••

1••

1."

1••

1••

J••

6Sl 52J 5 YUI 1

The maximum stress at node 40 on the 40-50 element is 5539 psi. The stress at node 40 due to mode #4 was 3969 psi. Mode #4 was the largest contributor to the stress at node 40.

m

2 TIll 1417 lt2 1 YUI 1417 191 J'UI 1424

1

114 Il'111 1••

1••

U If!

1."

1••

1••

1••

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a '111

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1

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1••

26lI

1••

1••

1••

1••

1171 C "" 2311

ma

1••

1."

221

lIJ 312 Il YU) III 312 Il YlJI

The maximum stress at node 5 is 1470 psi. The stress at noc.le 5 due only to mode !il was 1294 psi.

lm l 'III

1 TIll

...

For example:

2271 2121 J YU)

,(JI

11'111

1

1.15

l 'III 4213

411

l12

C5

III Il 'Ill 1,. 51 III Il ml

•

"1

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161l

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674

. , ...

464 4 ml

1.15

675

Il:52

25

%.21

'(1)

314 214

ml 1lS2 l 'III

I~

ml

1148

ne

Element stresses due to the c.lynamic relief loads. The top value is the maximum stress due to the interaction of aIl of aIl of the system modes. The second value is the largest stress due to any one mode. The bottom number on the left tells which mode.

ISiI J'UI

J 'III

l 'Ill

Cl 27

1••

'(1)

lm)

22

1••

97J J

l "" lm

1764

lm

ne 1 m)

1.15

l 'CIl

1'14

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1294

1547

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J6625 l..zs

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l..zs J662S

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1•• l.lS

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Support reactions due to the combination of the static sustained and the dynamic solutions .

alti' 1Ifte_

:m2

"J

lItJ

Stresses due ta the combination of the static sustained and the dynamic solutions. This stress combination can be compared to the B31 code allowables for occasional stresses.

l~

S.A

-448-

WATER HAMMER

~

6-8" "JUMP"

DUE

10 'lUlUs lHE TRIP AT NODE 4S - )

-

---..

..

Wa~er hammer is ~hat effec~ where mechanical loadings occur primarlly due ~o a single ~raveling pressure wave in a piping sys~em. For example, in ~he model above, while under s~eady sta~e condi~ions ~he pump at node 5 is arawing fluid thru it at a steady flow velocity of 6 f~./sec. The turbine driving the pump a~ 5 automatically Kicks

offline because of an overheating problem. The pump slows to a s~op. As this happens the fluid column moving toward the pump lS stopped. A compressive pressure builds at the inlet ~o the pump. The wave that separa~es ~he compressed fluid on the pump side from ~he fluid upstream moves bacK thru the fluid with the speed of sound ln the fluid of 3250 ft./sec. As this wave moves past successive elbow-elbow pairs a force imbalence is experienced at each elbow-elbow pair until the wave has completely past both elbows in the pair. This phenomenon is illustrated below:

-449-

~-~"::Hh. :. . ",- _

,- i''~

- l~.~EI,. .~t=t,-

'\I-t~ . !.h4M fh~-wvu WAl4:.)

, /;

iLLoù\l.,)C,

AT T~ ~?er~

.. OP'5ov,.,ai) "-h.J r~ F,,~.,p) G~IS1S ~"'7kIalJ . iwo a.G~-I,;;l,.1Jo~ PAI/tS. r~E"

~1~~l'S TlI'~ ,ho..., M

T,.1'-\04 -E' -é;,o.J l

Ll::'n

,

ln.J 0

CoI44?~--r;c;.I....,e

PI1.'-.-ÇWrk:-i....IJf..,,~ .

lsAt.J· VrJ'6A-U:--C,E"D

'FDii~

·~,:-rll..~(

Er"1)~<.... n; {fitI9.f--l1D~;~JAIIJSiPE

PIPE

THE.UJ.J8J4LlJ.Jeln h~&€ Ac-'T5 UlVr/" 7H-E P/U:;'f{1/f4E WA-VE PA-SSe-s 7H-E Si."...zt>JO

l:;.7,.6()w --tlJ'TtI'/; PA-//l.

-~~ ~~frz.D

IH\;:-rV",," ~

FIG. 5.20. Metaphorical shock propagating into a fiowing medium which bas been suddenly brought to rest. (Reprinted courtesy R, Courant and K. 0, Friedrichs, Supersonic Flow and Shock Waves, Interscience Publishers, New York, copyright 1948,)

WH~-r

Ta TH-E

\'5 "t\~ M..oW 1)t~c::::.TLOtJ

1z,~~,?1..1:;, '1

-450-

1...,

"5tiAj)l~/5

NORMALLY WATERHAMMER PROBLEMS OCCUR AT ELBOW-ELBOW PAIRS WHERE THE DISTANCE BETWEEN ELBOWS IS THE GREATEST. At these particular pairs the duration of the loading is the longest. producing the largest displacement of the piping system. ln addition to the compressive wave that emanates away from the pump on the supply side. there can also be a rarefaction (or low pressure) wave that moves away from the pump on the discharge side. This low pressure region moves away from the pump in the discharge piping. and is due to the energy in the moving fluid column that has already gone thru the pump. and that wants to keep moving. The magnitudes of the pressure waves can be computed conservatively as shown below. ln addition a surge fluid program can be run to get the exact magnitudes of these loads.

L'" '.JI""

p.~./ ~,~

~ ;;>/ OL .. t:;

w~L.7ZC.·

C~'E ~,...~..

~p - IS

1'/2.. ~$r_u. /2.-,' ~~ II::'

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Is

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r#E

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rHE

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lAI rf.f€ Fl.~/J} fH~ p,,,,,.; lkt.ot'ITY. (USl.llflu y (HL FL.o",; YBotl'ry)

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F,r:-rJ./ (;.i>rrïOAJ) fi)~ J4 /llADa,.C _ A:,I+/t. fI) 1

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-452-

There are two different types of waterhammer analyses the user should be aware of: 1 - General 3-D waterhammer impact and response 2 - Axial, 1-0 waterhammer impact and response

An example 3-D waterhammer problem is the turbine trip examp1e given above. The system being ana1yzed is tru1y three dimensional, and modal components are primarily bending of the legs adjacent to the elbow-elbow pairs. Axial 1-0 waterhammer problems are those associated with long, straight sections of pipe where the modal components that make-up the solution displacements are primarily the axial lIextension" modes of the long straight runs of pipe. The two solutions can be differentiated as low frequency component solutions and high frequency component solutions: GENERAL 3-D WATERHAMMER -- Low Frequency Component Solution AXIAL 1-0 WATERHAMMER -- High Frequency Component Solution. (Axial extension modes of the pipe.) Individua1 characteristics of each type of waterhammer problem are described be10w: AXIAL, 1-0 WATERHAMMER: 1 - ln these cases the primary concern is either the displacement of the 10aded end of a long straight 1ength of pipe, or the 10ads on a thrust bloCk at the end of a long straight run of pipe. These response properties can usual1y be obtained by computing the first few axial modes of the straight piping section. The procedure here is to compute, say, the first la, and record the displacement at the 10aded end; then compute the first 15 and see if there is any change in the disp1acement. If not then the displacement solution can be assumed to have converged. (Providing at 1east sorne of the extra 5 modes were axial modes of the long straight run.) 2 - There are two typica1 types of prob1ems where long axial runs are concerned: i) Oisp1acement 1imited, where the design must prevent excessive 10ading on equipment nozz1es, valves or anchors, and ii) Thrust bloCk design for underground piping. Examp1es of each are shown in the illustrations be10w:

-453-

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-454-

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3 - It is very important for the user to emphasize to the eigensolver, in these types of problems that he is primarily interested in the axial modes of a certain section of pipe. (Otherwise al1 of the lower, non-contributing modes will be ca1cu1ated first, before the modes of interest, causing the user to sit and wait for a long time.) To 00 this the user shou1d oe1ete a11 masses not in the straight run, and he should de1ete a11 masses in the straight run not in the axial direction. The CAESAR II input for de1eting a11 masses in the Y and Z directions in a mode1 wou10 be: (Providing a11 of the nodes in the mode 1 were between the node numbers 1 and 1000 !!) Mass,

Direction,

Start Node,

Stop Node,

Increment

o y 1 1000

o

Z 1 1000

4 - Friction in these problems can be very important. Remember that the dynamic friction stiffness is taken from the static solution results. Knowing the approximate value of the friction from the static solution the user can adjust the friction multiplier to approximate the actua1 value he wishes. An example should make this clearer:

-455-

CONTROL PARAPETER .. IST -

(----- Static a.o.d Ca.e fer Non11near R. .traint Statu. (----- Stiffn. . . Factor for Friction (O.O-Not u.~) No. of Ei9~Yalu. . calcul.t~ (o-Not u.~) (----- Frequ.ncy cutoff (HZ)

7

0.0

(----- "-x.

o

33

<----- Cl 0 . .1 y Spac8d

0.1 20 0.03

/~

Criteri. (sec.)

<----- Structural Da.pin9 (X of critical) ( - Zero Prlod Acceleration (FOI"' ReQ. Guide 1.60) (9'.'

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GENERAL 3-D WATER HAMMER: Given the example above, (reproduced below) the following observations can be made: 6-." "Jutor' DUE 1'0 1VRBlNE TRIP AT NODE cs)

-

1 - The displacements at the ends of the large elbow-elbow pairs converged quickly.

