ACKNOWLEGEMENT:

First of all thanks to Allah Almighty who has guided us and made difficult thing easier for us throughout our project and secondly to our respected teachers and dear parents for their help and prayers…

MAIN CONTENTS

CHAPTER NO 1: introduction to pressure vessels1.1 INTRODUCTION.1.2 TYPES OF VESSELS.1.3 STRESSES IN PRESSURE VESSELS.1.4 PRESSURE VESSEL SAFETY1.5 VESSELS IN REFRIGRATION SYSTEM1.6 FACTOR OF SAFETY1.7 STRESS ANALYSIS.1.8 STRESS/FAILURE THEORIES.1.9 FAILURES IN PRESSURE VESSELS.

1.10 LOADINGS. 1.11 ASME SECTION VIII DIVISION 1 1.12 TYPES AND CLASSES OF STRESS.

1.13 DEFINITIONS.1.14 WEIGHTS OF PRESSURE VESSEL COMPONENTS 1.15 DESIGN PRINCIPLES.

CHAPTER NO 2: stresses and their effects 2.1 STRESS 2.2 TYPES OF STRESSES 2.3 TENSOR 2.3 (a) DUAL SPACE 2.3 (b) STRESS ENERGY TENSOR 2.3 (c ) CAUCHY STRESS TENSOR 2.4 FORMS OF DEFORMATION DUE TO APPLIED LOAD 2.5 MECHANICAL FAILURE MODES 2.6 TYPES OF STRUCTURAL ELEMENTS 2.7 TYPES OF CROSS-SECTIONS USED 2.8 TYPES OF SUPPORTS USED 2.9 STATICAL DETERMINANCY 2.10 STRESS STRAIN DIGRAM ULTIMATE TENSILE STRENGTH YIELD STRENGTH FRACTURE STRESS NECKING 2.11 STRESS CONCENTRATION2.12 VON MISSES CRITERION2.13 PLANE STRESS 2.14 PLANE STRAIN2.15 MOHR STRESS CIRCLE

2.16 PRESSURE VESSEL DESIGN MODEL FOR CYLINDERS

CHAPTER NO 3: materials for pressure vessels

ASME pressure vessel codes ASME section viii division 1

CHAPTER NO 4: design of pressure vessel

Summary Diagram Material Properties Shell & Plate design Head design N-1 4`` sch-160 N-2 4`` sch-160 on head N-1&N-2 Flanges N-3 1``NPT6000# H.cplg N-4&N-5 4`` process conn M-1 12``x16`` MWY on shell M-2 12``x16`` MWY on Head Vessel Weight & Volume Lifting Lugs

CHAPTER NO 1

INTRODUCTION TO PRESSURE VESSELS

CONTENTS:-

1.1 INTRODUCTION.1.2 TYPES OF VESSELS.1.3 STRESSES IN PRESSURE VESSELS.1.4 PRESSURE VESSEL SAFETY1.5 VESSELS IN REFRIGRATION SYSTEM1.6 FACTOR OF SAFETY1.7 STRESS ANALYSIS.1.8 STRESS/FAILURE THEORIES.1.9 FAILURES IN PRESSURE VESSELS.

1.10 LOADINGS. 1.11 ASME SECTION VIII DIVISION 1 1.12 TYPES AND CLASSES OF STRESS.

1.13 DEFINITIONS.1.14 WEIGHTS OF PRESSURE VESSEL COMPONENTS 1.15 DESIGN PRINCIPLES.

1.1 INTRODUCTION:-

 What is a pressure vessel?

Definition:

A pressure vessel is any container that has pressure that is different than atmospheric pressure.  Also, any container...vessel that has the potential to pressurize should be considered a pressure vessel.

figure1.1 horizontal pressure vessel

figure 1.2 horizontal pressure vessel

1.2 How many types of pressure vessels are there?

There are many types -

Thinned wall Thick walled

Storage tanks

Transportable Containers

Propane bottles

Gas cylinders

A pressure vessel is a container that holds a liquid, vapor, or gas at a different pressure other than atmospheric pressure at the same elevation.

You could even say that a carbonated soda bottle is a pressure vessel.  In fact, there is over 15-psi of pressure in a high carbonated soda bottle.  We must be practical too.

