CAESAR II Training_original

CAESAR II Statics Training

Contents

Introduction ... 6

Interface ... 6

Default Data Directory ... 6

Units ... 7

Create Custom Units File ... 7

F = Kx Example ... 9

Model Input ... 12

Load Case Editor... 16

Hand Calculation ... 18

Output Processor ... 19

Axial ... 21

Theory and Development of Pipe Stress Requirements ... 25

Basic Stress Concepts ... 25

Example ... 30

3D State of Stress in the Pipe Wall... 33

Failure Theories... 36

Maximum Stress Intensity Criterion ... 38

Code Stress Equations ... 40

B31.1 Power Piping ... 40

Pipe 1 ... 42

Input Model ... 43

Error Checking ... 52

Review Load Cases ... 54

Review Results ... 55

Supt 01 ... 64

Locating Supports ... 65

Adding Supports to Model ... 69

Analyse ... 75

Fix Model ... 80

Turbo ... 92

Manifold ... 111

Tutor ... 128

Fix Model – Part 1 ... 139

Include WRC 297 flexibilities ... 139

Review Results ... 141

Fix Model – Part 2 ... 143

Build Steel Structure ... 144

Combine Pipe and Steel models ... 154

Review Results ... 157

Fix Model – Part 3 ... 158

Loop Optimisation Wizard ... 161

Fix Model – Part 4 ... 166

Model the Expansion Joint Assembly ... 170

Document the Analysis ... 174

Custom Reports ... 174

Filters ... 178

Generate Report ... 179

Cool H

0 – FRP Piping... 194

Basics of Fibreglass Piping Analysis ... 195

CAESAR II’s Orthotropic Model for Piping Systems ... 198

Requirements of the ISO 14692 Code... 199

Allowable Stress Data for this Model ... 202

Configuration Options for FRP Piping: ... 204

Model the system ... 206

Load Case Setup ... 213

Review Results ... 213

Solving Expansion Problems ... 214

Static Seismic Using the ASCE 7-05: ... 214

Inertial Loads ... 214

Seismic Data Input ... 219

Load Case Setup ... 223

Review Results ... 225

Gas Transmission Pipeline - Buried Pipe ... 228

B31.8 Design Code ... 229

Modelling the system ... 229

Material Database ... 234

Buried Pipe Modeller ... 237

Restraint Parameters ... 237

American Lifelines Alliance Clay Model ... 238

American Lifelines Alliance sand model: ... 240

Meshing Parameters ... 241

Virtual Anchor Length (VAL):... 243

Bury Pipe ... 244

Load Case Setup ... 249

Results Review ... 250

Fatigue ... 251

Riser ... 254

Load Case Setup ... 277

Analysis ... 278

SUPT01 – Water Hammer ... 280

Edit the Model ... 281

Determine the Load ... 282

Define the Occasion Load ... 283

Load Case Setup ... 283

Check the Results ... 284

Reset the Load Cases ... 284

Check Results ... 285

Introduction

CAESAR II is pipe stress analysis software which uses beam theory to evaluate piping systems to numerous international standards. CAESAR II is not Finite Element Analysis (FEA) software, but instead uses a stick model built up of elements connected by nodes.

This course will introduce CAESAR II and demonstrate various modelling and analysis methods in order to evaluate and correct piping systems.

Interface

When starting CAESAR II, the Main Window appears. This is the window where all tasks are started from. This includes opening/creating an input file, reviewing Load Cases, reviewing results or accessing any auxiliary modules such as WRC 107/297 processor or the ISOGEN stress isometrics module. All modules open up in their own separate window.

When opening a new file, the file will open, but you will return to the Main Window. You can then choose to go to the Input processor, Output processor or the results for this file.

These modules (and other auxiliary modules) and their interfaces will be introduced as they occur throughout the training.

Default Data Directory

CAESAR II has the option to specify the default working directory – that is all files working with will be saved/opened from this default location. Of course it is still possible to navigate using windows explorer functions, this setting is just the default location when selecting New/Open.

Select File > Set Default Data Directory from the CAESAR II Main window

Units

CAESAR II performs all internal calculations in English units. To enable the entry and review of data in alternative units (such as SI), units files are used by CAESAR II. These units files simply convert the internal CAESAR II English units to the user’s preferred unit. Each CAESAR II file (referred to as a “Job File”) uses a particular units file which is specified on creation of the job. Files can be converted from one units file to another if required. The units files have the extension *.FIL and are located in the CAESAR II System directory, or in the same directory as the job file. The file to use is specified in the Configuration file.

Create Custom Units File

Throughout this course, we wish to use specific units for various parameters such as Pressure, Density etc. As such we require a units file which is different to the supplied default files. So we will create our own CAESAR II units file.

Select Tool > Make Units Files from the Main Window

This allows the creation of new units files, or the review of existing units files, useful if you receive a units file from a colleague and wish to check the units in use in the file.

Choose to Create New Units File and for the template file to use as a start point, select *MM.FIL. Give the new file a name and click View/Edit file.

In the Units File dialogue box, change the following units from the MM defaults:

Stress  N/sq.mm.

Pressure  bars

Elastic Modulus  N/sq.mm. Pipe Density  kg/cu.m. Insul. Density  kg/cu.m. Fluid density  kg/cu.m. Transl. Stiffness  N/mm Uniform Load  N/mm

Ensure Nominals is set to ON. This allows the entry of pipe nominal sizes and schedules into the input, which will be converted to actual diameters and wall thicknesses (e.g. enter 4 into the diameter field and CAESAR II will convert this to 114.3mm).

F = Kx Example

As CAESAR II uses a stick model, and beam theory, it is easy to prove this using a simple cantilever example. This example will introduce the basic modelling methods in CAESAR II and introduce the Input Spread Sheet, Load Case editor and the Output Processor. In addition, we can check the CAESAR II results against some simple hand calculations.

