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CAESAR II Dynamics TrainingInstructor: Chris BradshawTuesday, 09 April 20131WelcomeWelcome to the CAESAR II Dynamics Training courseExitsToiletsMobile PhonesAgenda

09:30 Course Start11:00 Break13:00 Lunch15:00 Break16:30 17:00 Course Finish

Developed since 1984Worlds most common pipe stress analysis software (80%+ Market share)Static and Dynamic CapabilitiesDeveloped by COADE; based in Houston, TX USAInterface with Intergraph Smart Plant 3D & CADWorxOver 30 Design codesStructural steel modeller includedAutomatically generate ISOGEN stress isometricsAdditional modules for local stress analysis such as WRC 107/297, API 610 and more all included

CAESAR IICOADEDevelopers of CAESAR II since 1984Based in Houston TXOther COADE Products:CADWorxAutoCAD based 3D Plant DesignPlant and P&ID CapabilitiesCAESAR II Industry standard pipe stress and flexibility analysis softwareOver 30 design codesPVElitePressure Vessel analysis softwareASME VII Div 1 and 2, EN-13445, PD5500TANKTank Design and analysis softwareDesign and evaluate tanks to API 650/653

Intergraph CADWorx & Analysis SolutionsCOADE was acquired by Intergraph in January 2010Renamed Intergraph CADWorx & Analysis Solutions (ICAS)All COADE Staff retainedNew staff brought in for support to allow developers to concentrate on developing softwareeCustomer system introduced for technical support

Virtual MachinesThe course is run on a Virtual Machine. This is to ensure that all machines are identicalVirtual Machine runs inside the host machine.CAESAR II is licensed by the green ESL plugged into the USB port of the machine.Generally you will see no effects, except:If you accidentally close the Virtual Machine you will return to the host machineIf you wish to copy any of your training files (e.g. onto a flash drive) you may need to connect the flash drive to the Virtual instead of the host (this may be automatic)The ESL may need to be connected to the Virtual Machine. Your instructor will show you how to do this if necessary

Dynamic analysis of piping systemsA Brief IntroductionDynamic Analysis of Piping SystemsDynamic Load Changes quickly with timeForces and moments not always resolvedUnbalanced loads = pipe movementSum of forces and moments != 0 means the internally induced loads can be different (higher or lower) than applied loads

Static LoadsDynamic LoadsLoad varies slowly, or does not vary, with time(Weight, thermal expansion, settlement, spring loads etc.)Load varies quickly with time(Earthquake, fluid hammer, vibration, relief valve etc.)System (internal forces and restraint loads) always had time to fully react to the applied loadSystem (internal forces and restraint loads) may not have time to fully react to the applied load before it changesSystem is always in equilibrium (sum of forces and moments on system are zero)System is not in equilibrium (sum of forces and moments on system are not zero)With no unbalanced forced, system remains at restWith unbalanced forces, system moves, according to F = MAInduced system reactions (internal forces and restraint loads) are equal to applied loadsInduced system reactions (internal forces and restraint loads) are not equal to applied loads, and may be much higher or lower.Dynamic Loads on a system depends on the relationship between the timing of the load vs. the timing of the systemTypes of LoadRandomWind

Seismic

Types of Load

Types of LoadHarmonicEquipment VibrationPulsationFlow Induced Vibration

Types of LoadImpulseRelief ValveFluid HammerSlug Flow

ExampleConsider the following System:

Restraint Loads respond fully to any imposed load in 25msExampleapply a load P to this system for duration of 5ms, and apply this load every 10ms For this applied load profile, the restraint loads would follow a force vs. time profileSystem Requires 25ms to fully respondLoad is applied for 1/5 of that timeTotal system response is 1/5 imposed loadbefore restraints can react, load has been removedInduced loads 9and therefore forces, moments and stresses are also 1/5 than a static load of same magnitude

this system requires 25ms for the restraint loads to respond fully to the imposed load. We are applying the load for just 1/5 of that time, so the total system response is 1/5 of the imposed load, due to the fact that before the restraints can fully react to the applied load, the load has been removed. The induced reaction loads (and by extension, the member forces and moments, and stresses) are much lower (1/5 in this case) than would occur under a static load of the same magnitude (each restraint under a static load would see a reaction equal to P/2, for a total of P).14ExampleNow consider identical system, except the restraint loads respond fully in 1ms rather than 25ms.For the same imposed load (duration 5ms occurring every 10ms) the system response would be

This system response is close to the applied load.ExampleThe pertinent question then is what is fast and what is slow. There is no answer here the important thing is the relative response time of the system vs. the rate of change of the applied load.