-457-

2 - Stresses in parts of the plplng system not associated with the two elbow-elbow pairs studied (45-75, and 90-110), converged very quicKly, (i.e. after only the basic low frequency modes were computed). 3 - Stresses in the axial runs of pipe converge very slowly, (i.e. after

a large number of modes are calculated). This is because th~ stress wave propagation in the pipe is being monitored. ThlS value will converge to the same number for all pipe elements in the axial run, (providing friction stiffnesses are not great). See the notes on waterhammer for computing the recommended lengths to accurately monitor stress wave propagation. (Note that stress waves MOye at the speed of sound in the pipe material !!)

General notes on the waterhammer problem follow:

* On the pump or valve supply side the magnitude of the pressure wave is calculated as shown in this example using:

dp = r * c * dv.

* On the pump or valve discharge side the maximum magnitude of the pressure wave is the pressure.

difference between the fluid vapor pressure and

the

line

* On the supply side a positive pressure wave moves away from the pump at the speed of sound in the fluide 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 speed of sound in the fluide The maximum magnitude of this "negative" pressure wave is the difference between the pump discharge pressure and the fluid vapor pressure. Once 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 fluid adjacent to the pump flashes. As the negative pressure wave moves away from the pump these vapor bubbles collapse instantly. This local vapor "implosion" can cause extremely high pressure pulses. In addition there may be a fluid bac~flow created due to the rapid drop in pressure. In this case the bac~flow "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 are usually not cyclic. The pressure wave passes

thru the system once at full strength. Reflections of the wave May then cause secondary pressure transients. Without a transient fluid simulation or field data, the usual procedure is to assume a single significant pass of the pressure wave.

-458-

* Where critical p1p1ng is concerned, or where the maximum loads on

snubbers an~ 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, combinations of different force sets are usually not rune This approach has proved satisfactory when applied to large, hot steam piping systems that have very few fixed restraints, and a high number of low modes of vibration. Extrapolation to other types of piping systems should be made at the designers discretion.

* CAESAR-II does not check the integrety of the piping system due to the local increase in hoop stress that occurs as the fluid pressure wave passes each pipe cross-section.

* The magnitude of the water hammer loads can be reduced by slowing the mechanism that tends to reduce the flowrate. In the case of valve c10sing, this means slow1y closing the valve. In the case of a pump going off line, this means slowly removing power from the pump. "Sl owl y" in each of these instances can be estimated from: T =

2L/c

where T = time of one wave cycle, sec. L = Characteristic length of the piping system. Usually taken as the length between the pump or valve and the source or sink. c = Speed of sound in the fluide the pump or valve stops in a time shorter than "T" then the waterhammer shou1d be analyzed as shown in this example for "instantaneous" closure. Calculations for this problem are given below: If

Of primary interest is the largest time that must be used 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 refl ect i ng systems wi 11 be made to get a "fee 1" for the variation of the computed "T's". The longest time will be 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 ft. T

= (2)(270)ft./{4281)ft/sec = 126

milliseconds

The length thru which the wave passes that causes the most trouble is the length between nodes 45 and 75: T = (2){90)/(4281) = 42 milliseconds So, if the pump or valve can slow down in greater than 126 milliseconds the tendancy for water hammer in the piping system will probably be abated. If the pump or valve can slow down in greater than 42 milliseconds then the tendancy for water hammer in the 45-75 length will be abated. -459-

* Water hammer excitation initially produces axial acoustic waves in the

steel pipe wall that can induce locally very high, very 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 or material due to water hammer loads is a concern the following rules should be followed: 1 - A very high number of natural frequencies must usually be included in the analysis. Cutoff frequencies of 300 Hz are not unusual. These are the axial natural modes of the pipe between the excited elbow-elbow pairs. Higher modes must be computed until the inclusion of extra modes doesn't produce an appreciable change in the force/stress response. The maximum frequency cutoff can be estimated from: SQRT( E/r ) / L where: E = Pipe material modulus of elasticity, r = 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 elbowelbow pair for the 20 ft. pipe lengths is given below: f cutoff

= SQRT( E/r ) / L SQRT( (30E6) (32.2)(12) / = (202388 in./sec) / (20 ft.

(0.283) )/ 20 12 in/ft) = (843.3 rad./sec) / ( 21r rad./cycles) = 134.2 Hz. =

2 - Design rules should be based on material endurance limit curves with appropriate safety factors applied. An example endurance limit curve for steel is shown in the CAESAR user's guide on page 3-69 (u), and was taKen from Section VIII Division 2 of the Pressure Vessel Code. 3 - The length of any element in the primary axial runs should not be greater than about ct/4. where c = the speed of sound in the pipe and "t" is the duration of the water hammer load. Calculation of the greatest element length for the 45-75 elbowelbow pair is given below: Lmax

= ct/4 = (16000) = 84 ft.

ft/sec (0.021) sec / (4)

and so, 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 80 ft. Shorter duration loads require shorter elements to monitor the passing of the stress wave. Applications in pile driving analysis confirm this approach.

-460-

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

The example problem from the user's guide, (shown above and reproduced below) is illustrative:

-461-

CAESAR- Il

PROBLEM:

VERS-ION 2.1e

DYNAMIC ANAl YSI S OF WATER HAMMER LOADS

The cooling water suppl y line shown in the following page 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. Design an alternative support scheme to eliminate the large field displacements associated with the turbine trip. Fluid Properties: 250 psi @ 140 degrees F Flow Velocity: 6 fps Water Bulk Modulus: 313000 psi

CONTENTS:

Geometry and CAESAR Pulse Table Generation •••••• 1.B Water Hammer Load Calculations ••••••••••••••••••• 3.B Dynamic Input Echo ...•...•...••....•.........••.. 6.8 General Notes for Analyzing Water Hammer Loads ••• 9.B Ouput Discussion •...•..•.....•...•••••.•••..•••.. 12.B CAESAR Output Reports •••••••••••••••••••••••••••• 13.B

-462-

WATER HAMMER LOADS 6-S" "Jur.IP" DUE TC nJRI:IINE TRIP AT NODE 45)

r

Z" . . . . . .x

i:~MQa 3-_UEFIIlTl. 4 - l'lUE TIlLE __TI. 5 - na IIIST TIlLE IElEMTlm• • ,- BIEF a..1III1B1I 7-f'IlS[.mllllFllŒsm • -~ LDIIS

Define the type of shock: e.g. "#45-75 FREQ FORCE LOG LOG" (The # si gn tells CAESAR to read the shock table from an ASCII file named "45-75". The ASCII file contains FREQUENCY vs FORCE data points, and each axis is to be logarithmically interpolated. See the dynamic input listing for more details).

---1__~

, - ilia QSEJ.... 1 - alTltllYIIIIIIC aIIIIMTI. .

step

Define where the thrust loads act. (The significant elbow-elbow pairs in this jOb are 45-75 and 90-110. Unbalanced thrust loads will act at 45 in the X, and at 90 in the X.

C - SIVE JTIIMIC IlftT

t - DIEU JYIIIIIIC I..vr E - PDfIltll JYIIIIIIC __!SIS f - 1ft," t. aEW lIil _

onvert TIME HISTORY data i nto shock data. (Thi s creates the files "45-75" & "90-110" - see below)

([sc)

Define the dynamic shock load cases. (A different shock load case will exist for each significant elbow-elbow pair in the model). ~ell

CAESAR what type of dynamic analysis te run.

LB -463-

CAESAR PULSE TABLE GENERATION

'45-75 FREQ FORCE LOS LOS

FORCE SPECTRUft DATA FILE Jobnale = HAftftER + USER ENTERED TI"E HISTDRY f TI"Ellilliseconcs) FORCE!lb.l f

f

t

t t f

.00000 .16000 21.16000 21.32000

FREQ IHZl .00062 .01000 .05063 .16000 .39063 .81000 1.50063 2.56000 4.10063 6.25000 9.15063 12.96000 17.85062 24.01000 31.64062 40.95999 52.20061 65.él00l 81.45065 100.00000

f++

Time History Data Input

.00 } 17289.00 17289.00 .00

"ULTIPLIER .00696 .00046 .00825 .02139 .05195 .10763 .19914 .33873 .53847 .80726 1.14294 1.51707 1. 85421 1.99998 1. 99996 1. 99993 1.99988 1.99982 1.99972 1.99958

Computed frequency ana lysi s) •

NOTE +++ EDF on spectrui file

2.B -464-

response (used in

the

force-shock

SOLUTION:

The magnitute of the pump suppl y side pressure wave which emminates from the pump discharge at node 5 can be estimated from: dp dp

where:

r * C * dv

=

is the pressure rise due to the "instantaneous" stopping - is the fluid density - is the speed of sound in the fluid - is the change in velocity of the fluid

r c dv

pump's

The speed of sound in the fluid can be estimated from: * c = [ Ef / (r + r ( Ef / E) ( d/ t ) ) ] ** O. 5

where: Ef E d t r

-

-

the the the the the

is is is is is

bulk modulus of the fluid (313000 psi) modulus of elasticity of the pipe (30E6 psi) pipe mean diameter pipe wall thickness fluid density (62.4 lbm/ft3)

r + r (Ef /E)( dit) = 62.4 lbm/ft3

= 79.1875 313000 lbf c =

1 + (313000/30E6) (8.625 - 0.322)/0.322)

(

l bm/ft3 1/2 lbm ft

32.2

ft3

144

i n2

---------- -------------- -------------- -------------- = 4281 i n2

79.1875 lbm

ft2

1bf sec2

ft/sec

Apply the equation above for the magnitude of the water hammer pressure wave. dp

=r *

(62.4) l bm

C

* dv

(62.4) l bm

= --------ft3

=

=

---------ft3

(6.0) ft

(4281)ft

--------

--------

sec

sec

(4281)ft

(6.0)ft

--------

-------

sec

sec

lbf

sec2

----------

( 32 .2 )l bm ft

ft2 --------

(144)in2

345.6 psi

* See the PIPING HANDBOOK,

Crocker & King, Fifth Edition, McGraw-Hill pp. 3-189 thru 3-191 for a more detailed discussion and evaluation of the speed of sound.