CAESAR II calculates forces using . Using the example below, we will create a simple

cantilever model, fixed at one end, and apply a displacement of 2mm at the other end. We can then calculate the force required to generate this 2mm displacement – and see this in the results.

First we will create the model in CAESAR II. Create a new file in CAESAR II, called FKLateral

After creating the new job file, the first time it has opened, the units will be displayed to the user for confirmation. You will notice that the units file displayed here for our file is English (CAESAR II

Click OK on the units review screen and the input spread sheet will open. To confirm/check the units, hover over any field in the input – the units used in this field will be displayed in the tooltips. For example, we changed the pressure units to bars, but the pressure field displays the units as lb./sq.in.

Close the input screen – we will change the units and return to the input with the correct units displayed.

In the CAESAR II main window select Tools > Configure/Setup

In the window which appears, select Database Definitions from the categories tree on the left.

Save and exit.

Return to the Piping Input. Again, the units file to be used will be displayed – this should now be your custom units file.

Model Input

We will now create the simple cantilever model and apply a 2mm displacement at the free end. The model will be as follows:

One element 10m in length going from node 10 to node 20 in the X direction, anchored at one end. 8” pipe with Standard wall thickness.

The input spread sheet will have defaulted to nodes 10 to 20, so simply enter 10000 in the DX field. We are in mm units already.

Enter the pipe diameter and wall thickness – this is 8” NS and STD wall thickness. As we have “nominals” set to ON, simply type in 8 in the diameter and hit enter. The actual OD for 8” pipe will be inserted. Repeat for the wall thickness, simply type in “S” and press enter.

Now we must fill in the pipe properties. We need to know the material properties to carry out the analysis. Select A106 – B from the list of materials. Notice that all materials have a number to identify it, you can simply type in the material number here - in the case of A106-B this is 106. Selecting the material will fill in the Elastic modulus and Poisson ratio and various material

That is our pipe itself. We now need to anchor it at one end (node 10) and apply a displacement at the other end (node 20).

Place the anchor by double clicking the Restraints check box. All the check boxes shown in the middle column on the spread sheet must be double clicked to check/uncheck.

To define a restraint you must specify a minimum of the node that the restraint will be attached to, plus the type of restraint. Press F1 for more information on the different restraint types. We need an anchor, so select ANC and locate it at node 10.

Now we will apply the 2mm displacement at the opposite end. Double click the Displacements check box to apply a displacement.

Specify the displacement at node 20, and specify a 2mm displacement downwards in the Y direction – i.e. enter -2 in the DY row.

Leave the remaining rows empty – do not specify 0. Specifying 0 fixes the node in the specified direction. Entering 0 in each row would be the same as an anchor. Leaving the values blank leaves the remaining directions free.

Finally to complete the analysis we must specify a design temperature and pressure. In our case these are not really relevant as we are only concerned with displacement, so just enter 21°C in T1 and 1bar in P1 fields.

The file can now be analysed.

Before analysis the input must be error checked in order to identify any issues which may prevent the analysis running (such as specifying both an anchor and an applied displacement at the same point), or anything which may provide incorrect results (such as Stress Intensification factors not present at a geometric intersection).

Run the error checker to check the model.

You should see only one note in the error checker report – the C of G. This can be useful for identifying problems such as incorrect densities applied – giving an incorrect weight for example.

If you receive anything other than this C of G, review the model for any issues. A common error on this exercise is the following:

This indicates that the displacement and the anchor have been specified at the same location. Check that the Anchor is specified at Node 10 and the displacement is specified at Node 20.

Load Case Editor

Once the error check is successful, we can create load cases to analyse the system. Access the load case editor. This button is only available after a successful error check.

The load case editor will be shown.

The default load cases are the Operating, Sustained and Expansion cases, as required by the design codes such as B31.3. Remove all these load cases, as we are only concerned with the displacement.

Add one new row.

Into the load case we can add any of the loads defined in the input into the load case. As we are only concerned with the displacement, drag in D1 – Displacement Case #1 into the L1 row.

Also select the stress type as SUS(tained).

The analysis will now take into account only the displacement reaction.

Hand Calculation As we know, The stiffness K is

; Where D = Pipe OD and d = Pipe ID

E being Modulus of Elasticity and L being the length.

So if we wish to know what force is required to displace the cantilever 2mm, we can calculate this quite easily.

So for a 2mm displacement, the force required is

Output Processor

Back in CAESAR II, run the analysis by clicking on the “Running Man” icon from within the load case editor.

You will see the following message explaining that certain loads have been defined in the model but are not included in any of the load cases to be analysed – this is OK in our case, but can serve as a useful warning if you have may loads/load cases defined. Select OK as is…Continue and click OK to analyse.

Once the analysis is complete, the Output Processor will be shown. We can view various results for any load case from here, plus general model reports such as the Input Echo. These reports can be viewed on screen, or output to Word/Excel/Text or straight to a printer.

In addition Custom report templates can be created, and any available report can be selected and added to the Output viewer Wizard, and exported/viewed to create/view a comprehensive report very quickly.

For now we will just check the displacement at node 20 to verify that it is 2mm, and the force at node 20 to check against out hand calculation.

The DY at Node 20 is -2mm, as we specified.

Now to check the force at node 20; view the Global Element Forces report.

Axial

We can repeat this exercise for axial forces. This is a simple change in the model changing the displacement from the Y to the X direction.

The analysis can be quickly re-run in cases where a change such as this has been made by using the Batch Run “Double Running Man” icon. This will run the error checker followed immediately by the analysis (providing there are no Errors).