 

Pressure vessel types?

Thinned wall -

These pressure vessels are the most categorized.  A thinned walled pressure vessel is any cylinder [shell] ratio  that is 10% or less the ratio of the thickness to the diameter.  Another way of saying this is a pressure vessel is thinned walled if the diameter is 10-times or more the thickness.

  t     < 0.1 d           

figure1.3 stress distribution in thin walled pressure vessel

Thick walled -

These pressure vessels are the least common.  A thick walled pressure vessel is any cylinder [shell] ratio  that is 10% or more the ratio of the thickness to the inside diameter.

  t     > 0.1 d           

figure 1.4 thick walled pressure vessel

Storage tanks -

Storage tanks are a category of thin walled pressure vessels except that are typically under 15-psi and are super thin when compared to the ratio above.

figure 1.5 hot water storage tank

Transportable Containers -

These are the most common pressure vessel and potentially the most ignored.  These are mass produced and require testing every 10-years for propane and gas.

Propane bottles - Fork trucks, barbecues, 

Gas cylinders - CO2, O2,...

Other - Containers, gas cans, bubblers,...

1.3 STRESSES  

A pressure vessel has to retain to pressure.  In doing this the pressure applies two types of stresses in a pressure vessel.  They are circumferential and longitudinal.  

 

figure 1.6

What is important to remember is longitudinal stresses are half as much as the circumferential stresses.  Therefore, we can say that longitudinal strength is twice as strong as circumferential strength.  This is only true for illustration purposes.    

FACTOR OF SAFETY:-

Factors of Safety are used because no manufacture can guarantee 100% quality. Every pressure vessel has a factor of safety. A factor of

safety accounts for uncertainties in materials, design,…,and fabrication.

Factory of Safety [FS] = Actual Breaking Strength                                                Load

To believe in that a F.S. makes a PV [Pressure Vessel] safe is DANGEROUS and unwise.  Putting this in another way, a factor of safety compensate for imperfections in the pressure vessel; therefore, every pressure vessel should be treated the same regardless of the factor of safety.

 1.4 PRESSURE VESSEL SAFETY

There are four types of over pressurization devices:

Rupture Disks.  Relief Valves.

Safety Relief Valves. 

Safety Valves.

 

 

What should you expect on an over-pressurization device?

Every device must have a name tag. The name tag must have one of the following ASME symbols ‘UV’ [spring loaded over pressurization device] ‘UD’ [rupture disk]. The name tag will have the set pressure and capacity. The set pressure should never be greater than the pressure vessels MAWP [maximum allowable working pressure].

Here are some special rules to the set pressure-

Relief Valves, Safety Relief Valves, and Safety Valves should be set at or below pressure vessel ASME nameplate MAWP.

Rupture Disks should be set not higher than the PV [pressure vessel] nameplate. Under special circumstances the rupture disk can be set up to 110% of the MAWP. In addition to this special condition, whenever there is a possibility of internal fire in the pressure vessel the rupture disk can be set not higher than 160%, but a pressure vessel engineer will need to be consulted.

Remember There should never be a shut off valve between the PV and over pressurization device.

 

How should the over-pressurization device be installed?

Always in an Upright installation. Installed the over pressurization device a few pipe diameters

away from the PV, but consult the code and manufacture for maximum distance.

Make sure the exhaust discharges safely away, so no one could be injured.

 

TYPES OF OVER PRESSURIZATION DEVICES:-

 Safety Valves - 

Safety valves are strictly for vapor or gas service. The vapor or gas should be relatively clean to ensure continued and successful operation. A typical vapor is steam, an example for gas would be compressed air. These are not meant for liquids. These valves pop open at a set pressure and reset at a lower pressure called blow down.

Safety Relief Valves - 

These valves differ from safety valves in that they are meant to handle fluid streams that have liquids and vapor. These valves pop open at a set pressure and reset [blow down] at a lower pressure [very much like a safety valve].

 Relief Valves - 

Relief valves open at a set pressure and re-close at the same pressure. These devices are suitable for liquid service.

 Rupture Disks - 

Rupture disks are probably the most versatile over- pressurization device. These can only be used once. They are the only device that can be used in conjunction with other over-pressurization devices.

