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SEMINAR ONPiping Stress Analysis Part-II

– G.Palani

“KNOWLEDGE IS NOTHING, UNLESS ITIS SHARED”

Piping Stress Analysis Part-II – G.Palani

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Table of Content:

1. How are piping system classified by stress? 32. Advantage in using accurate restraint stiffness 33. Piping Nozzle Evaluation 5

3.1 Methods for analyzing equipment nozzle loads 63.2 Centrifugal Pumps and Pressure Vessel nozzle load details 6

4. Expansion Loops 144.1 Expansion Loop Examples 19

5. Thermal Expansion 265.1 Equipment – Anchors 265.2 Using Coordinates to find free expansion 265.3 Expansion influencing vessel anchor end 275.4 Different Expansion Coefficients – Piping System 325.5 Line Spacing Requirement 335.6 Locating Friction Balance 37

6. Attachments 396.1 Force Nomograph 396.2 Stress Nomograph 406.3 Stress Pipeway Layout (Attached Separately)


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1 How are piping system classified by stress?

A. Critical Service Piping Systems – By computer analysis

Pump, turbine, blower and compressor pipingPiping designed for 500°F or greaterPiping designed for 1000 psig or greaterPiping greater than 24 inch diameterPiping connected to sensitive equipment such as fired heaters, fin-fan coolers, reactors and boilersPiping supported or guided from stress-relieved vesselJacketed piping

B. Intermediate Service Piping Systems – By manual calculation

Piping designed for 250 to 499 °FPiping designed for 500 to 999 psigPiping from 6” to 24” diameterNonmetallic pipingVacuum linesPipeway and yard piping

C. Mild Service Piping Systems – By Visual Inspection

Other piping not included in Critical and Intermediate PipingSystems

2 Advantage in using accurate restraint stiffness

Caesar II default restraint stiffness is in the range of 1E12 lb/in.

Case-1:

For a rigid piping configuration, considering both ends anchor, thefollowing are the data for the piping configuration:

12” Sch STD, Low carbon steel pipe @ 350 °F as shown below,


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According to beam theory (Guided Cantilever Method):

= Pl3 / 12EI = elM = Pl/2

Solving for P: P = 12EI / l3Solving for M: M = 6EI / l2Solving for Se: Se = 6EI / l2Z = 6ER / l2

The above Se equation shows that the stress range decreases with thesquare of the length of the absorbing leg, so longer the leg absorbingthe displacement, the lower the stress range.

For the above configuration (Fig. 2-20)

= 1.88 E -3 x (10 x 12) = 0.23”Se = 6 x 29 E6 x 6.375 x 0.23 / (10 x 12)2 = 17,720 psi

Note: This calculation does not consider the SIF at the elbow at the topof the leg. If SIF is considered (for long radius elbow SIF is 2.8) thestress range would result in 49,600 psi, which is excessive.

Case-2:

If the restraint has a lateral stiffness of 10,000 lb/in (instead of 1E12)the thermal growth is partially absorbed by the pipe and partiallyabsorbed by the restraint:


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For 12” std wall pipe, I = 279.3 in^4 & Z = 43.8 in^3

Se = 2675 psi

This significantly reduces the stress range from the previous value of17,720 psi.

Vessel nozzle stiffnesses can be calculated manually using WRC 107/297or some equivalent.

3 Piping Nozzle Evaluation:

Piping loads on nozzles of equipment such as pumps, compressors,turbine, and heat exchangers have the tendency to deform or overstressequipment casing, overload bearings or cause shaft binding.