3.B -465-

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 wave form for both types of water hammer pulses is below:

shown

f'.,.1 LU

11)

t:/J-f"

Ci

J

~

.}

P s _..

~

w J. P '

III

tI! Q..

AL P.CH• L p..

l'v

T"/,.,e

T'hE'

PUMP DISCHARGE PRESSURE VARIATION (Measured at the Discharge Nozzle)

Pod Ps Pos dp

PUMP INTAKE PRESSURE VARIATION (Measured at the Pump lnlet)

-

Discharge pressure Source (tank or static) pressure Suction pressure (while running) Pressure fluctuation due to the instantaneous stoppage of flow the pump pv - Liquid vapor pressure at flow temperature

thru

There will be an unbalanced load on the plplng system due to the time it takes the pressure wave to pass successive elbow-elbow pairs. The magnitude of this unbalanced 10ad can be computer from: F unbalanced

= dP *

Area

The duration of the load is found from t = Lie; where L is the length of pipe between adjacent elbow-elbow pairs. For this problem the elbow-elbow pairs most likely to cause the large deflection at node 45 are 45-75 and 90-110.

The rise time for the unbalanced dynamic loading can be estimated from: t rise = die where d is the pipe diameter

4.B -466-

CALCULATIONS:

L 45-75 = 7 + 4(20) + 4 = 90 ft. 90-110 = 3(20) + 15 = 75 ft. Area = 7r /4 di2 di = 8.625 -(2)(0.322)

L

Area = 1T /4 (7.981)**2 F unblanced

= dp

t duration = LIe = (90)/(4281) = (75)/(4281) t rise

* Area

=

in.

50.0 in2

= (345.6)

(50.0)

= 21 milliseconds = 17.5 milliseconds

= d/c = (8/12)/4281

= 7.981

=

= 17289

lbf

: on leg from 45 to 75 : on leg from 90 to 110

0.16 milliseconds

Because the p'p,ng in this example is ductile low carbon steel, the major design variable will be the large displacement; i.e. the problem will be assumed to be solved when the restraint system is redesigned to limit the large displacements due to water hammer without causing any subsequent thermal problem due to over-restraint. The Dynamics input data listings for this job, with notes follow:

5.B

-467-

CAESAR II DYNA"ICS INPUT DATA CHECKING JOBNA"E = HAftftER UNITS: l!ftgth (in.) Force !lb. 1 lIass IIbI) SUU lIb. lin. )

CONTROL PARAKETERS SPECTRU" (----- Dynalic Analysis Type 1 0.0

(HARftONIC/SPECTRUft/"ODES/RA~1

20

{----- Static load Case for Nonlinear Restraint Status (----- Stiffness Factor for Friction (O.O-Not usedl { t~.~ (0 N t ,----- "aXeNO. OT1 E"1genvaI uesI ca I cu a~ - 0 used) (----- Frequency cutaH ŒZ! ---

0.1 20 0.03 0.5

{----- Closely Spaced lIade Criteria (----- Earthquake Duratian (For DSRSS lethod) (sec.1 (----- Structural Dalping Il of critical) (----- Zero Period Acceleration (For Reg. Guide 1.601 (g's)

N N

(-----

O

"ODAl (----- Spatial or Kodal COlbination first SRSS (----- Spatial COlbinatian "eth ad ISRSS/ABS) SRSS (----- "odal COlbination "ethod IGROUP/I01/DSRSS/ABS/SRSSI y (----- lntlude Pseudostatit COlponents (YIN) ABS {----- Pseudostatit COlbination Kethod (SRSS/ABS) ABS (----- Direttional COlbination "ethad (SRSS/ABS) N

(----- 6 (----- Estilated no. of significant figures in eigtnvalues lE-12 {----- Jacobi S.eep Tolerance 1EI0 {----- DecDlposition Singularity Tolerance

y

{-----

o

(----- Subspace size !O-Not Used) {----- No. to Converge before Shift Allo.ed (O-Not Usedl (----- No. of iterations per shift 10-PgI cOlputedl {----- l of iterations per shift before arthogonalization <----- Force orthaganalizatian after convergence (YIN) {----- Use out-of-core eigensalver (Y/NI (----- Frequency Array Spaces

2

o o N N 100

6.8 -468-

CONTROL PARAMETERS input data echo. Only three entries are required for th1s job. the ,...;n1ng values shown are the original defaults.

The most conservative approach would use the AaS (absolute) summation method for modal cOlllbinations.

FORCE SPECTRUR EDITIN6 Forc!, Direction, Stirt, End, Inc FORCE SET Il CORRESPONDS TD THE ELBOW-ELBOW PAIR 45-75

f

-17289 X 45 1 f

FORCE SET 12 CORRESPONDS TO THE ELBDM-ELBOW PAIR 90-110

-17289 X90 2

Define the point of application, IIllIil11U11\ IUgnitude, and direction of the wate.. hllAer 10iding. Each separlte elbow-elbow pli .. hls a uni que force set llUlllber. For the el bow-e1bow pat .. 45-75 the force set nlilDe.. 15 1. For the elbow-elbow plir 90-110 the force set nUilber is 2.

DYNARIC LOAD CASE SHDCK CONTRIBUTIONS Nale, Factor, Direction, Stirt, End, Inc DYNAIIIC LOAD CASE 1 1 f THIS SHOCK CASE WILL DESCRIBE THE PRESSURE lAYE IIIPACT IN THE f 45-75 ELBOW ELBOW PAIR ONlY. 45-75 1 1 1 DYNAIIIC LOAD CASE 1 2 f THIS SHDCK WIll DESCRIBE THE SHOCK IN THE 90-110 ELBOW-ELBDW PAIR f ONLY. 90-110 1 X2

DYNARIC LOA» CASE 1 3 f THIS SHOCK CASE SHOWS WHAT HAPPENS IF THE ftAllRUftS DUE TD THE f DISPLACEIIENTS OF THE INDIVIDUAl SHOCK CASES HAPPEN TD OCCUR AT f THE SARE lIftE. I.E. THIS COULD HAPPEN IF THE 1I0000000T DUE THE f PRESSURE WAVE'S PASSINS THE FIRST ElBOW-ElBOM PAIR KAS NOT DIED f DUT BY THE TlftE THE PRESSURE MAVE ENTERS AND PASSES THE SECOND f LARSE ELBOW-ELBDW PAIR. 45-75 1 X1

C..pute the dynillic response due to the unblllnced water hamme.. 10ld in the 45-70 elbow-elbow Pit ...

Coapute the dynillic response due te the unbllanced water hammer 10ld in the 90-110 elbow-elbow pair.

CGllpute the clynllli c response due te the COIDination of the two unballnced 10ids. Each 101d set ..epresents a different DIRECTIOtIAL cGllpOnent.

90-110 1 X2

7.B -469-

SPECTRun DEFINITIONS Nale, Range, Ordinate. Range Interpolation, Ordinate Interpolation '90-110 FREg FORCE LOS LOS • FORCE SPECTRUn DATA FILE • Jobnale = HAnnER • USER ENTERED TInE HISTORY • TInE tlill iseconds) FORCE (lb. ) • .00000 .00 + .16000 17289.00 t 17.66000 17289.00 t 17.82000 .00 t

ttt

FREg (HZ) "ULTI PLI ER .10245 •00062 .00123 .01000 .00470 .05063 •01796 .16000 .04337 .39063 .08987 .81000 1.50063 .16630 .28314 2.56000 . 45110 4.10063 .07969 6.25000 .97231 9.15063 1.31730 12.96000 1.67243 17 .85062 1.94325 24.01000 1.99996 31.64062 1.99993 40.95999 1.99988 52.20061 1.99982 65.bl001 1.99972 81.45065 1.99958 100.00000 NOTE t •• EDF on spectrui file

.......... This

l1ne deflnes the type of spectrum table read ln. (Reml!lllber. the' slgn means read the actual table values ln from an ASCrI fil e of the same 1IclIIe). to

Echo of the user entered tlme hlstory data •

-

... OLF (Oynamic Load Factor) data generated automatlcally fra. the tlme history by the PULSE TABLE GEHERATOR •

8.B -470-

be

WATER HAMMER LOAOING -- OUTPUT DISCUSSION MASS PARTICIPATION REPORT - This report illustrates how sensitive each of the piping system's modes are to the water hammer dynamic loading. High modal participation factors indicate that the mode is easily excited by the applied dynamic forces. If subsequent displacement reports indicate high dynamic responses then the modes having high participation must be damped or eliminated. Once a particular mode is targeted as being a problem it may be viewed tabularly via the mode shape report, or graphical1y via the animated mode shape plots. (4-0UTPUT from the CAESAR Main Menu). DISPLACEMENT REPORT - This report gives the maximum possible pOSitive or negative displacement that may occur at some time during the event. Values in this report are a1ways positive. RESTRAINT/FORCE/STRESS REPORTS - If high modes are inc1uded, as discussed in the notes in this section, then these reports give the maximum 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 inc1uded, then these reports give the maximum values of forces and stresses in the system due to gross deformation alone. COMBINATION CASES - The force spectrum approach to the water hammer problem does not inc1ude consideration of the time relationship between modal or directiona1 maximums. Complete1y conservative results can be guaranteed by taking the absolute summation of both the modal and directiona1 response properties. Running one load case for each main piping run, and a final load case including a11 of the individua1 load cases typically gives the ana1yst a "good feel" for where prob1ems existe 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 including the 45-75 and 90-110 contributions together yie1ded litt1e extra information. PROBLEM SOLUTION - A guide and axial 1imit stop at nodes 65 and 105 produces little increase in thermal stresses, (which were low to begin with), and serves to attenuate the large axial disp1acements in the 1ine due to the water hammer load. Loads on this support due to the low mode disp1acements are seen to be sma1l. Local, very short duration loads may not be so smal1. The restraint should be designed with this in mind. A few simple design rules are usua"y sufficient: 1 - Flexible is better. The restraint shou1d only be stiff enough ta sufficiently attenuate the low frequency gross deformation.