1.5 VESSELS IN INDUSTRIAL REFRIGERATION SYSTEMS

HIGH PRESSURE RECIEVERS:-

Figure High pressure receiver

LOW PRESSURE RECEIVERS:-

figure liquid level maintained in a low pressure receiver

OIL POTS:-

figure OIL POTS WITH THEIR TYPICAL CONNECTIONS.

SEPARATION ENHANCERS:-

The fig (a) shows drawing inlet flows downward and drawing vapor from the top.

(b) installation of a metal mesh for mist elimination.

THERMOSYPHON RECEIVERS:-

figure combination of a thermosyphon receiver with a system receiver [1]

1.6 STRESSES IN PRESSURE VESSELS:-

Structural elements must be considered).

1.7 STRESS ANALYSIS

“STRESS ANALYSIS IS THE DETERMINATION OF THE RELATIONSHIP BETWEEN EXTERNAL FORCES APPLIED TO A VESSEL AND THE CORRESPONDING STRESS.” [2]

1.8 STRESS/FAILURE THEORIES

They are1. MAXIMUM STRESS THEORY.2. MAXIMUM SHEAR STRESS THEORY.

MAXIMUM STRESS THEORY:-

MAXIMUM SHEAR STRESS THEORY:-

GRAPH OF MAXIMUM STRESS THEORYQUADRANT 1: BIAXIAL TENSION.QUADRANT 2: TENSION.QUADRANT 3: BIAXIAL COMPRESSION.QUADRANT 4: COMPRESSION.

GRAPH OF SHEAR STRESS THEORY.

COMPARISON OF TWO THEORIES:-

1.9 FAILURES IN PRESSURE VESSELS

CATEGORIES OF FAILURES:-

TYPES OF FAILURES:-

1.10 LOADINGS

1.11 STRESS ASME CODE , SECTION VIII, DIVISION1 VERSES DIVISION 2:-

1.12 TYPES AND CLASSES OF STRESS

Simultaneously are called stress categories.

TYPES OF STRESS:-

CLASSES OF STRESS:-

1.13 DEFINITIONS

mum design temperature would be the MDMT.

1.14 WEIGHTS OF VESSELS AND IT’S COMPONENTS

1.15 DESIGN PRINCIPLES

DESIGN LOADS:-

the design.

[3]

CHAPTER NO 2

STRESSES AND THEIR EFFECTS

CONTENTS:

2.1 STRESS2.2 TYPES OF STRESSES 2.3 TENSOR 2.3 (a) DUAL SPACE 2.3 (b) STRESS ENERGY TENSOR 2.3 (c ) CAUCHY STRESS TENSOR2.4 FORMS OF DEFORMATION DUE TO APPLIED LOAD2.5 MECHANICAL FAILURE MODES2.6 TYPES OF STRUCTURAL ELEMENTS TYPES OF CROSS-SECTIONS USED2.7 TYPES OF SUPPORTS USED2.8 STATICAL DETERMINANCY 2.9 STRESS STRAIN DIGRAM ULTIMATE TENSILE STRENGTH YIELD STRENGTH FRACTURE STRESS NECKING 2.10 STRESS CONCENTRATION2.11 VON MISSES CRITERION2.12 PLANE STRESS 2.13 PLANE STRAIN2.14 MOHR STRESS CIRCLE2.15 PRESSURE VESSEL DESIGN MODEL FOR CYLINDERS

2.1 STRESS: The concept of stress was introduced by Cauchy around 1822 as;

Stress is a measure of the average amount of force exerted per unit area of the surface on which internal forces act within a deformable body. In other words, it is a measure of the intensity, or internal distribution of the total internal forces acting within a deformable body across imaginary surfaces. These internal forces are produced between the particles in the body as a reaction to external forces applied on the body. External forces are either surface forces or body forces. Because the loaded deformable body is assumed as a continuum, these internal forces are distributed continuously within the volume of the material body, i.e. the stress distribution in the body is expressed as a piecewise continuous function of space coordinates and time.