The force should be as follows:

We are still using F = Kx, but we are using the Axial stiffness. Therefore:

The forces calculated such as in the previous example produce bending moments throughout the piping system. Bending moment is produced when a Force is applied at a distance – MB = F x L

Once the bending moment has been calculated, beam theory is used in order to calculate the stress at this point. rearranges to

is the section modulus Z. So this reduces further to

The stresses are calculated using this basic theory and compared to the allowable stresses in the design codes. CAESAR II has many design codes available, all of which have evolved separately over time, thus the way the stresses are calculated for each specific code are slightly different. However, looking at one of the most common piping codes – B31.3 – it can be seen that the equations used are based on the basic bending as detailed above.

B31.3 Chemical Plant and Petroleum Refinery Piping Sustained: Expansion:

As can be seen, the equations essentially use bending stress M/Z. The equations are a little more complicated than the basic cantilever example for the following reasons:

 To address piping systems in 3 dimensions

 To address areas in a piping system where particular geometry/components, such as at a branch connection or a bend, can increase the stress, and therefore the likelihood of failure. At these points, the stress is increased by a Stress Intensification Factor (SIF) known as i. The design codes contain formulae to calculate these SIFs.

 Stresses can also be caused by Pressure and Axial Forces

 The Stresses are categorised into Sustained, Expansion and Occasional, as detailed below. Sustained Stress: This is primary stresses caused by primary loadings such as the weight and pressure of the piping system.

Expansion Stress: Expansion stresses are secondary stresses caused by secondary loadings such as the thermal expansion and applied displacements.

Occasional Stress: Combines sustained stresses with those produced by an occasional loading such as earthquake of relief valve operation. As these are occasional loads, the allowable can be increased by a scalability factor, k. k is usually dependant of the duration or frequency of the occasional load.

Theory and Development of Pipe Stress Requirements

Basic Stress Concepts

Normal Stresses: Normal stresses are those acting in a direction normal to the face of the crystal structure of the material, and may either be tensile or compressive in nature. In fact in piping, normal stresses tend more to be in tension due to 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 number of different types of loads. For a piping system these are:

Longitudinal Stress: Longitudinal or axial stress is the normal stress acting along the axis of the pipe. This may be caused by an internal force acting axially in the pipe.

Where:

Longitudinal Stress

Internal axial force acting on cross section

Cross sectional area of pipe

( )

Outer diameter Inner diameter

A specific instance of longitudinal stress is that due to internal pressure:

⁄ Design pressure

Internal area of pipe ⁄

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

⁄ Or: ⁄

For convenience the longitudinal pressure stress is often conservatively approximated as

Bending Stress: 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 fibre to the maximum tensile outer fibre. Calculating the stress as linearly proportional to the distance from the neutral axis:

⁄ Where:

Bending moment acting on cross section

Distance of point of interest from neutral axis of cross section Moment of inertial of cross section

The maximum bending stress occurs where c is highest – the maximum value c can be is equal to the radius of the pipe.

⁄ =

Where:

Outer radius of pipe. Section modulus of pipe

Summing all components of longitudinal normal stress (for axial and bending):

Hoop Stress: Hoop stress is another of the normal stresses present in the pipe and is caused by internal pressure. This stress acts in a direction parallel to the pipe circumference.

The magnitude of the hoop stress varies through the pipe wall and can be calculated by Lame’s equation as:

( )

Where:

Hoop stress due to pressure Inner radius of pipe

Outer Radius of pipe

Radial position where stress is being considered

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 is caused by internal pressure and varies between a stress equal to the internal pressure at the pipe’s inner surface, and a stress equal the atmospheric pressure at the pipe’s external surface. Assuming that there is no external

pressure, radial stress is calculated as:

Where

Radial stress due to pressure

Note that radial stress is zero at the outer radius of the pipe, where the bending stresses are maximised. For this reason, this stress component has 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 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.

Where:

Maximum shear stress

shear force

shear form factor. Dimensionless quantity (1.333 for solid circular section)

These shear stresses are distributed such that they are at the maximum at the neutral axis of the 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 Shear stresses are usually small, shear stresses due to forces are traditionally neglected during pipe stress analysis.

Shear Stresses may also be caused by torsional loads.

⁄

Where:

Internal torsional moment acting on cross-section

distance of point of interest from torsional centre (intersection of neutral axes) or cross section torsional resistance of cross section

( )

Maximum torsional stress occurs where c is maximised. Again at the outer radius.

⁄

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

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

Longitudinal Stress: Shear Stress:

Hoop Stress:

Example

This example calculation illustrates for a 6” nominal diameter, standard schedule pipe (assuming the piping loads are known):

Cross sectional properties

Outside diameter

Mean thickness

Inside diameter 154.076mm Cross sectional Area ( – )

( _{– )} Moment of Inertia ( – ) ( _{– )} Section Modulus _⁄ ⁄

Piping loads Bending Moment Axial Force Internal Pressure Torsional Moment Stresses Longitudinal Stress Shear Stress Hoop Stress Bending Component of Longitudinal stress

is the radius where the stress is being considered. This will be at a maximum value at the outer surface where

⁄

( )

Torsional Stress The maximum torsional stress occurs at the outer radius where again

3D State of Stress in the Pipe Wall

During operation, pipes are subject to all these types of stresses. Examining a small cube of metal form the most highly stressed point of the pipe wall, the stresses are distributed as so:

There are an infinite number of orientations 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 maximised and another for which one normal stress is minimised – in both cases; all shear stress components are zero. In orientation 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 S1 (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.:

The converse of these orientations is that in which the shear stress component is maximised (there is also an orientation in which the shear stress is minimised, 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 in a three dimensional state of stress is equal to ½ the difference between the largest and smallest of the principal stresses (S1 and

S3).