12.8 -471-

2 - 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 restaints in a problem area should probably also be "beefed up" to withstand the local passing of the impact stress wave.) Portions of the CAESAR output reports for this jOb follow.

Existin9 Syst •• Analysis FILE: HAllIER DATE:JUN 19,19B7 Runnin9 throu9h 20 Hz SHOCK PARTICIPATION SCAlE -----cosin.s---FACTOR CI CY CZ DESCRIPTIIIl FREI! FACTOR

CAESAR II "ASS PARTICIPATION

"ODE

DYNARIC SHOCX LDAD CASE 1 2 3

4 5 Il

7

B 9

10 11

12 13 14 15 16

...,

-5.31195 -28.23074 .5 4.33605 .B .7Un 1.2 .117557 1.11 2.1 . 1.087411 -1.38042 2.2

., " .... J

3.1 3.4 3.9 4.1 5.9 Il.1

17

B.O B.7 B.9

18 19 20 21 22 23

12.4 14.B IB.3 18.B 21.4

11.11

.3~294

.144115 -.19210 .098311 1.02645 -.021aS5 .05112 .05765 -.00729 -.21473 .00306 -.04182 -.00693 .00145 -.02195 .03429

1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

.0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0

.0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0

45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 4~75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1 45-75 1

13.8 -472-

Participation Factors for water hanner load in el bow-elbow pair 45-75. (Original Cortfiguration)

CAESAR Il "ASS PARTICIPATION

l'IODE

FILE:HAIIER2 DATE:JUI 19,1987

"odifl.d Syst •• AnilYSIS Sto~s ~ 6uid.s , 65 ~ Ils

SHOCX PARTICIPATION SCALE -----Cosin.s---Cy CZ DESCRIPTION FREI! FACTOR FACTOR CI DYMAIIIC SHOCX LOAD CASE

1 2 3

4 5 6 7 8 9 10 11

12 13 14 15 16 17 18 19

.7 1.3 1.7 1.9 2.4 2.5 3.2 3.9 5.3 5.8 7.7 8.0 8.8 11.6 12.3 14.8 18.3 18.7 21.2

-.00109 -.00024 .00385 -.00892 .01217 -.00080 .00057 -.02517 -.00911 -.00028 .00000 .00646 .03249

-.00104 .01319

.oom

.00015 .02419 -.00093

1.00 1.00 1.00 1.00 1.00 1.00 1.00 I.CIl

1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

1.0 1.0 1.0 1.0

1.0 1.0

1.0 1.0 1.0 1.0

1.0 1.0 1.0 1.0

1.0 1.0 1.0 1.0

1.0

14.8

.0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0

.0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0 .0

45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75 45-75

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Participation Factors for water hamer loads in el bow-elbow pair 45-75 (Modified Configuration)

-473.

CAESAR 1J DISPLACEftENT REPORT FI LE: HAIlftER Shock ~ 45 on Run 45-75 DATE:JUN 19,1987 --------Tr.nslitionslin.)-------DI DY DZ

NODE

".J 10 15 20 25 26 30 35 40 4S

SO 55 110 6S 70 75 80 . 85 88 89 90 95

.0000 .0545 .1353 .2397 1.5554 3.7803 ~.5717

0.1226 0.41148 0.8459 6.8459 0.8459 6.8458 6.8455 6.8449 6.6320 3.0379 1.9507

109 109 !10 Ils

120 125

.3204

.2607 .2447 .2440 .2432 .2424 .2418 .23n

.1213 .0164 .0057 .0000

Existlng Systf. Anilysis Running throuqh 20 H:

--------Rotitionsldfg.)--------RI RY RI

.0000 .0479 .0479 .0479 .0501 .0000 .0614 .0539 .0000

.0000

.0000

.0359

.0872

.0439 .0621 .3310 .3312 .3369 .3880 .3880

.0972 .0971, .0290

.0767

.4082

.0000 .0000 .0000 .0000 .0000 .1136 .0000

.6124 1.886t 2.9898 2.9582 l.lI622 1.5800 1.5800 1.5748 1.7123 1.7310 1.7386 2.1793 1.9565 .9155

.1102 .1104

.0979 .0874 .0000 .0000 .0000 .0142 .0060 .0001 .0000 .0000 .0000

.14n

.1192 .1011 .00llé .0023 .0000

.0244

.0583 .0558 .0732 .1218 .1175 .1012 .0874 .0772

.0722 .0337 .0977 .2615 .3234 .3095

.3047 .2345 .1653 .1102 .0945 .0823 .0679

.0168 .0163 .0000

.0000 .1842 .216-4 .2179 .r'ill .8747 .7020

.5188 .4713 .2401 .2673 .3l196 .2483 .3048 .7575 1.1465 1.2581 1.1051 .7204 .6805 .5n4 .1263

.1988 .1931

.2231 .1725 .1526

.0000 .3722 .3980 .3991 .5592 .5240

.4780 .3394 .21193 .0998 .0497 .0125

.00ll2 .0260

.0993 .2494 •52311 .11116 .5042 .3372 .1675 .0451t .0092 .0121 .0313 .0827

.1264

.0192

.0414

.0185

.0401 .0000

.0000

15.8 -474-

oISPLACEMENT REPORT for water hal11l1er load in el bow-elbow pair 45-75 (Original Configuration)

:AESAR Il DISPLACEftENT REPORT FILE:HAIIER:? 3hock @45 on Run 45-75 DATE:JUN 19,1987 --------Trinslitionslin.)--------

NODE 5 10 15 20 25 26 30 35 40 45 50 55 60 65 70 75 80 85 88 89 90 95 100 105 108 109 110 115 120 125

DI .0000 .0006 .0014 .0024 .0131 .0139 .0153 .0287 .0176 .0003 .0003 .0002 .0001 .0000 .0000 .0003 .0013 .0015 .0001 .0000 .0000 .0000 .0000 .0000 .0000 .0000.

DY .0000 .0005 .0005 .0005 .0006 .0000 .0093 .0080 .0000 .0094 .0000 .0000 .0000 .0000 .0000 .0004 .0000 .0007 .0007 .0007 .0007 .0000 .0000 .0000 .0000 .0000

.0000

.0000

.0000 .0000 .0000

.0000 .0000 .0000

Dl .0000 .0006 .0007 .0010 .0036 .0036 .0029 .0121 .0121 .0092 .0153 .0060 .0065 .0000 .0021 .0025 .0025 .0023 .0011 .0012 .0012 .0022 .0020 .0000 .0003 .0003 .0003 .0000 .0000 .0000

IIodified Syste. Anilysis Stops ~ Suides @6S &115

--------Rotitions(deg.)--------RI RY RI .0000 .0011 .0012 .0012 .0013 .0031 .0100 .0088 .0098 .0102 .0096 .0075 .0054 .0032 .00l! .0004 .0007 .0008 .0010 .0010 .0010 .0007 .0005 .0003 .0002 .0002 .0002 .0000 .0000 .0000

.0000 .0026 .0030 .0030 .0086 .0059 .0048 .0125 .0147 .0225 .0164 .0073 .0028 .0018 .0017 .0009 .0008 .0007 .0006 .0006 .0006 .0004 .0004

.0003 .0000 .0000 .0000 .0000 .0000 .0000

16.8

.0000 .0038 .0040 .0040 .0076 .0175 .0264 .0209

.0176 .0106 .0067 .0018 .0005 .0001 .0004 .0006 .0004 .0005

OISPLACEMENT REPORT for water hammer load on elbow-elbow pair 45-75 (Modified Configuration)

.OOOS .0003 .0002 .0001 .0000 .0000 .0000 .0000 .0000 .0000 .0000 .0000

-475-

SLUG FLOW: Slug flow is a phenomenon whereby ~ short cylinder of fluid, moves independantly thru an otherwise gaseous piping system, as shown roughly in the illustration below:

--

ua

. -_____ ---- - _____________--..r SL.U~

<=-(L.lto..lllf:ll

L",.", C, T j.4.

I,E. ~ - Lo,-,-, W~1.L

ST'=:;~IN\. fia""" /10./ 7"1:Cnor-J

G:N Pr\. l T"i

lba.-

'Z - IwO - P~E EL.OW S"'(Sn;;~ ~ '3 - /;..1 A iJ V/:;-77..T A-iVT Co<-LI:!CTliJI:,J (A! 12t~/~F4,I'-!eS ,-

To analyze this problem the user must know, or be able to conservatively estimate: L - The length of the fluid cylinder. (The longer the more conservative.) Di - The inside diameter of the pipe thru which the fluid is moving. V - The estimated velocity of the fluid cylinder. (If no other value is available then taken as between 0.5 and 1.0 times the gas flow velocity.) Slug flow impacts each elbow, or change of line size independantly, as shown in the figure below:

- L'ft~T

-~--_ S~T"_ l'><'r \ïot"1~

cr~os.~ .~~~

~n~--oÇ

'$~~ _~ _LQ~S_

FUMO-C-<~-~

__ ~,I.,_\"___ L"",,- _i> ~ï__

__ :L\-b,s.~~o.w ._rJoo,l'j)I:o-~~~_

DF _____ .

..~~ ~-P~~ill_

.__ Nr_ -476-

~_~~_E!.;aol..o"l_·

The slug flow time history is estimated as shown in the figure below:

Sl.o~4 Wl+W~

l~l"iIr(..ï I-O~

_--r-------.,.