UNITS:

The SI unit for stress is the pascal (symbol Pa), which is equivalent to one newton (force) per square meter (unit area). The unit for stress is the same as that of pressure, which is also a measure of force per unit area. Engineering quantities are usually measured in Megapascals (MPa) or gigapascals (GPa). In imperial units, stress is expressed in pounds-force per square inch (psi) or kilopounds-force per square inch (ksi). [4]

figure 2.1 Stress in a loaded deformable material body assumed as a continuum. 40 41

The most general definition of stress is;

“ force per unit area”

mathematically;

σ = F / A

A = UNIT AREAF = FORCE

Structural and solid body mechanics are concerned with analyzing the effects of applied loads. These are external to the material of the structure body and result is internal reacting forces. These internal reacting forces are termed as stresses, together with the deformations are displacements, conforming to the principles of Newtonian mechanics.

Both the analysis and design of a given structure involves the determination of stresses and deformations. [5]

DESCRIPTION:

Equilibrium:

A particle is in the state of equilibrium if the resultant force and moment acting on it is zero.

Hence according to Newton’s law of motion it will have no acceleration and will be at rest. This hypothesis can be extended to the clusters of particles 42 that interact with each other with equal and opposite forces but have no overall resultant. Thus it is evident that solid bodies, structures, or any subdivided part, will be in equilibrium if the resultant of all external forces and moments is zero.

A material body can be acted upon by external forces, which are of two kind: surface forces and body forces. Surface forces or contact forces act on the bounding surface as a result of mechanical contact between bodies, and their intensity is proportional to the area of contact. Body forces, such as gravitational and magnetic forces, are forces distributed over the volume of a

body, and their intensity is proportional to the mass of the body. Surface forces can also occur within internal surfaces of a body.

These acting external forces are then transmitted from point to point within the material body, leading to the generation of internal forces. The transmission of such forces is governed by the conservation laws of linear and angular momenta Newton's Second Law of motion. For bodies in static equilibrium, these laws are related to the principles of equilibrium of forces and moments, respectively.

The measure of the intensity of this internal forces acting within the material body across imaginary surfaces is called stress. In other words, stress is a measure of the average quantity of force exerted per unit area of the surface on which these internal forces act. For example, if we compare a force applied to a small area and a distributed load of the same resulting magnitude applied to a larger area, we find that the effects or intensities of these two forces are locally different because the stresses are not the same.

2.2 TYPES OF STRESSES:

Following are the basic types of stresses;

(i). normal stresses(ii). shearing stresses(iii). bearing stresses

(i) Normal Stresses: “ this types of stresses occurs in the member under axial loading”

normally denoted by “σ”

figure 2.2 Normal stress is the intensity of forces acting perpendicular to infinitely small area dA with and object per unit area. If the normal stress acting on dA pulls on it then it is called as tensile stress whereas if it pushes on the area then it is called as compressive stress.

The plane of a tensile or compressive stress lies perpendicular to the axis of operation of the force from which it originates.

figure2.3 tensile stress

Compressive stress:

“Compressive stress is the stress applied to materials resulting in their compaction (decrease of volume).”

When a material is subjected to compressive stress, then this material is under compression. Usually, compressive stress applied to bars, columns, etc. leads to shortening.

Loading a structural element or a specimen will increase the compressive stress until the reach of compressive strength. According to the properties of the material, failure will occur as yield for materials with ductile behaviour (most metals, some soils and plastics) or as rupture for brittle behaviour (geomaterials, cast iron, glass, etc).

In long, slender structural elements -- such as columns or truss bars -- an increase of compressive force F leads to structural failure due to buckling at lower stress than the compressive strength.

Compressive stress has stress units (force per unit area), usually with negative values to indicate the compaction.

figure2.4 compressive stress

(ii) Shearing Stresses:

“ shearing stresses are caused by the application of equal and opposite transverse forces”

normally denoted by “τ”

figure 2.5

shear stress is applied parallel or tangential to the face of the material as opposed to the normal stress which is applied perpendicularly…

The plane of a shear stress lies in the plane of the force system from which it originates.

figure 2.6 description of planes in tensile compressive and shear stresses

figure 2.7 description of tensile and shear stress

(iii) Bearing Stresses:

“ bearing stresses are created by bolts and pins in the members they connect”

normally denoted by “σ”

σ = P / tdwhere;

P = loadt = thickness of memberd = diameter of pin of bolt

figure 2.7 bearing stress

The applied load divided by the bearing area. Maximum bearing stress is the maximum load in pounds, sustained by the specimen during the test, divided by the original bearing area.