The values of the principal and maximum shear stress can be determined through the use of Mohr’s circle. The Mohr’s circle analysis can be simplified by neglecting the radial stress component, therefore considering a less complex (i.e. 2D) state of stress. A Mohr’s circle can be developed by plotting the normal vs. shear stresses for the two known orientations (i.e. longitudinal stress vs. shear and hoop stress vs. shear), and constructing a circle through the two points. The infinite combinations of normal and shear stresses around the circle represent the 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 bending and torsional stresses are maximised and the radial normal and force-induced shear stresses are usually zero) is subject to 2D plane stress and thus the principal stress terms can be computed from the following Mohr’s circle:

The centre of the circle is at and the radius is equal to(* + )

. Therefore the principal stresses S1 and S2 are equal to the centre of the circle, plus or minus the radius

respectively. The principal stresses are calculated as: *( ) + and *( ) +

As noted above, the maximum shear stress present in any orientation is equal to or :

Continuing our example: Mohr’s Circle of Stress

Centre of circle Radius of Circle √( ₎ √( )

Maximum Principal Stress S1

_{√( )}

√( )

Or from the Mohr’s circle above, S1 = 78.07 + 50.59 = 128.66 MPa

Maximum Principal Stresses S2 and S3

√( )

√( )

Or from the Mohr’s circle, S2 = 78.07 – 50.59 = 27.48 MPa

Failure Theories

The calculated stresses are not much use on their own, until they are compared to material

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

There are three generally accepted failure theories which may be used to predict the onset of yielding in a material:

 Octahedral Shear or Von Mises theory  Maximum Shear or Tresca Theory  Maximum Stress or Rankine Theory

These theories relate failure in an arbitrary 3D stress state in a material to failure in 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; that is, release of the load does not result in the specimen returning to its original state.

The three failure theories state:

Von Mises: “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:

√

In a uniaxial tensile test specimen at the point of yield:

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

√( ) ( )

√

Therefore under the Von Mises theory:

Tresca: “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:

In a uniaxial tensile test specimen at the point of yield:

Therefore

Therefore, under Tresca theory

Plastic deformation occurs in a 3-Dimensional stress state whenever the octahedral shear stress exceeds √ 𝑆𝑦𝑖𝑒𝑙𝑑

Plastic deformation occurs in a 3-Dimensional stress state whenever the maximum shear stress exceeds 𝑆𝑦𝑖𝑒𝑙𝑑

Rankine: “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, S1 (by definition, S1 is always the

largest of the principal stresses.)

In a uniaxial tensile test specimen at the point of yield:

Therefore, under Rankine theory:

Maximum Stress Intensity Criterion

Most of the piping codes use a slight modification of the 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 , the maximum shear stress existing at failure during the uniaxial tensile test. Recapping, the maximum shear stress in a body is given by:

For a differential element at the outer surface of the pipe, the principal stresses were computed earlier as: √( ) √( )

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:

√

Plastic deformation occurs in a 3-Dimensional stress state whenever the maximum shear stress exceeds 𝑆𝑦𝑖𝑒𝑙𝑑

Multiplying both sides by 2 creates the stress intensity, which is an artificial parameter defined simply as twice the maximum shear stress. Therefore the Maximum Stress Intensity Criterion, as adopted by most piping codes, dictates the following requirement:

√

Note that when calculating only the varying stresses for fatigue evaluation purposes, the pressure components drop out of the equation. If an allowable stress based upon a suitable factor of safety is used, the Maximum Stress Intensity criterion yields an expression very similar to that specified by the B31.3 code:

√

Where:

longitudinal normal stress due to bending shear stress due to torsion

allowable stress for loading case

Continuing our example for the 6” diameter, standard wall pipe, in which longitudinal, shear and hoop stresses were calculated:

Assuming that the yield stress of the pipe material is 206 MPa (30,000 psi) at operating temperature, and a factor of safely of 2/3 is to be used, the following calculations must be made:

√

√

The 101.185 MPa is the calculated stress intensity in the pipe wall, while the 137.33 MPa 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.

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 calculate exactly 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 calculation much more difficult. When considering the hoop and radial stresses, it is no longer clear which of the principal stresses is largest and which is the smallest. Additionally the subtraction of S1 – 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 moment loading only. This provides an equally easy to use equation and sacrifices little as far as accuracy is concerned.

The explicit stress requirements for the B31.1 piping code addressed by CAESAR II follows. Note that most codes allow for the exact expression for pressure stress to be using in place of in the sustained stress calculations.

Note also that there are many additional piping codes addressed by CAESAR II. B31.1 Power Piping

The B31.1 code requires that the engineer calculate sustained, expansion and occasional stresses, exactly defined as below:

Sustained Where: = sustained stress = intensification factor

= resultant moment due to sustained (primary) loads

√

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

1. ¼ of the ultimate tensile strength of the material at operating temperature 2. ¼ of the ultimate tensile strength of the material at room temperature 3. 5/8 of the yield strength of the material at operating temperature (90% of the

yield stress for austenitic stainless steels)

4. 5/8 of the yield strength of the material at room temperature (90% of the yield stress for austenitic stainless steels)

Expansion

Where:

= expansion stress range

= resultant range of moments due to expansion (secondary) loads

= √

= Allowable expansion stress

= basic allowable material stress at the cold (installation) temperature, as per Appendix A of

B31.1 Code Occasional: Where: = Occasional Stresses

= resultant moment due to occasional loads

= √

= occasional load factor

= 1.2 for loads occurring less than 1% of the time = 1.15 for loads occurring less than 10% of the time.

Pipe 1

This exercise will provide further practise with the piping input, and introduce alternative editing tools which may increase productivity in creating models. We will also investigate and review the results to see what to look for and see how the piping system is behaving, and how to correct any issues which may arise during the design.

The first stage of this exercise is to input the model. The model is below; you will also have the same isometric printed on a separate hand-out in a larger format.