The following are the standards to calculate the allowable nozzle loadsdue to piping for the below equipments,

Steam Turbine - NEMA SM-23Centrifugal Pumps - API 610Centrifugal Compressors - API 617Air Cooled Heat Exchangers - API 661Pressure Vessels - WRC 107 / 297


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WRC Bulletin 107 (Local Stresses in Spherical and Cylindrical Shells dueto External Loadings) / 297 (Local Stresses in Cylindrical Shells due toExternal Loadings on Nozzles – Supplement to WRC Bulletin 107)

3.1 Methods for Analyzing Equipment Nozzle loads:Performing a testFinite Element AnalysisInternational Codes and Standards

3.2 The following are the International Codes and Standards forevaluating the nozzle loads due to piping for Centrifugal Pumps andPressure Vessels,

3.2.1 API 610 – Centrifugal Pumps for Petroleum, Petrochemical andNatural Gas Industries (Tenth Edition, Oct-2004)ISO 13709:2003, (Identical) – Centrifugal Pumps for Petroleum,Petrochemical and Natural Gas Industries


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3.2.2 WRC Bulletin 107 Analysis:

Based on the work done by Bijlaard, WRC 107 was prepared.This Bulletin uses the finite element analysis program toexamine the stresses in vessel nozzles due to external loadattachments.WRC Bulletin 107 (Local Stresses in Spherical and CylindricalShells due to External Loadings)Note that WRC 107 computes stresses in the vessel shell at thenozzle/vessel interface. Stresses in the nozzle wall are notcomputed. WRC 107 is used to analyze attachments tocylindrical or spherical vessel attachments.WRC 107 method should not be used when the nozzle is verylight or when dimensionless parameters fall outside the limitsof their respective figures.WRC 107 should not be used if either of the followinginequalities is not satisfied:

d/D < 0.33D/T > 50

Where:

D = mean diameter of vessel, ind = outside diameter of nozzle, inT = thickness of vessel wall, in

WRC 107 nomenclature and orientation of loads is shown below,


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Based upon various dimensional ratios of the vessel/nozzleconfiguration, the engineer selects 12 parameters from 12different curves.These parameters are used in 15 equations to calculate 48different stresses – Circumferential membrane and bending,longitudinal membrane and bending and shear stresses (in twodirections) at each of eight locations in the vessel (6*8 = 48stresses).These eight locations are the inner and outer edges (subscript land u respectively) of the vessels at 0, 90, 180 and 270 degreesaround the nozzle.

3.2.2.1 ASME Section VIII – Div. 2 Requirements on WRC-107:AD-160 Fatigue EvaluationAD-160.1 Operating Experience:

When the user is considering experience with comparableequipment operating under similar condition as related to thedesign and service contemplated (Synonyms: Consider as apossibility), fatigue evaluation shall not be performed.


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AD-160.2 Rules to determine need for Fatigue Analysis ofIntegral parts of Vessels:

A fatigue analysis need not be made provided all of Condition Aor all of Condition B is met. If neither Condition A nor B is met,a detailed fatigue analysis shall be made in accordance with therules of Appendices 4 and 5 for those parts which do notsatisfy the conditions.

AD-160.3 Rules to determine need for Fatigue Analysis ofNozzles with separate reinforcement and non-integralattachments such as pad type reinforcement, fillet weldedattachment etc.:

A fatigue analysis of pad type nozzles and non-integralattachments need not be made provided all of Condition AP orall of Condition BP is met. The rules of Condition AP areapplicable only to vessels constructed of materials covered byFigs. 5-110.1 to 5-110.4. If neither Condition AP nor BP is met, adetailed fatigue analysis shall be made in accordance with therules of Appendices 4 and 5 for those parts which do notsatisfy the conditions.

3.2.2.2 Caesar WRC107 Program Table Format:

The table shown below contains 72 values – eight entries oneach of nine lines. The eight entries represent the stresses atthe eight locations Au, Al, Bu etc. The nine lines are threegroups of three lines – the groups representing radial(circumferential), tangential (longitudinal), and shear stresses,while the three lines in each represent the loadings from thepressure, sustained and expansion load cases.


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The stress terminology shown above is:

Pm - Stresses due to internal pressure in the vessel. The Hoopand Longitudinal values of these stresses can be readilycalculated by hand. Due to the discontinuity of the nozzle cut-out; there can be no hoop stresses at location C or D and nolongitudinal stresses at locations A or B. Also shear stresses inthe vessel wall due to pressure are negligible.