For the two force member under axial loading; stress analysis is done by estimating the normal and shearing stresses in an oblique plane. Secondly ultimate strength of the material is determined and finally by using the factor of safety of any component the allowable load for the structural component is determined.

DISCONTIUITY STRESSES: Discontinuity stresses occurs in case of compound cylinders…compound cylinders are used to increase the range of pressure that can be used inside a cylinder.

Membrane stresses:“ Membrane stress in mechanics means the average stress across the cross section involved”

Thermal stresses:“ Thermal stresses arises in the material when they are heated and cooled”

Principal stresses:“Normal stresses along principal directions are called as principal stresses”

Tangential stresses:“ tangential stresses occurs in the direction perpendicular to the circumference”

Radial stresses:

“ radial stress is a stress towards or away from the central axis of the curved member”

Circumferential stresses:

Longitudinal stresses:

“ Longitudinal stresses occurs along the longitudinal axis”it usually occurs in case of pipe shaped objects [6]

2.3 TENSOR

“An element as a result of tensor product of vector spaces is called as tensor”

Given a finite set { V1, ... , Vn } of vector spaces over a common field F. One may form their tensor product V1 ⊗ ... ⊗ Vn. An element of this tensor product is referred to as a tensor.

figure 2.8 stress tensor

2.3( a) Dual Space:

“In mathematics, any vector space, V, has a corresponding dual vector space (or just dual space for short) consisting of all linear functionals on V”.

Dual vector spaces defined on finite-dimensional vector spaces can be used for defining tensors which are studied in tensor algebra.

2.3 (b) STRESS ENERGY TENSOR:

“The stress-energy tensor (sometimes stress-energy-momentum tensor) is a tensor quantity in physics that describes the density and flux of energy and momentum in spacetime”

figure 2.9 stress energy tensor

2.3 ( c) CAUCHY STRESS TENSOR:

In general, however, the stress is not uniformly distributed over a cross section of a material body, and consequently the stress at a point on a given area is different than the average stress over the entire area. Therefore, it is necessary to define the stress not at a given area but at a specific point in the body .

figure 2.10 a point in an object under stress

According to Cauchy, the stress at any point in an object, assumed to be a continuum, is completely defined by the nine components of a second order tensor known as the Cauchy stress tensor

9 components of a second order tensor

The Cauchy stress tensor obeys the tensor transformation law under a change in the system of coordinates. A graphical representation of this transformation law is the Mohr's circle for stress.

According to the principle of conservation of linear momentum, if the continuum body is in static equilibrium it can be demonstrated that the components of the Cauchy stress tensor in every material point in the body satisfy the equilibrium equations (Cauchy's equations of motion for zero acceleration). At the same time, according to the principle of conservation of angular momentum, equilibrium requires that the summation of moments with respect to an arbitrary point is zero, which leads to the conclusion that the stress tensor is symmetric, thus having only six independent stress components, instead of the original nine.

By continuum body we mean anybody which undergoes gradual transition from one state to another state without abrupt changes example ductile materials..

2.4 FORMS OF DEFORMATION DUE TO APPLIED LOAD

Forces results into four basic forms of deformations or displacement of structures or solid bodies

1. TENSION2. COMPRESSION3. BENDING4. TWISTING OR TORSION

TENSION:

is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object.It is the opposite of compression. As tension is a force, it is measured in newtons (or sometimes pounds-force) and is always measured parallel to the string on which it applies.

COMPRESSION:

Due to applied load there is a decrease in length or volume of the material

It is opposite to tension. It is subjected force applied by one object on another object

Since compression is also expressed in terms of force so it is measured in Newton and pound.

BENDING:

In the force analysis of frame works the members were only subjected to the axial force, namely tension or compression then due to the effect transverse loads acting on the structure, the resulting deformation is called as bending.Bending is very common in structures and machines i.e, floor joists, railway axles, aeroplane wings, leaf springs etc.The external applied loads which cause bending give rise to internal reacting forces.