As before with the cantilever example, the model will be input using the node numbering system. Each section between two nodes is called an element. i.e. node 10 to node 20 are linked together by an element, referred to by ‘element 10 to 20’. Prior to entering geometry, it can be very useful and is a good idea to mark up the isometric drawing with the intended node number sequence.

We will use a slightly different method of inputting the data, which will allow us to maximise the graphics area during input. In the main “Classic Piping Input”, on each area, notice the “>>” symbol in the top right corner:

Double click this symbol to “tear off” the particular section of the input spread sheet. This will allow the Classic window to be minimised for the most part thus maximising the graphics.

Tear off the Node Numbers, Dimension Deltas and Pipe Sizes areas. As the material temperatures and pressures do not change throughout the model we can enter these on the first element and then we will not need them again.

Input Model

Enter A106-B as the material, 330°C as the temperature and 17 bars as the pressure

In this model we also require insulation; 65mm thick Calcium Silicate.

The rest of the information we will need to enter for our model can be done via the three windows we have “torn off”. Minimise the Classic piping input (of course this can always be maximised at any point if needed).

Finally we can enter the pipe size and schedule, along with the densities and corrosion allowance, as per the isometric.

The fluid density can be entered as 0.72SG and CAESAR II will convert this specific gravity to the correct units. As before the pipe size can be entered as 10 for 10” and S for STD schedule piping.

We will begin at the bottom “right” pipe where it is connected to a pump. This will be node 10. Note that this is an anchor, a fixed point in our system. Element 10 to 20 is 400mm in length, in the --Z direction. Enter DZ as -400mm

Node 10 is also fixed so we need to specify an anchor. Use the toolbar on the left hand side of the graphics window (default location) to specify a restraint.

The Auxiliary Data – Restraints window will appear. Specify that the anchor is at node 10. The auxiliary data window can now be closed.

Our first element is complete, and should look like the one below:

Use the Continue button to create a new element:

This next element is a 300# flanged gate valve. We could enter this in a number of ways. The valve will be rigid relative to the surrounding piping, so must be specified as a “rigid element” with a weight. This can be done either as 3 separate elements (flange – valve – flange), or as one overall element with the total length and combined weight specified. This can be done manually or by using the valve flange database to obtain the length/weight automatically from CAESAR II’s catalogue, which we will do. Select the Valve flange database button and select a gate valve with flanged ends, class 300.

The element will appear in node 20 to 30.

The correct length will be inserted (and the element will continue in the same direction as the previous element). Also note that the Rigid check box is checked and the rigid weight has been entered with the relevant weight for a 300# gate valve and flanges. (Hover briefly over the Classic piping input where it is docked).

Continue to the next element .

Enter the DZ as -825mm. This element also leads into a bend, so press the Bend button on the right hand toolbar. If using the classic piping input we could check the bend check box to achieve the same result.

The bend auxiliary data window will appear.

The default bend type is a long radius (1.5D) bend This radius can be changed. Common bend radii are available in the drop down, alternatively any radius required can simply be typed in here.

Continue to the next element .

This time we are now continuing in the –X direction. DX is -1050. The bend will now be visible in the graphics.

Continue to the next element .

This element is a 10”x12” concentric reducer and is 203mm in length. Enter DX as -203mm and specify that this is a reducer.

The Reducer Auxiliary will appear and we can specify further data, including the second end size. A s before, entering a nominal size in here will be converted to the actual OD. Enter 12 in the diameter 2 and S in the thickness 2 fields, which will be converted to the actual values.

Continue to the next element .

Finally continue from the end of the reducer to the centre of the tee, 254mm as shown on the isometric. DX is -254mm

We can now take advantage of the fact that the model is symmetrical and use the functions in CAESAR II to mirror the piping to create the opposite leg.

Use the Select group function to activate the graphical selection mode and draw a window around the model.

All elements will turn yellow to indicate that they are currently selected. Ensure all components are selected.

The Duplicate function can be used to copy, and mirror if required, selected elements.

Duplicate the selected elements and choose to mirror about the Y-Z plane.

We also need to increment the node numbers so that we do not have duplicate nodes. Currently our model goes from node 10 through to node 70.

If we increase the node numbers by 70, node 10 will become node 80, 20 becomes 90 and so on. Therefore the second leg will be node 80 through to node 140. The only issue with this is that there are no common nodes, so the piping will not actually be connected. This can easily be fixed by chaging node 140 (the centre of the tee on the second leg) to become node 70 (the node at the centre of the tee on the first leg). This will connect up the piping at the common node, 70 – the centre of the tee.

Click OK and the pipe will be duplicated, but as already stated there is no common node so CAESAR II does not know where to place the pipe. As such it locates it at the origin. The resulting model looks like the following.

All we need to do is connect element 130 – 140 to element 60 – 70. This can be done by changing 140 to become node 70. Select element 140. There are various ways of doing this – either double click in the graphics area, or user the navigation buttons to navigate to the correct element (as this is the last element the end button will quickly take you to the correct element).

The Edit Node numbers window should now read 13 to 140 and the element will be highlighted in the model.

Simply change the “To” node from 140 to 70. The model will now be connected as should look like the one below:

We can now complete the model by adding the vertical leg and connection to the vessel. Skip to the last element. This can be done by again using the Last Element navigation button or using the Ctrl + End buttons on the keyboard.

Click “Continue” to move to the next element . The node numbers will default to 70 to 80. We need to change this to 70 to 140.

This element is the vertical leg, and is 7m in the Y direction. DY is therefore 7000. This also leads into a bend so select the Bend icon as well. .

Click “continue” and place the final element 140 to 150 in the –Z direction, 2000mm. The final element connects to the vessel, so we will place an anchor at this point. Click the retsraint button and specify an anchor at node 150

Notice in the isometric that at the vessel connection, there are DY and DZ displacements. These are due to the thermal expansion of the vessel.

Select the Displacements button and enter in the required values 3mm in DZ and 12mm in DY.

