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II Training_original

Description

CAESAR II Statics Training

CAESAR II Version 2011

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

Contents Introduction

6 Units

19 Axial

Theory and Development of Pipe Stress Requirements

25 Example

Supt 01

69 Analyse

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

Manifold

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

Cool H20 – FRP Piping

Gas Transmission Pipeline

228 B31

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

260 Jacket

264 B31

SUPT01 – Water Hammer

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

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,

This is the window where all tasks are started from

This includes opening/creating an input file,

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,

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

Click on the ellipsis button at the end of the text field and browse to E:\Training\CAESAR II\Exercises

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

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,

Give the new file a name and click View/Edit file

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

In the Units File dialogue box,

change the following units from the MM defaults: Stress  Pressure  Elastic Modulus  Pipe Density  Insul

Density  Fluid density  Transl

Stiffness  Uniform Load 

N/mm 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

enter 4 into the diameter field and CAESAR II will convert this to 114

Give the file a label as well to easily identify the file

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

F = Kx Example As CAESAR II uses a stick model,

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,

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,

After creating the new job file,

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 default units) not the units file we have just created

By default,

CAESAR II uses the units file set in the Configuration/Setup as the default file for new jobs (and also as the units to use to display the output results)

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

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

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

Now change the Units File Name setting to the units file just created

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

Save and exit

Return to the Piping Input

the units file to be used will be displayed – this should now be your custom units file

Verify that the correct units are in use via the tooltips

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction 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,

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

Selecting the material will fill in the Elastic modulus and Poisson ratio and various material allowables under the Allowable Stress area,

depending on the design code selected (B31

3 default)

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

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,

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

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction Specify the displacement at node 20,

and specify a 2mm displacement downwards in the Y direction – i

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

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

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

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction 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

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

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

Also select the stress type as SUS(tained)

The analysis will now take into account only the displacement reaction

Before we analyse the piping system,

let us first perform the hand calculation in order to check

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction 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,

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction 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

Select the load case (SUS) D1 and the Displacements standard report and click to show on screen:

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction The DY at Node 20 is

Now to check the force at node 20

view the Global Element Forces report

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction 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:

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction

The CAESAR II results,

Global Element forces report should verify this:

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction 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

Expansion:

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Introduction 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,

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,

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 is usually dependant of the duration or frequency of the occasional load

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements

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

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements 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 ⁄

⁄ 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

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements 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 The hoop stress can be approximated conservatively for thin-wall cylinders by assuming that the pressure force applied over an arbitrary length of pipe,

l is resisted uniformly by the pipe wall over that same arbitrary length

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements

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

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements

Where: Maximum shear stress shear force shear form factor

Dimensionless quantity (1

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:

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements 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

Moment of Inertia

Section Modulus

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements 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 ⁄ (

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements Torsional Stress

The maximum torsional stress occurs at the outer radius where again

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements

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,

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

A Mohr’s circle can be developed by plotting the normal vs

shear stresses for the two known orientations (i

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:

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements

The centre of the circle is at

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: *(

As noted above,

the maximum shear stress present in any orientation is equal to

Continuing our example: Mohr’s Circle of Stress Centre of circle

Radius of Circle

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements Maximum Principal Stress S1 √(

Or from the Mohr’s circle above,

S1 = 78

07 + 50

59 = 128

Or from the Mohr’s circle,

S2 = 78

07 – 50

59 = 27

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements

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,

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:

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements Therefore the octahedral shear stress in a uniaxial tensile test specimen at failure is calculated as: √(

√ Therefore under the Von Mises theory: Plastic deformation occurs in a 3-Dimensional stress state whenever the octahedral shear stress

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 maximum shear stress exceeds

Copyright © 2011 Intergraph CADWorx & Analysis Solutions

CAESAR II Statics Training Theory and Development of Pipe Stress Requirements 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,

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

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

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: √

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CAESAR II Statics Training Theory and Development of Pipe Stress Requirements 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

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

In this case,

the pipe would appear to be safely loaded under these conditions

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CAESAR II Statics Training Theory and Development of Pipe Stress Requirements

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

1 and B31

combined the bending and torsional stress terms,

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

Note that most codes allow for the exact expression for pressure stress

the sustained stress calculations

Note also that there are many additional piping codes addressed by CAESAR II

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,

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

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CAESAR II Statics Training Theory and Development of Pipe Stress Requirements 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,

Occasional:

Where: = Occasional Stresses = resultant moment due to occasional loads =√ = occasional load factor = 1

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CAESAR II Statics Training Pipe 1

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

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”,

notice the “>>” symbol in the top right corner:

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CAESAR II Statics Training Pipe 1 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,

In this model we also require insulation

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,

The fluid density can be entered as 0

As before the pipe size can be entered as 10 for 10” and S for STD schedule piping

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CAESAR II Statics Training Pipe 1

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,

Element 10 to 20 is 400mm in length,

Enter DZ as

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

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CAESAR II Statics Training Pipe 1

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,

Select the Valve flange database button and select a gate valve with flanged ends,

The Flange – Valve – Flange check box can be used to split the component into 3 elements ifrequired

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CAESAR II Statics Training Pipe 1

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)

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CAESAR II Statics Training Pipe 1 Continue to the next element

Enter the DZ as

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

In addition,

further data can also be entered such as if the bend is flanged or mitred etc

Accept the default long radius bend

The graphics will not display the bend yet,

as there is no following element

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CAESAR II Statics Training Pipe 1 Continue to the next element

This time we are now continuing in the –X direction

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

The Reducer Auxiliary will appear and we can specify further data,

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,

note the node numbers in the image:

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CAESAR II Statics Training Pipe 1 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,

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,

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,

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CAESAR II Statics Training Pipe 1

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)

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CAESAR II Statics Training Pipe 1 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 need to change this to 70 to 140

The node numbers will default to 70 to 80

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CAESAR II Statics Training Pipe 1 This element is the vertical leg,

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,

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,

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

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CAESAR II Statics Training Pipe 1 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 geom