"--

~ i\..o$"S

r~t':)~T "'S~T> '-'~ L.~JtJ':;

(2.. t9 -

~.;e /U::~

2~AJO

~l..'-'~

L(;;.1*I>/u ç

'SLU"

D 1.~~Ptf"fI il=+1-S

E!7}G~

~~("

S~.c.'nO,-.)

.

sc.. U I1

.,-H-z,.S POIIW

r,

"zA-r) lut

IJ-€"-'l) Af'o-J~L.~

V ~. ~."". 7!7-~ VéV/.J.{Yl:ZO.cI'1Y

"

.

'" l

SI,;.I,.t~. lM.f~'T Pt-ta..~...; :;

..

,

--~_477-

(L:-l2.ë)

7V= .

Yv

...

In lieu of any more accurate momentum change calculations, the following may be used to estimate the characteristics of slug flow loadings at flow reductions, i.e. orifice plates, reducers, etc .•• Slug Impact Rise and Fall Time

=

0.0

Slug Impact Duration = 2L/V Sl ug Impact Dynami c Load where:

LVr A-

is is is is

the the the the

r~agni

tude = rAV

2

estimated length of the slug cylinder, velocity of the slug, fluid density, and inside area of the pipe.

The basic steps in the slug flow analysis become: 1) Estimate where the maximum problems due to slug flow impact are most likely to occur. 2) Compute the magnitude of the slug flow load at each of the elbows or flow restrictions of concern. 3) Estimate the time waveform of the slug flow load, and use the pulse table generator to get a response spectrum for each different slug flow time waveform. 4) Define the force spectrum shock load cases. For elbows that are very close, the user may wish to apply the loads from both in the same load case. 5) Run the spectrum analysis, and review the predicted displacements, forces and stresses. Typical slug flow analyses is concerned primarily with the excited low frequency modes having large displacement components in the direction of the slug flow loading.

-478-

.

CJH:+-N~E

lAI

Cff IM/c, E / /toi'

lAi

IUcM t::/t.,) 1ùM

;;e/2.'p/:::ÂJ /) leu c..-1't2.

C~

DflZ-l;;--c. no0;:::

. CIf7l77..Jc,€

c~ ç ç:

110./

1'"

IN~j)~/";~__ j)t(?é7!77()A.J =- f -c:.o (J) fAVZ.

Mo~~7Zt~

M.cMé:U7l1..11.1 lA! /lW)M fSNn",cA,.

~

) fA- tJ

7t>

-.(

~

$t..~C; ïTU.~

/;./ Cov/.-f/ ~ <;

2-

D(.,{74o/A/~ ~/1U:-.-c..T7DA.J ~(Î-~r )fArJ ~

Pf:;7ZI::te:~ .l) It;, '" e,A't2- :;0

J) 1f2.l;C not--J -= (~~) \ A V 2-

"DI(L~~OF

7/it

7i-rc 0 t-t rc, ill;../'7

DESIGNING A NEW COMPRESSOR PIPING SYSTEM TO AVOlD PRESSURE PULSATION Typically large compressor piping systems are analyzed using an analog model of the fluid/compressor/piping acoustics. The purpose of this analysis is to keep the pressure fluctuations due to acoustic resonance low. From the analog design of the piping system come basic geometry limitations, as well as size and position of pulsation bottles, acoustic natural frequencies, etc. The responsibility of the piping designer is to support the piping system such that mechanical natural frequencies are as far away as possible from acoustic natural frequencies. Rerouting the pipe in these situations is often not allowed as this would change the acoustic performance and the analog study (which is very expensive) would have to be repeated. From the diagram shown below, the user can see that a good estimate for "far away" is: 1.5 Wacoustic < Wpipe < 0.75 Wacoustic where:

Wacoustic is the acoustic natural frequency that the designer is cautioned to stay away from, and Wpipe is the natural frequency of the pipe.

All of the natural frequencies in the piping system should be kept away from all of the acoustic frequencies by the equation above. Unfortunately there is no cutoff as to how far, "high-frequency-wise" this rule should be taken. ln high pressure piping systems it is not unusual to generate mechanical response up to 100 Hz. (By high pressure we mean over 10,000 pSi.) ln lower pressure systems, the acoustic forcing function will be smaller and so the tendancy to vibrate will be smaller. These systems clearly don't need to have their frequencies separated into the higher frequency ranges. 6r-----r---~nr----r-----r-----~--~

5r---~r----r~----r-----r-----~--~

O~=---~----~----L---~2~----L---~3

w/w.

-479-

DESIGNING A NEW PIPING SYSTEM TO AVOlD EXCESSIVE LOW FREQUENCY VIBRATION Most piping analysts view dynamic analysis as a time consuming, and infrequent, but necessary evil. In reality the CAESAR "RANGE" check is quiCk and easy to run, and can give the analyst valuable dynamic information. ln a majority of steam piping systems, and in many petro-chemical applications involving large, hot piping, the systems are almost completely spring supported; from the boiler or tower to the first turbine, pump, exchanger or separator. Systems of this type have very little lateral stiffness, and as a result can have very low natural frequencies in the horizontal directions. It is a good design rule of thumb to keep all natural frequencies in systems of this type above about (5) Hz. If left below 5 Hz., any tendancy to vibrate the piping system may result in large, unrestricted movement of the line. These excitations can be due to wind, vehicular travel, slight pressure pulses in the line, etc •• (i.e. sources that would not usually cause vibration problems.) The white noise from the se random vibrations excite the low mechanical natural frequencies of the piping system. The RANGE check tells the user, (in the time it takes to perform a single static solution) how many natural frequencies exist in a particular frequency range. The design procedure would be: 1) Statically design the plplng system so that thermal loads are properly considered and dealt with. 2) Run the range check to make sure that the natural frequencies are above 5 Hz. 3} If not then extract the first modes of vibration and decide where to guide or install snubbers. NOTE THAT CONSTANT EFFORT SPRINGS ADD NO VERTICAL STIFFNESS INTO A DYNAMIC ANALYSIS !!!

-480-

OESIGNING A TABLE TOP TURBINE PIPING SYSTEM Piping systems on table top turbines typically have a geometry similar to that shown below so that the condenser can be placed almost immediately under the turbine exhaust.

~

'S'Tl$L ~""n:~.

C; 0 IV C,tl,/:;O--r-c. \'

1",...ô/'c TOP

"

The dynamic loading due to turbine rotation is usually specified by the manufacturer in the form of an eccentricity and mass:

-481-

These values are turned into dynamic forces as shown below.

lu. ::_._~_(P O.P...J4~~........

= v ::-

F,..;

=

~"QCI <::-(c,L...n """-1"

o,ooc'3

M

~'IIJ

C"o)s~

=-

~0

H~ ~

IIV,

w.1..,.., = ~zoo) L./)J4
~il..

{32,YLA Pt:

These are phased loadings. The loading along the X axis is zero wh en the loading along the Y axis is at its maximum. and the loading along the X axis is at its maximum when the loading along the Y axis is zero:

-482-

(/2.)1.4,.1

I••.:)\"t~ T~

(.u.~S

T~Lo~

ArT'

Cp....,)

JS

.J'V..A$S

,~

TI1'1-S l~

Jb~ ..no...,

T~

-x

D.(l~np~,

(..:) l "r~

T~

T }Tt

CF).)) 'D,~c..Tko~ ;

W

V€C7bR.

P~2~$ ;:

L.oA.--O

ICS.

Tt.t1-S /.....,

p~ lno~

T~ - "

M

("f;.; ~ wt ) ~

Ces (tjo".;..

f- (;:;; SlAJ

wt ) ~

wé)

t ?_~

SH1Fi

Turbines do not come up to their operating speed immediately, rather they are brought to speed slowly while the internal parts, boilers, etc. all come slowly to temperature and pressure. Because of this it is important to keep the turbine speed away from low mechanical natural frequencies that could be excited by the slight vibration due to the eccentric rotating masse

-483-

The design procedure is normally as shown below: 1) Run a simulated force spectrum analysis to compute the modal participation factors for (4) orthogonal configurations of the dynamic loads. Compute the natural frequencies of the piping system such that the high frequency cutoff is 1.5 times the turbine running speed. The (4) orthogonal configurations of the loads would be: (for a Z axis turbine) Force Force Force Force

set set set set

#1 #2 #3 #4

= (Fn) = (Fn) = (Fn) = (Fn)

in in in in

the the the the

X direction. Y direction. (1,1,0) direction. (-1,1,0) direction.

2) Take the modes having the largest participation factors and run individual harmonie analyses at approximately each frequency to determine the effect that mode will have on the systems performance. If the modes are determined to be potentially dangerous, and if they are far enough from the running speed (i.e. less than 0.75 times the running speed) then the operators can be instructed to keep the turbine away from that speed while coming on-line. If the modes are determined to be potentially dangerous and they are greater than 0.75 times the running speed then some redesign of the system supports is probably warranted. THE EXCITATION FREQUENCY FOR THESE HARMONIC RUNS SHOULD BE APPROXIMATELY 1.1 TIMES THE COMPUTED NATURAL FREQUENCY FOR THE PARTICULAR MODE OF INTEREST.

-484-

MAÎc.cnl~(.

'SH@100Cl:. 1(.,/500

~--05 / ).S' 00---'---.115 10

3'

./ I-Z'

T~ loCoF

P~ "'!'

(~"v.L..

125 -........-......~-....'QI~'O

----____

-~---.-__

'3 s'

le

1 .j::>

-----_____.

Cot>€ Ai'P'NÙI)( Â Fou.OWf/!JC,:}

ROTATE ZOOM PAN RANGE SHOW RESET HRDCPY PENPLT

p~,

Fl.l..iE 4A~

STO. WlA-u•.