Pl - WRC 107 calculated stresses due to sustained loads

Q - WRC 107 calculated stresses due to expansion loads

Sm - Allowable stress intensity for the material at operatingtemperature

The three calculated stress intensities must now be comparedto:

Pm < SmPm + Pl < 1.5 SmPm + Pl + Q < 3.0 Sm (Sm is the average Sm at cold and

hot temperatures)

3.2.3 WRC Bulletin 297 Analysis:

297 (Local Stresses in Cylindrical Shells due to External Loadingson Nozzles – Supplement to WRC Bulletin 107)


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Conditions of WRC 297 are:

d/D < 0.5d/t > 2020 < D/T < 2500d/T > 5

Where:

d = outside diameter of nozzle, inD = mean diameter of vessel, int = thickness of nozzle, inT = thickness of vessel, in

4 Expansion Loops

Need for Expansion Loop:

One of the device used to improve the flexibility of piping are expansionloops.Piping systems with high temperature expands. The objective in pipingdesign is not to restrain this expansion but to redirect, absorb andcontrol its direction without overstressing the system. Loop absorbspiping expansion.Loops provide the necessary leg of piping in a perpendicular direction toabsorb the thermal expansion.Expansion loops may be symmetrical or non-symmetrical.

Disadvantage:

They are safer but take more space and piping. In some cases, it mayrequire additional supports.


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Points to Remember:

Hotter and larger lines are placed outside as outer loops because thelonger absorbing length (H) is needed. Smaller lines with lowertemperature are placed as inside loops.Guide is used on both sides of the loops for proper functioning, becauseguide directs the expansion into the bend along the axis of the pipe,which avoids shifting the lines sideways as shown in Fig. 5.6 below.

Three dimensional loops are widely used because this arrangement doesnot block the routing of low temperature lines under the loop. The usualraiser height is about 3 feet.

Vertical loops are placed at road crossing and sometimes are non-symmetrical due to the location of the road.


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To control the expansion of the piping, anchor must be located at themiddle of the piping so that the expansion will be equal at the bend ends(if the expansion is less than 10”) – Refer Fig. 1-29 below.


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Guides must be placed on two sides of the anchor to prevent the linefrom “snaking” and wandering into adjacent lines – Refer Fig. 1-29 above.To reduce the structural forces:

a) Place the heavy lines near column, not near the center of a long spanas shown in fig. 1-30.

b) The anchor forces due to piping expansion must be balanced or mustbe within an allowable limit as shown in fig. 1-30.

Expansion Loop Requirements:

Generally, if the total expansion in any direction on the pipeway is lessthan 10”, the loop could be avoided by locating the anchor in the middleof the run.

The total expansion between the loop anchors should not exceed 12”

Locating Pipeway Loops:

Ideally loops shall be located centered between anchors with equal legson either side of anchor as shown in fig. 1-37 below.


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When this is not practical make legs on either side of anchor as equal aspossible. By making these legs equal, the forces at the anchor shouldremain nearly balanced.

Multiple Loops:

More than one loop may be required when:

Spacing between braches and neighboring lines or steel is limited.It is impossible to make branch connections flexible enough.When loop becomes too large to support or fit into space available.Anchor forces between too unbalanced and steel cannot be economicallybraced.


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Fig. 1-41 shows a poor arrangement, since the unbalanced forces are moreand the total expansion between loops to absorb (14”) exceeds allowablelimit of 12”. The alternate way of approach is to make use of multipleloops as shown in fig. 1-42 below,

4.1 Expansion Loop Examples:

Example 1:

Find the size of the loop to absorb expansion in 200 ft of 12” carbonsteel pipe at 400°F. Assume the height to width ratio.


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Answer:

Total Expansion = 200 (0.027) = 5.4”

Using the nomograph, to determine loop size as shown below,

Read L2 (Bend length required to absorb expansion) in ft as 50 ft.

Assume H = W, then L2 = 2H + W = 50 ft. Thus H = W = 17 ft, makingL2 = 51 ft.