What if we had applied the load for 25ms duration?The system would have sufficient time to respond to the applied loadThe response would be the same as a static load

In fact, a static load is simply a dynamic load with a sufficient duration so that all systems can respond fully to it.

This leads us to two observationsWhen a system responds slowly to the applied loads, the induced reaction loads are much lower than the applied loads.When a system responds rapidly to applied loads, the induced reaction loads are approximately the same as those which would occur under the same static load.

But what happens when the system response is somewhere in between?

ExampleConsider the systems described previously (i.e. with response times of 1ms and 25ms), this time loaded with a harmonic load, cycling between P and P, with a frequency of 1 cycle per 25ms

The system with 25ms response time lags behind and fails to fully develop response loads

The second system with 1ms response time responds almost instantly and just about responds fully

ExampleNow consider a system which has a response time somewhere in between about 12.5ms.

Upon initial loading, the system initially attempts to respond to the load P, with restraint loads equal to -P/2. Since the system lags, it does not fully develop these restraint loads, but after 12.5ms, will have a total system response of somewhere around -0.7P. Considering the cyclic load, the applied load on the system will be P (at 12.5ms). This results in a net load of (-0.7P) + (-P) = -1.7P:

ExampleThe system now attempts to resolve the net load of -1.7P with two restraint loads of +0.85P. Assuming that at time T=25ms, these loads have actually only reached +0.6P (due to the response lag), or a total of +1.2P, the external load will now be +P, so the net system load will be +2.2P as shown below:

ExampleThis net load will then be resisted by total restraint loads (system response) of -2.2P, which will have reached approximately -1.5P by T=37.5ms, At which time the load will have reversed again, creating a net load on the system of -2.5P.

Continuing in this way, the net load on the system will be approximately 2.8P at 50ms, -3.0P at 62.5ms, 3.1P at 75ms, and so forth. The total developed load (total restraint loads) is shown as a function of time in the figure below.

This may continue until the developed load spirals out of control and the structure failsExampleFrom this example, it is clear that there is a third possibility for a system response under dynamic loading the induced load may far exceed the applied load.

As can be seen, system response loads (reactions, internal forces and moments, etc.) are not necessarily equal and opposite to the applied dynamic loads, so the net sum of the forces and moments acting on the system are something other than zeroThis means that the system does not meet the laws of static equilibrium, and cannot be solved using traditional static solution methods.

Since the sum of the forces and moments acting on a dynamically loaded system may not be zero, there may be unbalanced loads acting on the system.According to Newtons laws of motion, an unbalanced force on a system results in motion, due to the acceleration expressed by F=MA.

This information is useful in identifying problems due to dynamic loads a simple rule of thumb states that if the piping system is moving, it is being subjected to a dynamic load.

The system response loads may be quite different to dynamically applied external loads. The response can be classified by calculating the ratio of system response to applied loads (or expected response for a static load of same magnitude)This is called the Dynamic Load Factor (DLF)Dynamic Load Factor (DLF)M

DLFNote: The force in the spring is KX which, equal to 2W is exactly twice the static load, for a DLF of 2.0. Remember that this is for an instantaneously applied constant load, not a harmonic load.1 Load: Arm swing

3 Systems:

RopeFishing RodCricket BatDLF3 Options:Flexible SystemDLF < 1Rigid SystemDLF = 1Resonant SystemDLF > 1

1 system - Swing3 Loads:

Slap on the back : FlexibleSlow Walk behind: RigidProper push: ResonantRope = Flexible response DLF 1Bat = Rigid ResponseDLF = 1The important thing here is the timing, or more accurately, The RELATION of the system characteristics vs. the load characteristicsImagine a single Degree of Freedom oscillator

Harmonic responseHarmonic response

Harmonic responseResonant Response