vessels subject to external pressure · pdf file© intergraph 2014 presented by: ray...

© Intergraph 2014

Presented by: Ray Delaforce

Vessels subject to External PressureBefore After

The result of just air pressure !

Basic principles of compressive force

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Consider a simple bar subject to a tensile force

Failure can be predicted with fair precision knowing: The Tensile Strength UTS Cross-sectional area A The tensile force F

The force to promote failure is: F = UTS x AThat is a simple prediction

For a tensile, the force to promote failure is: F = UTS x A

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For a tensile, the force to promote failure is: F = UTS x ANow, consider a compressive force applied to the same barIt bends like this – important - It changes shape !

Does not change shape

Look at the consequences of changing shape - bending

Fx

There is both a bending stress and a compressive stressIn the case of the bar subjected to tensile – there is one stress

Here is the change in shape !

Stable

Now we bend the column (or plate) into a cylinder

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Fx

Stable

Subject it internal pressure P

P

, it becomes a stable circleNow to external pressure P , it becomes less stable

Un-Stable

Subject to just Membrane stress

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Stable

Subject it internal pressure P

P

, it becomes a stable circleNow to external pressure P , it becomes less stable

Un-Stable

Subject to just Membrane stress Subject to Membrane and Bending stresses

Failure is predictable Failure is un-predictable

Consider the rolling process to form cylinder from plate

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Consider the rolling process to form cylinder from plateIt is passed through roller to form the cylindrical shapePassed back through the rollers until the cylinder is formed

This process does not form a perfect cylinder, it is slightly oval

D1D2

Codes limit the difference between D1 and D2 to about 1-1/4%

Some shapes subject to external pressure are very un-round

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Some shapes subject to external pressure are very un-round

This has to be subject to very special analysis

That is why deep sea submersible are spherical –it is the most stable shape

Large thin tanks are very prone to vacuum collapse

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Large thin tanks are very prone to vacuum collapse

These tank have a very large D/t ratio, which makes them very weak when subjected to vacuum conditions

We learn that the D/t ratio largely determines the ability to withstand even a partial vacuum

A shorter cylinder is also better to withstand a vacuum condition

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A shorter cylinder is also better to withstand a vacuum conditionA long cylinder can be made shorter by adding a vacuum ring

Now we have learned two important facts: A large D/t ratio makes a cylinder weaker A large L/D ratio makes a cylinder weaker

In every pressure vessel code, these ratios are importantFor internal pressure, the Pressure it can take can be predicted:

P = 2.S.tD

That formula does not work for cylinders subject to external pressure:

Because there are bending as well as membrane stress present

Theoretical work has been done on cylinders subject to vacuum

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Theoretical work has been done on cylinders subject to vacuum

Cylinder subjected suffer Lobing as the pressure increasesIncreasing external pressure

No PressureNo lobes

More Pressure2 lobes




In practice, this is not so predictable

To withstand external pressure two metal characteristics are important Young’s Modulus E Yield strength SY

These characteristics are not important for internal pressure

We look a little more closely to the Effective Length of a cylinder

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We look a little more closely to the Effective Length of a cylinderThis is the effective length of the cylinder as it stands alone

L

However, when heads are added, the effective length changes

L

Effective length exists between points of support

Now, suppose we add a vacuum stiffening ring

There is now another point of support

Making the effective length shorter

L

Consider a Cone instead of a head

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Consider a Cone instead of a head

This now becomes the Effective Length

L

Why is the point of support not here ?

It has to do with the Shell to Cone junction

We now take a short detour to discuss the cone junction

Cone Junction Analysis

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First, we consider the basic principles

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First, we consider the basic principlesFirst, we consider the basic principles

Apply internal pressure – see what the cone wants to do

P

The cone wants to separate from the cylinder

It cannot because it is welded to the cylinder here

This is what it does instead


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First, we consider the basic principlesFirst, we consider the basic principles

Apply internal pressure – see what the cone wants to do

The cone wants to separate from the cylinder

It cannot because it is welded to the cylinder here

This is what it does instead

P

Notice the movement

Let us examine the forces that are acting in this region

Consider the free body diagram (to see the forces that are acting)


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Consider the free body diagram (to see the forces that are acting)

Treating this point as a hinge: Resolving components

The cylinder must have the reaction forces – here they are

This is the force causing the problems

There is a compressive hoop stress here

Let us take another view of the forces acting on the junction


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Let us take another view of the forces acting on the junction

The forces trying to collapse the junction can now be seen

By analysing a small piece of the junction we can see the forces

There must be a balancing force – here it is

A compressive hoop stress is trying to collapse the junction

F

F

Remember, Stress = Force / Area


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Remember, Stress = Force / Area

F

F

We now consider the Area of the junction , and Effective Area

Excessive stress can be reduced by Increasing the Area

If necessary, we can add a compression ring – increases area

Full details can be seen in: ASME Appendix 1-5 and 1-8

Cone Junction AnalysisConsider a cone subject to internal pressure

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Cone Junction AnalysisConsider a cone subject to internal pressureLet us revisit our illustration of the Cone-Shell junction

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Cone Junction AnalysisLet us revisit our illustration of the Cone-Shell junction

L

This is the situation if the cone-shell junction is not reinforced

If reinforced by self reinforcement or a ring added, this happens

Maybe a ring was required to give sufficient reinforcement

We also have to consider the small end junction of the cone-shell

L L

If the small end is not reinforced, the effective length changes

L

The lengths L are for the cylinders only – not the cone itself

The Cone is treated as a completely separate element

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The Cone is treated as a completely separate elementThe cone is turned into a equivalent cylinder

With transformed dimensions

Le te

Do

, only DO is the same

Recall, these dimensional ratios are important for cylinders subject to external pressure: A large D/t ratio makes a cylinder weaker A large L/D ratio makes a cylinder weaker

Let us see how this works in practice

Let us consider this vessel as an example of the foregoing

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We first consider the design where the cone is not reinforced (demo)

4 516mm

The large cylinder fails under external pressure

When both end of the cone are considered as reinforced (demo)

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When both end of the cone are considered as reinforced (demo)

4 516mm1900mm

The large cylinder withstands the external pressure with a short L

Did you notice the cone junction at the large end of the cone failed ?

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Did you notice the cone junction at the large end of the cone failed ?Reinforcing ring required here

There is another effect when external pressure exists

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There is a compressive axial stress induced in the shell

This does not present a problem, because: The axial stress is half the hoop stress It becomes important when there is a moment present

when there is a wind load or, when there is a seismic load

The moment can produce a problem in the shell

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One side there is an increased compressive stress

Increased compression

Increased tension

Compressive stresses added

together

The combined compressive stress could buckle the shell

Let us look at the concept of Load Cases

Consider what constitutes Load Cases

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Stress from Pressure:

σP = ± P.D2.t

From weight:

σW = - Wπ.D.t

From the moment:

σM = ± 4.Mπ.D.t

The final equation depends on only: Effects from the pressure Effects from the weight Effects from the applied moment

So the final equation is:

σ = ± P.D2.t

- Wπ.D.t ± 4.M

π.D.t

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Stress from Pressure: From weight: From the moment:

Design pressureHydro pressureNo pressureVacuum

Operating weightHydro weightNo weight

Seismic momentWind momentHydro momentNo moment

Any combination can apply, for example

Or perhaps this

From what we have above, there are 48 load cases in all

We can see this in PV Elite (demo)

Thank you for watchingAny questions ?

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vessels subject to external pressure · pdf file© intergraph 2014 presented by: ray...

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