By Calculation (Guided Cantilever Method),

L2 = (3ED / 144 Sa) ^ ½

Therefore,

L2 = (3*29*10^6*12*5.4 / 144*20000) ^ ½


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L2 = 44 ft

The estimated loop size is shown in fig. 5.13 below.

Example 2:


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Kellogg Charts C-11 & C12 for Symmetrical Expansion Loop sizing andCalculating Forces & Moments


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Example 3: Determining which of the lines requiring loops need thelargest loop, second largest, etc., by the following process:

Multiply the total expansion of each line between its proposed anchorsby the pipe’s moment of inertia (E). (The stiffness of a line ismeasured by its “Moment of inertia”)The line with the largest of these calculated numbers will require thelargest loop, the next the smaller number, the next smaller loop etc.

The above calculation shows that the 16” line should be berthed wherethe 6” line is, the 10” line should be where the 16” is, and the 6” lineshould be where the 2” line is. Note, on longer than normal span, loopbowing may cause the pipe to lift off intermediate support causingoverspan.


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5 Thermal Expansion

The thermal loads that arise when free thermal expansion or contractionis prevented by supports or anchors, loads due to temperature gradientsin thick pipe walls, and loads due to difference in thermal coefficient ofmaterials as in jacketed piping.

The coefficient of linear expansion of a solid is defined as the incrementof length in a unit length for a change in temperature of one degree. Theunit is 10E-06in/in/°F.

The unit for mean coefficient of thermal expansion between 70°F(installation temp.) and the given temperature is given as in/100ft of pipelength.

To convert from 10E-06in/in/°F to in/100ft of pipe length, the followingrelation may be used:e, in/100ft = (coefficient) * 12 * 100 (Design temp. – install temp.)

5.1 Equipment – Anchors:

Most equipment is anchored to a foundation. Therefore equipmentnozzles are also anchors. Generally they are full anchors.The anchors are mechanically rigid but may have additional expansionwhen the equipment is hot.Even if the equipment is not bolted down, the weight may be great enoughto make the equipment an anchor point.

5.2 Using coordinates to find free expansion:

The algebraic combination of lengths in any direction is the same asthe difference in anchor coordinates (In all three dimensions, i.e.north, east and elevation). The above condition satisfies only when thepiping system temperature is uniform or same throughout theconfiguration considered for finding free thermal expansion.


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The fastest way to find the free thermal expansion is to multiply thedifference between the anchor coordinates times the coefficient ofexpansion. This is where the method has its greatest advantage.

Example 4:

Carbon Steel @ 300°F, e = 0.0182 in/ft.

North Expansion = 190’ * 0.0182 = 3.46”East Expansion = 65’ * 0.0182 = 1.18”

5.3 Expansion influencing Vessel Anchor End:

Points to Remember:

When finalizing the layout and plot plan the location of anchor needs tobe considered in relation to the major piping systems (large diameterpipe, pipe coming from underground, etc).The free thermal expansion does not depend on the piping arrangement.The free thermal expansion depends only on the relative location of the


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anchor points. It is very useful for the layout man who is locatingequipment.Stretching a pipe even a small amount takes a very large force.Preventing pipe from expanding thermally takes an equally large force. Inother words, the force required to prevent the pipe from expanding isthe same as the force required to stretch it an equal amount.The force found using the Nomograph-A can differ greatly from acomputer output, but is good enough for piping study purposes.

The following are the examples for the above points,

Example-5: (Fig. 1-11 & Fig. 1-12)

Case-1: Note that the Anchor End is at the right side ofHorizontal Vessel.

Answer:

In N-S direction the expansion to be absorbed is:

= eL (Coefficient of expansion * Length)= 0.046 * 30’= 1.38”


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In E-W direction:

= eL= 0.046 * 20’= 0.92”

Case-2: Change the anchor end of the drum as shown in figurebelow, (i.e.) the anchor is shifted to left side of Vessel

Answer:

In N-S direction the expansion to be absorbed is:

= eL (Coefficient of expansion * Length)= 0.046 * 10’= 0.46”


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In E-W direction:

= eL= 0.046 * 20’= 0.92”

Conclusion: The E-W expansion did not change from the previousexample. The N-S expansion was reduced considerably by just shiftingthe anchor end of the drum. Thus the case-2 requires less flexibilityand has the potential of saving pipe and fittings.