Error Checking

The model is now complete, so run the error checker.

We will receive a fatal error and three warnings. We must correct the errors before we can analyse the model. The warnings may be acceptable but we should check to confirm that the input is as intended.

So our error is mentioning that we have both an anchor and displacements speciified at node 150. This cannot be possible as the anchor fixes the point, but the displacements move the same point. We cannot have both at the same time. Remove the anchor and edit the displacements.

Double click the error message to go straight to the area of concern. Now click the restraints button to remove restraints. Click OK in the message which appears. Now edit the displacements and fill in 0 in all other field (DX,RX,RY,RZ). A displacement of zero will fix the node in that direction, so now our node is fixed in all directions, except for DY and DZ where the relevant displacements are applied.

Re run the error checker and investigate the warnings. The second two warnings are regarding the reducer alpha angle which is not specified. CAESAR II is therefore using a default computed value. This is acceptable here for us.

The first warning is stating that there is a geomtric intersectaion at node 70 (the tee) but we have not specified a type of tee, and therefore a SIF. This can sometimes be correct but is most often the result of an oversight, as in this case. Return to the input and locate node 70. The Find tool can be used to do this:

The Zoom to Node if found check box will also zoom into that node/element if it is found, useful on larger models.

On node 70 use the SIFs/Tees button to specify a SIF at this point. This only needs to be done on one of the elements connecting to node 70, it is not necessary to do this on all three.

Select an unreinforced tee.

Re-run the error checker. All should now be OK, only the reducer alpha warnings will remain, plus the C of G report.

Review Load Cases

Access the load case editor

Recall from earlier the design code (we are using B31.3) addresses the stresses produced by the various loads. In our model we have the following loads applied:

 Weight  Pressure  Temperature  Displacement

B31.3 requires that two checks are performed – Sustained and Expansion Sustained – Weight and Pressure

These load cases are defined by CAESAR II as the default (recommended) load cases, shown in rows L2 and L3.

Row L1 is an operating case (OPE) and is the “Hot” case consisting of the ‘real world’ loads. This case is not required by B31.3 (although some codes do require this case also). However as this case is a “real world” scenario it is used to estalish restraint loads and loads on equipment conections. In addition, it is used to derive the Expansion case. The expansion case is the algebraic difference between L1 and L2 (L1 – L2).

Accept these load cases and run the analysis by clicking the “Running Man” icon.

Review Results

After the analysis has run, the output processor will appear. The first thing to notice is that the EXP case is coloured red. This indicates that this case has failed to code stress check. That is, the computed stresses in the system at some point are greater than the allowables published in the code.

We need to fix this.

The report shows that the code stress check failed and highlights in red where the check failed. Double clikcing on any column will order the report by that coluumn. Double click on the Code Stress column header to order by highest stress.

First we will check node 70. This is the tee. Look at the SIFs here. The in-plane SIF is 4.625 and the Out-Plane is 5.833. The stresses at this point are therefore being multipled by 4 and 5 times. If we can reduce these SIFs then the stress will reduce and can easily be reduced below the code

allowable.

Return to the input and return back to node 70. Pick the Intersection SIF scratchpad and choose node 70.

Change the unreinforced tee to a reinforced tee. Specify a pad thickness of 10mm

Click the Recalculate button and notice the SIFs reduce dramatically. Now the stresses will be multplied by 2.887 and 2.415 rather than 4 and 5.

Re-run the analysis using the batch run feature. The Expansion case is still shown in red, indicating the system is still overstressed. But check the Stress report for the Expansion case and notice that only nodes 10 and 80 are overstressed.

So what is happening at nodes 10 and 80?

Nodes 10 and 80 are the initial anchor locations, so we need to find out what is causing the overstress. This is in the expansion case, so if we recall the code equation for the expansion case:

From this equation it can be seen that the dominant factor in the code equation is the bending moment – it is the only factor in the expansion case. So which bending moment is this, Mi, Mo or Mt?

Mt is torsion, Mz and Mi and Mo are inplane and outplane – so vary dependant on the location. What

we can see from the results though, is which bending moment is the highest in terms of our axes. View the Expansion case, Restraint Summary report.

We can see from this report that at nodes 10 and 80, the highest bending moment is the MY moment, at 116 kN.m. The MX is also rather high at 88 kN.m.

So we know what is causing the overstress, but how do we correct this and reduce the bending moment (and therefore the stress)?

Let us look at the 3D plot to view what is causing the bending moment. Close the report and view the 3D plot

In the 3D plot window which appears, ensure that the Load case we are viewing is the expansion case and select to Show the deflected shape. You may need to Adjust the deflection scale to get a more exaggerated deflected shape.

We are looking upwards at the pipe from below. The Y axis is pointing upwards (away from us). The pipe is undergoing thermal expansion and causing the pipe to bend at the anchor points.

Looking at the model from the side view will also explain the MX moment. As can be seen, the riser is expanding causing the MX bending moment.

If we could add some flexibility in to the area where the pipe is expanding we can absorb some of this expansion, and so reduce the bending moment. We have the top leg which is flexible, but the bottom leg is not flexible enough.

If we can transfer some of the flexibility in A to B then we may solve the problem.

To do this, increase the length of B by 1m and so consequently reduce the length of A by 1m. This should give us more flexibility at the bottom, hopefully without removing too much flexibility at the top.

Repeat for element 100 to 110.

We will have to also consequently reduce the length of 140 to 150 by the same amount (1m). Change this from 2000mm to 1000mm.

Re run the analysis and check the results. The expansion case is now no longer coloured red and the highest code stress is 91% of the allowable at node 10. We have successfully reduced the stresses and the model now passes the code stress checks.

Verify the sustained stress report that this is still acceptable – it should be around 37% of the allowable.