.,

'2" ~.",c,~ Su. lc!"'-TG

'"

"'<

A:r' 'Z 0

'Sv.Pi'OtL.'l'

\o",,"·"ll..

F(l.OkA

QUIT

.

---.--------): 3 0 ~ --

i

.•...., .•.....•.••..••..•,

Towon1.

'3 S '

00 U1 1

-----_____..

-------.---r! 3 5 Il --'-.-._ u

CO..-A PtATE:O

ÂT .A

A!oOé

TH~l'1jH..

V~~~El.

OF tl-#E

r

('S~Ë

1'2.5 'P~r

Dli). '- 20"

î.20~

B

rs'

oS:: @10 o ~ ZO, OOC

\S'

1

A 10ft:, &,.l

/s

...._-...._-....

t~PA,..,SIOAJ

/7.2(.,.8

7CJt-1P/;:nlfrui2l

100

-.-" .

IS 200. 0 8

OF

Fr.

(IO·")IAI/IJJ/f)E~F

820

"3s ,..---.---.--...._

oF.

A!J>ove Vt;-fS(-"

<; ~ ~Il..ï

."---.-,..,\ 40

g. _-- - -._z;;. . - . . . _ _[ 4~ \-Z

.Z---·- --..-~---X D,sI', @IOO

= (82.0-7 0 )OF (17."Zf.8)(IO-"'),ullu/o~ ('20.08)( 12) Fr é:!.

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59 1

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COADE® Engineering Software

PAGE _ _ _ _ OF _ __ DATE _ _ _ _ _ _ __

CUSTOMER _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ By _ _ _ _ _ _ _ __ PROJECT _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __

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CAESAR Il STRESS REPORT FILE:SSEI!I STArIC RUN WITH CASE 6 (OCC) 56 !OCC) =55+52 DATE: "AR 9.1988 LARGE ROTATION RODS -------------Str esses (! b./so. in. )------------~AX II!UI! ELmNT AXIAL BENDIN6 TORSION HOOP OCTAHEDRAL NODES STRESS STRESS STRESS STRESS SHEAR STRESS

m.

STRESS INTENSIFICATION IN-PLANE OUT-PLANE

----Stress!lb./so.in.)---CODE ALLOI/ABLE ! STRESS STRESS

105 108

1545. 1547.

138.

158. -158.

32118. 32118.

1525. 1527.

UIlIL Uill

1.9118 !.Il81

1675. 1685.

21945. 21945.

8. 8.

108 189

1547. 1562.

4113. 379.

158. -161.

3288. mB.

1598. 1581l.

3.41111 3.41111

2.834 2.B34

1951. 1941.

21945. 21945.

9.

189 111l

1562. 1575.

379. 11l61.

168. -83.

32118. 328B.

1588. 1759.

3.41111 3.4118

2.834 2.834

1941. 2636.

21945. 21945.

9. 12.

118 113

1575. 1575.

357. 182.

93. -83.

3288. 328B.

1531. 14B5.

1.11111 1. 11118

1.UII 1.Il811

1932. 1756.

21945. 21945.

9. 8.

113

1575. 1557.

532. 76B.

83.

114

-II.

328B. 32IlB.

1599. 1653.

3.4111 3.m

2.834 2.834

2187. 2325.

21945. 21945.

Ill. Il.

114 115

1557. 1541.

76B. 76B.

56.

321B. 32Il8.

1653. 1579.

3.4111 3.411

2.834 2.B34

2325. 2319.

21945. 21945.

Il. Il.

115 1211

1541. 1491.

247. 177.

-56. 56.

321lB. 321!B.

1465. 1438.

1.111 1.81!8

1.118 1.211.

1788. 1668.

21945. 21945.

8. 8.

12a 123

163B. 1636.

234. 24B.

-76. 76.

3218. 3218.

1462. 1465.

1.118 UII

ulla 1.111

1872. 1884.

21945. 21945.

9. 9.

123 124

1636. 1613.

793. 234.

-76.

92.

32118. 321B.

1545. 1468.

3.411 3.411

2.834 2.834

2429. 1847.

21945. 21945.

11. 8.

124 125

1613. 1575.

m. 819.

-92. 71.

3218. 321B.

1468. 15511.

3.411 3.411

2.834 2.834

1847. 2384.

21945. 21945.

Il.

125 131

1575. 1575.

254. 1645.

-71. 71.

mB. 32118.

1462. 1694.

1.111 1.1111

1.11111 1.111

1829. 3221.

21945. 21945.

8. 15.

1311 135

1575. 1575.

1645. 1461.

-71. 71.

m8. 3218.

1694. 1687.

LIli 1. BIll

1.111 1.111

3221. 3136.

21945. 21945.

15. 14.

135 141

1575. !S75.

1461. 895.

-71. 71.

32IIB. 3218.

1687. 1513.

1.IBI !.III

1.111 1.111

3136. 2471.

21945. 21945.

14. 11.

141 143

1575. 1575.

895. 412.

-71. 71.

3218. 3218.

1513. 1545.

1.111 1.118

1.111! 1.1111

2471. 1987.

21945. 21945.

Il.

143 144

1575. 1559.

1225. 547.

-71. 169.

3218. m8.

1815. 1647.

3.411 3.411

2.834 2.834

28111. 2117.

21945. 21945.

13. 11.

144 145

1559. 1545.

547. 314.

-169. 187.

321B. 3218.

1647. 1572.

3.411 3.411

2.834 2.834

2117. 1849.

21945. 21945.

11. 8.

11.

-487-

9.

8.

9.

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CAESAR II STRESS REPORT SIATlC RUN mH FILE:SSE"1 CASE 6 (DCCl S6(OCCl=S5+S2 DATE:I1AR 9.1988 LARSE ROTATION RODS -------------Stresses (] b. / SQ. in. ) ------------"AmUI! ElEIIENT AXIAL BENDIN6 TORSION HOOP OC TA HE DR AL NODES STRESS STRESS STRESS STRESS SHEAR STRESS

STRESS 1NTENS 1F1CA TI ON IN-PLANE OUT-PLANE

----Stress (1 b. / so. in. ) ---CODE ALLOI/ABLE STRESS STRESS

-187. IB7.

32B8. ma.

1544. 1566.

UIlB I.IlIlB

UBB UBII

1647. 1757.

21945. 21945.

8. 8.

6.

138611. 138611.

5798. 5799.

l.m UII8

1.m

-6.

6852. 6825.

21945. 21945.

31. 31.

169. 191.

11. -11.

1386B. 138611.

58114. 5811.

!.BIIII 1. Bill

1.m 1.8811

6964. 6941.

21945. 21945.

32. 32.

152. 213.

8. -8.

B.

-247.

Il.

132. 217.

I.BBII !.lIB

I.BBII !.IIBII

277. 4611.

21945. 21945.

1. 2.

3BIl 3111

1488. 1481!.

336. 485.

-IB7. 187.

32IIB. 321!8.

1588. 1688.

Ull8

I.m I.m

1823. 1885.

21945. 21945.

8. 9.

m

2344. 2344.

lB. IB6.

5. -5.

4725. 4725.

19811. 1983.

1.1111

31B

I.BII' I.III!B

2354. 24511.

21945. 21945.

11. 11.

31' 315

2344. 2344.

116. 363.

5. -5.

4725. 4725.

1983. 211115.

1. IIIIl 1.1l1l1l

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2458. 27118.

21945. 21945.

11. 12.

145 15B

1545. 15117.

1"2.

IBI 2115

6a5B. 6817.

3.

a.

2115 215

6795. 67511.

215 22B

-125.

m.

-488-

I.UI!

I.I!U

Uilil

I.UIl

CAESAR II NATURAl FREI2UENCIES FIlE:SSm DATE:i'lAR 9.1999 (Hz)

l'IODE

FREIlUENCY

5 6 7 8 9

.999 2.359 2.4117 3.8112 5.322 5.976 6.495 6.583 8.129

III 11 12 13 14 15

Il.813 11.881 19.417 26.726 41.116

3

9.m

(Radians/Sec) FREIlUENCY

6.211 14.824 15.123 23.887 33.436 37.547 41.811 41.361 51.441 61.433 74.222 74.646 121. '137 167.925 252.158

(Sec) PERIOD

1.012 __ 'Fn.\c::.-n.o~ kA.A-( \;;l.IIA-\IU .... ~ TI..\\'S w-aoe J A:w.;) \)tl Tl-!-€" OTI~ .424 .415 .263 - 12.1:;..~T(l..I'IrI~TS AcT 13)" " - ' i) .188 .167 <.SlJ'D~. c..l..."'Yt..'TAlo....J L ,( 13~ ~H-ov~i) .154 .152 .125 .112 .195 .184 .152 .1137 .125

r:so

-489-

CAESAR II i'!ASS PARTICIPATION

"ODE

FILE:SSEi'!1 DATE:"AR 9.1988

SHOCK PARTICIPATION SCALE -----Cosines---FREI! FACTOR FACTOR CX CY Cl DESCRIPTION DYNA"IC SHOCK LOAD CASE

2 )

4 5 0 7 8 9 H! 11 12 13 14 15

!.8 2.4 2.4

3.8 5.3 6.8 6.5 6.0

s.a 9.8 I1.S Il.9 19.4 26.7 41.1

41.48797 .8111119 2.39755 5.17366 2.76276

.allii

-l.mSI .11111 1. 73998 -.13916 -.21192 .1111112 .IIS79 .anll -.a8378

1.88 1.118 LIlIl

.8 •B .11

Lili

.8 .0

1.1l1 1.11 1.11 1.11 1.111 1.11 1.11 1. III 1••1 1.la I.II!