Example-6: (Fig. 1-19)

Both vessel and pipe CS at 300 F – 8” Sch. 40. Calculate the thermalforces at A & B as shown below,

Case-1: Let the Anchor end of Horizontal Vessel shall be as shownin Fig. 1-19 below,Note: Radial expansion of vertical vessel must be added for horizontalexpansion.


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Case-2: If the anchor end and slotted end were reversed, thenT.F.A would be:

Conclusion: Reversal of anchor end of horizontal vessel (Case-2)causes an increase in anchor force, (i.e.) as the expansion increases,the force required to restrain the expansion will also be increased.


SS material @ 350 F, 10” Sch. 20, with I = 114 calculate the thermalforce at A & B as shown in figure below,

Note: In this example, the guide acts as an anchor for forces in “B”direction, but not in “A” direction.


Refer to the Fig. 1-22 below; find the horizontal and vertical thermalforces.


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5.4 Different expansion coefficients have an effect in the pipingexpansion calculation:

Taking the difference between anchor coordinates does not workwhen portion of the system are different temperature and/or ofmaterials with different expansion coefficients.


Find the N-S and E-W thermal forces for the fig. 1-25 as shownbelow,


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5.5 Line Spacing Requirement:

When there are no flanges between the corner and first guide/anchoron the two adjacent lines, line spacing may be based on O.D of pipe orinsulation to O.D. of insulation plus expansion plus 1” clearance (ReferFig. 1-32) below.

= (D1+D2)/2 + thermal expansion + 1” (Clearance) + (Insu thk D1 + Insuthk D2 (if applicable))

Where, D1 & D2 are the O.D of two adjacent pipes.


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In expansion case, usually spacing is calculated with one line hot(operating) and one line cold (not operating).


Determine which line requires loops, based on the line spacing at theeast end of the pipeway under consideration. Assuming no extra spaceis available for thermal expansion.


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Answer:

Case-1: without considering the expansion,Refer to the Fig. 1-35 below,

A = 2.375” + 2” + 1” + 3” + 7” = 15.375” 16”B = 7” + 3” + 1” + 2.25” = 13.25” 14”C = Using Standard pipe spacing table = 12”D = Using Standard pipe spacing table = 19”


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Without considering the expansion, the total line spacing required fordistance A, B, C & D, Total = 15.5 + 13.5 + 12 + 19 = 62”

Case-2: Considering the expansion:

After the need for a loop has been established, locate the loopanchors

Let us locate the anchor for 4” SS @ 500 F – 2” IH line,

The coefficient of thermal expansion is 0.0501 in/ft.


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Allowable expansion at the right end, without increasing line spacing is= 16” – 7” – 3” - 1” – 2” – 2.25” = 0.75”

The maximum distance the anchor may be from the corner on bothends is found by dividing the allowable movement by the coefficient ofexpansion.

Left End L = 6 / 0.0501 = 120 ft

Right End L = 0.75 / 0.0501 = 15 ft

Referring to the figure 1-36, the expansion that the loop must absorbis given as,

=700 ft – 120 ft – 15ft = 565 ft,

Total expansion = 565 * 0.0501 = 28.3” of expansion.

Since the maximum expansion that the loop can absorb is only 12”,which is less than 28.3”, so multiple loops is required.

5.6 Locating Friction Balance:

Locating the friction balance of liquid headers that change size:


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The anchor is placed at the lines “Center of Gravity”.

The calculation is as follows, find the total weight and divide by two:

Locating the friction balance of vapour headers that change size:

Steam headers and flare headers should have their anchors locatedwithout considering water in the line. Include insulation, however ifpresent. The method of calculating is same as above,


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6.0 Attachments:


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nonmetallic piping

piping nozzle evaluation

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