In engineering mechanics, bending (also known as flexure) characterizes the behavior of a slender structural element subjected to an external load applied perpendicularly to an axis of the element. The structural element is assumed to be such that at least one of its dimensions is a small fraction, typically 1/10 or less, of the other two. When the length is considerably larger than the width and the thickness, the element is called a beam.

TORSION:

In torsion a solid or tubular member is subjected to torque about it’s longitudinal axis resulting in twisting deformation. The engineering examples of the above are obtained in shafts transmitting power in machinery and transport, structural members in aeroplanes, springs.

figure 2.11 torsion

2.5 MECHANICAL FAILURE MODES

Following are the failure modes that commonly exists in mechanical applications.

BUCKLING CORROSION CREEP FATIQUE FRACTURE IMPACT MECHANICAL OVERLOAD RUPTURE THERMAL SHOCK

BUCKLING:

In engineering,

“ buckling is a failure mode characterized by a sudden failure of a structural member subjected to high compressive stresses, where the actual compressive stress at the point of failure is less than the ultimate compressive stresses that the material is capable of withstanding”

This mode of failure is also described as failure due to elastic instability. Mathematical analysis of buckling makes use of an axial load eccentricity that introduces a moment, which does not form part of the primary forces to which the member is subjected.

figure 2. 12 buckling of a column

CORROSION:

“Corrosion can be defined as the disintegration of a material into its constituent atoms due to chemical reactions with its surroundings”

. In the most common use of the word, this means a loss of electrons of metals reacting with water and oxygen. Weakening of iron due to oxidation of the iron atoms is a well-known example of electrochemical corrosion. This is commonly known as rusting. This type of damage typically produces oxide and/or salt of the original metal. Corrosion can also refer to other materials than metals, such as ceramics or polymers. Although in this context, the term degradation is more common.

Most structural alloys corrode merely from exposure to moisture in the air but the process can be strongly affected by exposure to certain substances . Corrosion can be concentrated locally to form a pit or crack, or it can extend across a wide area to produce general deterioration. While some efforts to reduce corrosion merely redirect the damage into less visible, less predictable forms, controlled corrosion treatments such as passivation and chromate-conversion will increase a material's corrosion resistance.

fig 2.13 effect of oxygen concentration on the corrosion of a material in mm

CREEP:

“Creep is the tendency of a solid material to slowly move or deform permanently under the influence of stresses. It occurs as a result of long term exposure to levels of stress that are below the yield strength of the material”

Creep is more severe in materials that are subjected to heat for long periods, and near the melting point.

Creep always increases with temperature.

FACTORS:

The rate of this deformation is a function of the material properties, exposure time, exposure temperature and the applied structural load. Depending on the magnitude of the applied stress and its duration, the deformation may become so large that a component can no longer perform its function

example creep of a turbine blade will cause the blade to contact the casing, resulting in the failure of the blade.

Creep is usually of concern to engineers and metallurgists when evaluating components that operate under high stresses or high temperatures. Creep is a deformation mechanism that may or may not constitute a failure mode. Moderate creep in concrete is sometimes welcomed because it relieves tensile stresses that might otherwise lead to cracking.

Difference from brittle fracture:

Unlike brittle fracture, creep deformation does not occur suddenly upon the application of stress. Instead, strain accumulates as a result of long-term stress.

Creep deformation is "time-dependent" deformation.

Creep deformation is important not only in systems where high temperatures are endured such as nuclear power plants, jet engines and heat exchangers, but also in the design of many everyday objects.

FATIGUE:

In materials science,

“ fatigue is the progressive and localized structural damage that occurs when a material is subjected to cyclic loading”

The maximum stress values are less than the ultimate tensile stress limit, and may be below the yield stress limit of the material.

MECHANICAL OVERLOAD:

“failure or fracture of a product or component in a single event is known as mechanical overload”

It is a common failure mode, and may be contrasted with fatigue, creep, rupture, or stress relaxation. In structural engineering this term is used when analysing product failure. Failure may occur because either the product is weaker than expected owing to a stress concentration, or the applied load is greater than expected and exceeds the normal tensile strength, shear strength or compressive strengthof the product.