.8 .8

.a

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.0 .8 .8 .8 .11 .Il .8

. Il .11 .8 .1 .1

.8 .8 .8

.B

.1 .1

.a

.8

.8

l.0 !.B 1.8 1.8 1.8

LI !.8 1.11 1•• 1.1 1.8 !.I 1.8 1.1 1.8

1. 68H. 5 1.6IH.5 1.61H.5 !.6IH.5 1.6IH.5 1.6IH.5 1.6IH.5 l.oIH.5 1.6IH.5 1.0IH.5 1.68H.5 1.61H.5 1.6IH.5 1. 61H. 5 1. 61H. 5

-490-

l)'(UAvV\. \ CAESAR II STRESS REPORT (OCClCo.bination t 1

NODES

c.

.. "3

8

Q.I;S'Â.~ÎS

~PA

FILE:SSElll IlATE:IIAR 9.1988

----------5tress (1 b. j sa. in. l ---------AXIAL BENDIN6 TORSION liAI OCT

5IFI

SIFO

--(!b./sa. in. l-STRESS ALLON

liS 108

1545 1547

5143 35011

4772 4772

5552 5614

1.1111 1.111

!.IIII 1.1111

4688 5147

21945 21945

188 1119

1547 1561

9929

4772

8699

4873

7452 7874

3.41 3.41

2.83 2.83

11477 1112611

21945 21945

1119 Il\!

1561 1574

8699 22261

4873 2682

7\!74 11975

3.48 3.41

2.83 2.83

18261l 23836

21945 21945

Il\! 113

1574 1574

7838 3792

2682 2682

5647 42\!4

1.18 I.il

1.8\! 1.88

9413 5367

21945 21945

113 114

1574 1556

18763 13731

2682 1733

6868 7918

3.48 3.41

2.83 2.83

12338 15288

21945 21945

114 115

1556 1541

13731 li827

1733 1646

7918 6495

3.41

3.41

2.83 2.83

15288 12368

21945 21945

115 121

1541 1491

3797 4899

1646 1646

3582 3998

1." 1."

1.11 1.111

5338 6391

21945 21945

128 123

1637 1635

1794 1956

2274 2274

3411 3431

1." 1.81

1.\!8 1.88

3432 3592

21945 21945

123 124

1635 1613

5634 9347

2274 2163

4446 6114

3.41 3.41

2.83 2.83

7271 11961

21945 21945

124 125

1613 1574

9347 14172

2163 1399

6114 7891

3.41 3.41

2.83 2.83

18961 15646

21945 21945

125 138

1574 1574

4934 12978

1399 1399

3914 7144

1.18 1.11

1.11 1. ..

6589 14553

21945 21945

138 135

1574 1574

12978 14598

1399 1399

7144 7973

1.11 1.11

1.11 1."

14553 16173

21945 21945

135 141

1574 1574

14598 3525

1399 1399

7973 3147

1.11 1.11

1.11 1.11

16173 5111

21945 21945

141 143

1574 1574

3525 7222

1399 1399

3147 4934

1." 1.11

1.11 1.11

5111 8797

21945 21945

143 144

1574 1559

21524 11811

1399 3549

11977 7626

3.41 3.41

2.83 2.83

22199 13361

21945 21945

144 145

1559 1544

11811 4526

3549 4175

7626 5318

3.41 3.41

2.83 2.83

13361 6171

21945 21945

145 151

1544 1517

1592 4253

4175 4175

4795 5259

1." 1."

1.11 1.11

3137 5761

21945 21945

III 215

6849 6816

B4 457

173 173

5941 6151

1." 1.11

1.11 1••

6934 7274

21945 21945

-491-

'Dyp..,,)Cl,. CAESAR Il STRESS REPORT (OCC1Colbination • 1

NODES

lI'-A 1

c...

."3 ~ ~p,A, KeSuLTS

FILE:SSEIlI DATE: "AR 9.1988

----------Stress (1 b./ sa. in. 1---------AXIAL BENDIN6 TORSION l'fAX OCT

SIFI

SIFO

285 215

6794 675il

629 1364

148 148

6846 6374

!.BI! !.il8

Uil

215

125 246

1882 3684

117 117

579 1818

UI! 1. Il''

!.01!

22B 31!1! 301

1487 1479

5741 6945

4075 4075

5658 6"33

3115 318

2344 2344

172 353

97 97

318 315

2344 2344

353 3567

97 97

--(!b./so.in.)-STRESS ALLOW 7424 8114

21945 21945

Uil

121!7 3851

21945 21945

1. Ill! !.illl

UI! I.llil

7229 8425

21945 21945

2188 2121

UI! 1.31

UI! Uil

2516 2697

21945 21945

2121 3517

1.BII UI

!.0" 1. il"

2697 5911

21945 21945

!.il"

_492-

WIND INDUCED VIBRATION DUE TO VORTEX SHEDDING: The phenomenon of wind-induced vibration due to vortex formation and shedding is well documented in the literature. Th~ basic characteristics of vortex-induced vibration of cylindrical objects are that the excitation forces tend to be periodic, the resultant vibrations are resonant in characteristic, and the direction of vibration of the object is usually in the direction perpendicular to the wind. Excitation frequency:

fs

The frequency of excitation of a cylindrical object, such as a pressure vessel or pipe, caused by vortex shedding due to wind flow across itls length, can be calculated by the equation: fs

= (S)(V)/(L), in cycles per second.

where: S

= Strouhal number = 0.2 for the wind speeds of interest

v = Wind velocity, feet per second (fps) L

= Width

of frontal area, feet,

= Diameter for cylindrical shapes.

Converting wind velocity to miles per hour, diameter to inches and substituting in the above equation: fs

= (3.52)(Vmph)/(D), cps.

where Vmph

= wind speed, miles per hour.

D = Cylinder diameter, inches. This equation is used to calculate the excitation frequency for a wind speed of interest. Clearly the equation can be rearranged to calculate the wind speed which will cause excitation at a specific frequency of interest, i.e. a natural frequency of the piping system. Vmph Excitation Force:

= (D)(fs) / (3.52)

F(t)

The excitation force, as a function of time, due to vortex shedding associated with a specific wind speed is given by the equation: 2

F(t)

= (O.5)(p)(L)(V ) (C) sin{2*pi*fs*t)

where: p = Density of air, slugs per cu. ft. = 0.00234 slugs per cu. ft. L = Width of frontal area, feet = Diameter for a cylinder

v = Wind velocity, fps C = (1.414)(R) R=RMS lift coefficient = 0.5 for speeds of interest Since the sine function varies between +1 and -1, the maximum amplitude of the excitation force is: 2

Fmax = (0.5)(p)(D)(V )(C) , lbs per foot of length. Substituting in the numerical values and expressing V in miles per hour and D in inches gives: 2 Fmax = 0.0001481 (Vmph ) (D), lbs. per foot. This uniformly distributed force should be spread out over the span of interest and an harmonie analysis run for the natural frequency that is of concern. The natural frequency of the piping system is computed by an eigensolution. The frequency for the harmonie analysis should be different from the eigensolution frequency to account for structural damping. The modified frequency is computed from: w1 = (wn)sqrt[ 1 - 2c ] w2 = (wn)sqrt[ 1 + 2c ] where w1 = natural frequency lower than the system natural frequency that will simulate a damped system. w2 = natural frequency above the system natural frequency that will simulate a damped system. wn = calculated system natural frequency from CAESAR c = percent critical damping, usually 0.03. Either w1 or w2 may be used. The user is cautioned to pick the natural frequency that is farthest from any other system natural frequencies. A typical table generated for this type of analysis by COADE Engineering is shown on the following page.

~ATA6ORDA

COUNTY-PAlACIOS SEAIiAU. COIIIU SSI III ALUftlNUN HANDRAIl ANALYSIS

HRAIl_HZ

REVISIIa 2 JAIIIARY

~.

1989

FREIlUElCY - WIIIII SPEED - FORCE DATA PIPE OD ;

=

1.116 INCHES

:::z==:=:nrz=~&:aa:a:::a:::z:za

: sysm NAT

:: "ODE 1: MINER : FREII, Hz

: 1 1

EIIIIIV FREIIS AT IUND SPEED ElCITATlIIK : SPAN LEfI&TH : EICITATIOII FORCE : FœCE PEI NODE : 1 : 14 lIGIES), LBS :: 0.50 1 CR IIIF, Hzl IIPII KIOTS 1 FORCE, LBS/FT : INDES : PER SPM, LBS

1

Il

2 3 4 5

Il Il

Ô

Il Il Il Il

"

Il Il

7 8 9 10

Il Il Il Il Il Il Il Il

Il

11

Il

..

12 13 14 15 16

Il Il

...... .. "

"

Il

Il

. l7 .,, . la

,

....

19 20 21 ·., 22 Il 2l 24 .," 25 26 ·, 27 28 " :: 29 : : 30 , ·,,. 31 32

Il

Il

.

.j

34

35

. :6

. ,, ..

40.89938 41.28047 41.88889 42.118454 43.110973 44.11152 44.5~2

,';.00;37

Il

,, ,

.... , 4l._

39.9~34

,:.;4èIÎ

.

Il

~

32.60962 35.70562 39.52806 40.49241 40.86971 41.47207 42.25980 43.17579 43.67259 44.15167 45.10706 45.96583

80.2441 : 79.84187 80.114432 94.6303 : 94.15596 95.10227

Il

,

: : : : : : : : : : : :

39 40

Il

;

22.n576

1 23.60240

69.0467 : 68.70060

Il

.