Supt 01

This exercise is designed to demonstrate adding supports in CAESAR II and demonstrate the Operating case and Restraint load reports and give an indication of what the results mean on the restraint load report, along with a short example on how to combat issues with supports, such as “lift off”.

The model shown above will also be in your hand-out. Model the piping system as per this isometric. Anchor at nodes 10 and 90.

Locating Supports

The system is anchored at the termination points (nodes 10 and 90), but we also need to support the weight of the piping system as well.

If both supports are pinned (free to rotate), standard beam theory states that the max moment is at the centre of the span, l.

This moment is:

If both ends are fixed then the max moment is at the end of the span

This has a value of

Where

As piping systems are neither one nor the other and tend to be somewhere in the middle, a compromise therefore is reached with an approximation thus;

Taking into account the maximum moment could be somewhere between the ends and the centre – i.e. anywhere along the span.

This deals with continuous runs of pipe, however there are of course concentrated loads sometimes in the piping system, such as valves, flanges etc. The effects of these items on the pipe stresses can be estimated as well. For pinned connections, the maximum moment is located at the point of loading (P).

This maximum moment has a value of

Where a = longer portion of span b = shorter portion of span For fixed connections:

He maximum moment here is located at the end nearer the load, and has a value of

In either case (or in some case in between) the additional stress (M/Z) due to the concentrated loads should be added to the stress from the uniform load in order to determine the total stress.

Examining the formulas above, it can be seen that as the shorter span (b) approaches zero in length, the moment, and therefore the stresses approach zero as well.

So, if supports are located as close as possible to concentrated loads, the effects of these loads are reduced as much as possible.

The information on the preceding pages provides a simple rule of thumb to design for weight loading.

First support all concentrated loads in the system as closely as possible, reducing the stresses due to those loads to as close to zero as possible.

Next, we can use

Along with

If we knew the allowable stress, we could then use this information to determine a maximum allowable length of pipe – i.e. a distance between supports.

Rearranging the equations above, we can obtain

√

As this calculation will need to be done often, in order to save time calculating the Manufacturer

Standardisation Society of the Valve and Fitting Industry has calculated allowable piping spans for various configurations. These standard spans have been published and are shown on the next page. These spans assume:

 The pipe is standard wall with insulation  Maximum moment is Mmax = wl2/10

 No concentrated loads are present  There are no changes in direction

 Maximum allowable stress is taken to be approx. 10 MPa  Max deflection is approx. 2.5mm

 SIFs are not taken into account

It is rare that piping systems are only horizontal runs with no changes in direction etc.; therefore a caveat is taken in that changes in direction reduce the allowable span to ¾ of the standard span. In addition, the standard span does not apply to risers, since no moment (thus no stress) develops, regardless of length. However it is preferable to locate supports above the centre of gravity of the riser to prevent toppling.

Adding Supports to Model

Let us now return to our model SUPT01.

As there is an anchor at node 10, the valve here is supported.

The valve at 60-70 requires supporting. We will create a new node on element 50 to 60, called 57, and locate a +Y support here.

Use the Break command to split the element into two. First select element 50 to 60 and choose the break command.

We will locate this support close to the valve (node 60). Specify to break the pipe and insert the new node, number 57, 150mm from node 60:

The element will be broken and a new node inserted close to the valve. Locate a +Y support at this node 57.

The +Y support will support the pipe from below, and will allow movement in the +Y direction.

All the concentrated loads are now supported.

We can return and run through the piping system, placing supports as per the standard span. Check the table on page 48 to determine the maximum span for 12” pipe carrying water

The table on page 48 indicates that the maximum span for 12” pipe in water service is 7m. As discussed previously, for horizontal changes in direction, the support span is amended to ¾ of the standard span. 0.75 x 7 = 5.25m

The valve is supported at node 10. After node 10 the piping continues horizontally with a bend. The maximum span therefore is 5.25m. This places our support round the bend.

The piping after the bend is 13715mm before the riser. This can almost be split in two exactly with our 7m span spacing. Remember that the standard span does not apply to risers, and as mentioned before we will support the riser from the top, rather than trying to balance it from the bottom. As such we can locate a support near the middle of the 13715mm run, and one close to the bend (node 30).

Break element 30 to 40 and locate a new node number 33. Locate this 600mm from node 30. It is possible to add a support in at the new node location. We wish to add a +Y support at node 33, exactly the same support configuration as at node 57. So typing in 57 in the “Get Support from Node” field will place the same support as at 57 at our new node 33.

Repeat and break element 33-40. Break this 7000mm from node 33. Call the new node 37. Place the same +Y support from 33 (or 57) at this point too.

Continue on after the riser. There is already a restraint next to the valve, so we have fulfilled the minimum span up to the valve. After the valve we have horizontal pipe with a bend again.

Therefore our maximum span is 5.25m. The length of pipe is 4115mm and then 3640mm after the bend. There is an anchor at the end, so we just need one more support between the valve and the anchor at the end of the pipe.

Locate this support on element 70 to 80, close to the bend at node 80. Locate this 600mm from the bend.

Break element 70 to 80 and create a new node 77, 600mm from node 80 and with a +Y support the same as before.

Finally the piping riser needs to be supported. The length of horizontal run at the top and bottom of the riser is less than our span of 7000mm. There is no bending in the riser so in theory we can place a single restraint near the top of the riser.

Break element 40-50 and locate a +Y restraint 600mm from the tangent intersection point of the bend. Note that, although this support should satisfy our bending requirements on the horizontal sections, it may have a very large load since it will also support the whole of the riser.

The support located in the riser may be difficult to see, as it is probably hidden by the piping. To view the support either the size of the restraint symbol can be increased, or the pipe can be set to translucent mode.

The system is now supported as per the maximum span requirements. We can be sure that the sustained stress case therefore is acceptable, and should be in the order of approximately 10MPa. Error Check the model.