1

10.47247 11.214711 12.32979 13.711&:1 15.29209 17.04298 18.9S899 21.00937 23.004ô7 23.83961 32.93736 36.06448

'"7

Il

~

10.36827 11.103111 12.20710 13.57541 15.13993 111.87339 18.n034 20.80031

45.56041 46.42781 4I.UtD 47.9908 48.19800 51.48883 53.39239 54.17657 54.42982 63.19633 63.27482 63.54456 63.67601 64.3715b 64.56070 65.73975 69.39107

1.

. . ,, ., ,. ·:.

10.4205 11.1591 12.2686 13.6438 15.2162 111.9584 18.81149 20.9051 22.8905 23.7213 32.7739 35.8855 39.7272 40.69114 41.0756 41.1181 42.4727 43.3933 43.8926 44.3741 45.3343 46.1974

...

47.3529 : 47.9588 : 51.2333 : 53.1274 : 53.9077 : 54.1597 : 62.8827 : 62.9608 : 63.2292 : 63.36 : 64.0521 : 64.2403 : 6S.m5 : ,,,,310}

1 1

----

Il Il

::

..

z:zxr:pswszassssrzzz

aa.z=2:TS:STz:z:zzssz.............szzzs 1 1

47.11SS4 47.71840 50.97649 52.86110 53.63748 53.88822 62.56750 62.64521 62.91226 63.04240 63.73103 63.9182'1

65.08561

4.91 5.211 5.79 6.43 7.18 8.00 8.90 9.86 : 10.79 : 11.19 : 15.46 : 16.92 : 18.73 : 19.19 : 19.37 : lU6 : 20.03 : 20.411 : 20.70 : 20.93 : 21.38 : 21.79 lD.U : 22.33 : 22.62 : 24.16 : 25.05 : 25.42 : 25.54 : 29.65 : 29.69 : 29.82 : 29.88 : 30.21

: 30.30 : 30.85 : 32.56 •

,,;;::. ij:,s

Il

4.27 : 4.57 : 5.03 : 5.59 : 6.24 : Il.95 : 7.73 : 8.57 : 9.38 1 9.72 : 13.43 : 14.71 : 16.28 : 111.é8 : 16.83 : 17.08 : 17.41 : 17.78 : 17.99 : 18.18 : 18.58 : 18.93 : U.&I 19.41 : 19.65 : 21.00 : 21. 77 :

22.09 22.19 25.77 25.80 25.91 25.97 26.25 26.33 26.81 28.30

:v ..,o

: : : : : : : : : : 1

: 37.84 32.88 : : 44.63 38.7B :

0.005'1

0.0068 0.0082 0.0102 0.0127 0.0157 0.01~

0.0239 0.0287 0.0308 0.0587 0.0704 0.0863 0.09011 0.0923 0.0950 0.0986 0.1030 0.1053 0.10n 0.1124 0.1167

"ua 0.1226 0.1258 0.1435 0.1543 0.1589 0.1604 0.2162 0.2168 0.218b 0.2195 0.2243 0.2257 0.2340 0.2607

85 85

85 &:1 &:1 85 115 115

85 &:1 115

&:1 85 85 85 115

85 85 &:1 115

85

•as , 115

85

as

85 85 85 85 e5 85 e5 85

85 85 85

0.0421 0.0482 0.0583 0.0721 0.0897 0.1114 0.1378 0.1693 0.2029 0.2179 0.4160 0.4988 0.6113 0.11415 0.6535 0.&729 0.11987 0.7293 0.7462 0.7626 0.7960 0.8266

",'f 0.8685 0.8908 1.0166 1.0932 1.1255 1.1361 1. 5315 1.5353 1.5484 1. 5549 1.5890 1.5984

1.&573 1.8465

0.0084 0.0096 0.0111 0.0144 0.0179 0.0223 0.0276 0.0339 0.0406 0.0436 0.0832 0.0998 0.1223 0.1283 0.1307 0.1346 0.1397 0.1459 0.1492 O.lm 0.1592 0.1653 L,

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415'

PULSATION METHODOLOGY

Design: A typical "rule of thumb" to be used for designing plplng systems when a reciprocating pump or compressor is the driver is that all natural frequencies should be kept above approximately (6) Hz. When no other means for analysis is available, an eigenvalue extraction of the systems lowest modes should be performed and restraints inserted wh en modes exist lower than (6) Hz. Animated plots of the associated mode shapes can aid in the selection of restraint locations. Low horizontal modes may not need to be restrained if friction at (Y) restraints, which is ignored in the modal extraction, will in reality prevent any excessive vibration from occurring in horizontal directions.

Field Problems: Unexpectedly, and sometimes after support changes or process modifications an operating line will begin experiencing large amplitude, low frequency vibration. The first step in the solution is the construction of the dynamic model. Particular attention should be paid when modelling the piping system in the area of the field vibration. This might include accurately representing valve operators, in-line flange pairs, orifice plates and measuring equipment. lt is also a good idea to add extra nodes in the area where vibration is experienced. The extra nodes would be put at bend "near" nodes and at span midpoints. The extra points on the bend near nodes can be added using the ANGLE # and NODE # fields in the bend auxilliary spreadsheet. The next step is the eigenvalue/eigenvector extraction. If the system is large, then degrees of freedom far removed from the area of local vibration should be eliminated. Approximately (6) natural frequencies should be extracted. Natural frequencies and mode shapes define the systems "tendancy to vibrate". The mode shapes extracted should show how the system in the area of the local vibration problem is tending to displace. In most cases weak acoustic resonances are coupled with mechanical resonances to produce the large amplitude vibrations experienced in the field.

Very typica11y one of the first (6) mode shapes wi 11 show exact1y the shape di sp1 ayed by the pi pe vi bra ti ng in the field. If the mode shapes extracted do not show movement in the area of the local vibration then not enough degrees of freedom were removed from other areas. If the lowest mode shape in the area of the local vibration prob1em is above (15) Hz. then there is a good possibi1ity that either the vibration is mechanica11y induced or the f1uid pulsation peak pressures are very high. Either of these cases may represent critica1 situations which should be eva1uated by an expert. When the mode shape is identified which corresponds to the observed field vibration, the pulsation load model can be developed. Pulsation loads will exist at closed ends, at bends, and at changes in diameter. Harmonica1ly varying forces are put at these points in an attempt to get the mathematical model to vibrate 1ike the rea1 piping system. The driving frequency for the app1ied harmonie load should be equa1 to the frequency that pressure pulses are introduced into the 1ine. The magnitude of the harmonie load can be estimated within a range of to1erances. The actua1 design value is selected from this range such that resulting displacements of the model are close to those observed in the field. Output from the harmonie analysis can be processed in the static output processor and maximum restraint loads due to the dynamic forces calculated. It is critical when redesigning supports for dynamic loads that static thermal criteria are not violated by any new support configuration designed. Static thermal criteria and dynamic displacement criteria must be sati sfi ed simul taneously !!!!!! Given the current state of the harmonies processor the unba1enced pressure loads at e1bows can be mode1led on1y one e1bow-pair at a time. The ultimate objective of the harmonie analysis will be to find the elbow pair whose unbalenced load results in the observed field vibration. Unbalenced loads exist between adjacent elbows because the pressure peak in the traveling wave hits each elbow at a slightly different time.

417

If the pressure at elbow "a" is denoted by Pa(t) and the pressure at elbow "bU is denoted by Pb(t) then the unbalenced force which acts along the pipe connecting the two elbows is: F

=

A * Pa(t)

A

* Pb(t)

EQ. (1 )

where A is the inside area of the pipe. The expression for Pa(t) can be found assuming the pressure peak hits the elbow "a" at time t = 0:

= Pavg

PaCt)

+ 0.5{dP)cos wt.

EQ. ( 2 )

where: (Pavg) (dP)

average pressure in the line, a1ternating component of the pressure, (Pmax-Pmin) driving frequency.

(w)

If the straight pipe between the e1bows "a" and "bU is (L) inches long, then the pressure peak that has just passed e1bow "a" will get to e1bow "b" (ts) seconds later, where ( ts) =

(L)

/

c,

(c) being the speed of sound in the fluid. (Remember, pressure pulses travel at the speed of sound, not the speed of the fluid !!! ) The expression for the pressure at lib" can now be written: Pb(t)

= Pavg

+ 0.5(dP)cos (wt+Q).

Q is the phase shift between the pressure peaks at Q =

w

EQ.(3}

"a"

and

* (ts).

Combining equations l, 2 and 3 the unbalenced pressure force can be written: F(t}

= O.5(dP}A * [

cos wt - cos (wt+Q) ]

This function has a maximum: Fmax = O.5(dP)A sin Q,

"bU,

and a period of

l/w,

F(t)

and will be approximated with:

= O.5(dP)A

sin Q * cos wt.

The formulation of the harmonie loads can be summarized: 1. Decide which elbow-elbow pair is most likely to have an unbalenced force which could cause the displacements observed in the field. 2. Find upper and lower estimates for the following variables:

dP w c L A

-

Alternating pressure in the line (Pmax - Pmin) Driving frequency. Speed of sound in the fluid Length between the two elbows Area of the pipe.

3. Find the time it takes the pressure wave to get from one elbow to the other.

ts

=

L / (c-);

(c-) is the lower estimate for the speed of sound in the fluide 4. Find the largest estimated magnitude of the unbalenced pressure force: Fmax

=

(O.5)(dP+}A

sin [(w+)(ts)];

(dP+) is the upper estimate for the alternating pressure. (w+) is the upper estimate for the driving frequency. 5. Run a single harmonic analysis with a force of F

= Fmax

cos wt

acting along the axis of the pipe between the two elbows. If the pattern of the displacement approximates that seen in the field, and if the magnitude of the calculated displacement is greater than or equal to the magnitude of the displacement in the field then the harmonie load to be used in the design of the new restraints has been found.

COADE Pipe Stress Analysis Seminar Notes

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