Analyse

Access the Static load cases create only a single load case with weight only. We wish to check the support locations we have just placed are below the acceptable limits

Run the analysis.

The highest stress level is 10.7MPa, which is almost exactly at the allowable from the standard span limit (this limit is based on an allowable of 1500 psi ~10.3MPa). The system is supported from a purely weight induced stress perspective.

We can also now view the restraint loads to see how the weight loads are distributed. View the Restraint Summary report.

Again this look OK, all restraints are taking a downwards acting load (-FY), although the restraint at node 43 is rather large compared to the others; 30,000N vs. less than half that for the remaining supports.

These loads however are due to weight only. Let us run now the cases required by the piping code B31.3. Return to the Static Load cases and select the recommended cases.

Run the analysis and view the sustained stresses. These will have increased slightly due to the fact that we are now including the pressure term; however the stresses are still well within the

allowables determined by the code.

Similarly the Expansion Case stresses are also very low and well within the allowable – the system is flexible unlike the PIPE1 example.

Now we can check the restraint loads in the real world operating case. Remember the Operating case is not required by the code, but it does represent the actual loads in a “real world” scenario, for the purpose of designing restraints.

The loads are different to before as we have included the effects of thermal expansion.

The load on node 77 is 0 (in the FY). This shows that this restraint is not taking any load. What is happening here?

View the displacements report to see what is happening at this point.

At node 77 the pipe is moving upwards 2.3mm. Also notice that at node 50 (the bottom of the riser) the pipe is moving down 26mm.

Viewing the 3D Plot can confirm this:

View the deflected shape (you may wish to increase the deflection scale to exaggerate the deflected shape)

The 3D plot shows that the thermal expansion is causing the riser to expand downwards at the bottom (node 50). This in turn is causing the pipe to pivot at node 57 – giving the large operating load at node 57. The pipe pivoting at 57 causes lift off at node 77 – so we see a 0 load.

Fix Model

As we know, the riser is expanding due to thermal expansion. The datum point for this expansion is the support located at node 43, at the top of the riser, so all the expansion is going downwards – causing the lift off issue. To rectify this we can attempt to move the datum point of the expansion, so that the expansion is more evenly distributed.

Insert an additional restraint on the riser. Call this node number 45 and locate this restraint 6000mm below node 43. Insert another +Y support at this location. This should have two effects:

1. Give a better distribution of the weight loads of the riser

2. Cause less thermal expansion downwards at node 50.

Re-run the analysis, the batch run command can be used as we have only made a small change by adding a support.

Review the restraint report for the sustained and operating cases to see how the new restraint has affected the results.

The sustained case shows that we have a better distribution of the weight of the pipe on the riser, as the new restraint is taking some load. The load on node 43 at the top of the riser is now distributed between 43 and 45 (43 has dropped from 30kN to 9kN).

The operating case however still shows lift zero load at node 77, and now also zero load at node 43. Checking the displacements also confirms this; there is still a positive displacement at nodes 77 and now at 43. Node 50 is still moving downwards, although only 15mm now.

We will attempt to further distribute the thermal expansion of the riser by adding a third restraint, located near the bottom.

Break element 45-50 and create node 47. Locate node 47 6000mm below node 45 and locate a +Y support at this point.

Rerun the analysis and check that the stress levels have not been adversely affected.

As before, view the SUS and OPE loads on the restraints to see how the new restraint has affected the analysis.

The sustained report shows that we have further improved the weight distribution among the restraints.

The Operating case now shows a negative load on restraint at node 77. There is no more lift off here at 77. However the operating loads at node 43 and 45 are now zero. All the thermal expansion that was going downwards with only the one restraint at the top of the riser has now been forced upwards instead, causing the pipe to lift off at the top of the riser (43 and 45).

The pipe is lifting off 22mm at the top of the riser. The 3D plot can also confirm the situation:

So we have a situation where we are supporting the weight of the system adequately in the SUS load case, but we have an issue with the thermal expansion. If we replace the rigid +Y restraints along the riser with Variable Spring Hangers (VSH), these hangers should allow thermal growth whilst also supporting the required weight for the SUS condition.

Place Spring Hangers

Return to the input at locate element 40-43. Remove the restraint at node 43 by double clicking the restraint check box.

Place a hanger here instead. Double click the Hangers check box

Select the Hanger table as Carpenter & Paterson. Carpenter & Paterson are a UK manufacturer whose database of available spring hangers is programmed into CAESAR II. CAESAR II will automatically calculate the required load and movement at the location, and then review the database to select an appropriate spring for the calculated load and movements.

Notice that 2 hangers are located at this location also. The graphics will not change in CAESAR II, but this will locate 2 hangers at this location, and the selected spring will be based upon this shared load.

The first not simply shows the number of hangers in the job, and how many of these hangers will be designed by CAESAR II at run time.

The second message states that new load case combinations are required for the hanger design. A load case for weight loads is required, so that CAESAR II knows how much weight the hanger(s) need to support. Secondly, a hanger design operating case is required. This case determines the thermal expansion at each restraint location to determine the movement at the spring hanger locations as well. The values of weight from the previous weight load case are used. These two cases together are used by CAESAR II’s spring selection algorithm to select the appropriate spring from the in-built catalogue.

Access the load case editor to create these new load cases.

Click the Recommend button in the load case editor. CAESAR II knows that there are hangers present that require designing, so will recommend the correct load cases for hanger design – cases 1 and 2.

Accept these cases. Note the stress type for these cases are HGR type. The results for these hanger cases are supressed by default (theses are pre-analysis cases and the figures do not actually mean anything other than for the spring hanger selection).

Case 1 W (HGR)

This case performs a Weight Analysis only with all support locations as rigid restraints. This tells the spring selection algorithm how much weight needs to be supported at each location (usually in the Operating condition)