Examples include the many components which fail in car crashes, train crashes, and airplane crashes as a result of impact loading. The problem for the investigator is to determine which failures have been caused by the crash, and which may have caused the crash. It usually involves examining the broken parts for signs of fatigue crack growth or other damage to the part which cannot be attributed to the crash itself. For very large structural failures such as the collapse of bridges, it is necessarily a long and tedious process of sifting the broken parts.

THERMAL SHOCK:

“Thermal shock is the name given to cracking as a result of rapid temperature change”.

Glass and ceramic objects are commonly exposed to this form of failure, due to their low toughness, low thermal conductivity, and high thermal expansion coefficients. However, they are used in many high temperature applications due to their high melting point.

Thermal shock occurs when a thermal gradient causes different parts of an object to expand by different amounts. This differential expansion can be understood in terms of stress or of strain, equivalently. At some point, this stress overcomes the strength of the material, causing a crack to form. If nothing stops this crack from propagating through the material, it will cause the object's structure to fail.

Prevention from thermal shock:

Thermal shock can be prevented by:

1. Reducing the thermal gradient seen by the object, by: a) changing its temperature more slowly b) increasing the material's thermal conductivity

2. Reducing the material's coefficient of thermal expansion 3. Increasing its strength 4. Decreasing its Young's modulus 5. Increasing its toughness, by

a) crack tip blunting, i.e., plasticity b) crack deflection [7]

2.6 TYPES OF STRUCTURAL AND SOLID BODY COMPONENTS

1. TIE:

“A member that prevents two parts of a structure from moving apart is subjected to a pull at each end, or tensile force, and is termed as tie”

figure 2.14 tie 2. STRUT:

“A slender member which prevents parts of a structure moving toward each other is under compressive force and is termed as strut”

figure 2.15 strut 3. COLUMN:

“A vertical member which is perhaps not too slender and supports some of the mass of the structure is called as column”Column is capable of supporting axial loads

4. CABLE:

“A cable is generally recognized term for a flexible string under tension which connects two bodies”

It cannot supply resistance to the bending action.

5. BEAM:

“It is supported horizontally and carries transverse loading or vertical loading”

6. CANTILEVER BEAM:

“A common special case of beam is cantilever beam where one end is fixed and provides all the necessary support”

figure 2.16 cantilever beam geometry

7. BEAM COLUMN:

“As the name implies it combines the functions of beam and a column.”

figure 2.17 beam column

ARCH:

“The arch has the same function as the beam or beam-column, but is curved in shape”

the filling and carrying of load over an area or space are achieved by flat slabs or plates by panels and also by shells, which are the curved versions of the former.

8. SHAFT:

“The transmission of torque and twist is achieved through a member which is frequently termed as shaft”

figure 2.18 shaft [8]

\DIFFERENT TYPES OF CROSS- SECTIONS USED:

The members described above can have variety of cross-sectional shapes depending upon the particular type of loading to be carried. Some typical cross sections are;

1. angle

figure 2.19 angle

2. channel

figure 2.20 channel

3. I-section

figure 2.21 I-Section 4. T-section

figure 2.22 T-Section

6. Z-section

figure 2.23

7. Tubes: figure 2.24 [9]

2.7 TYPES OF SUPPORTS USED FOR STRUCTURAL MEMBERS

1. built-in or fixed support

one horizontal and one vertical reaction and in case of welded joint 2- reactions plus one moment in addition

fig 2.25 built on or fixed support

2. pin connection:

one horizontal and one vertical reaction

figure 2.26 pin connection 3. roller support

one vertical reaction

fig 2.27 roller support

4. sliding support

fig 2.28

one horizontal reaction

the applied loading on the structural component is transmitted to the supports which provide the required reacting forces to maintain the overall equilibrium. [10]

The separate members of the structure are joined together by bolting, riveting or welding. If the joints are stiff when the members of the framework were deformed under load, the angles between the members at the joint would not change. This would also imply that the joint is capable of transmitting a couple.

It is found in practice that there is some degree of rotation between members at a joint to the elasticity of the system. For the purposes of calculations, it is assumed that these joints may be represented by a simple ball and socket or pin in a hole. Even with this arrangement, which is of course cannot transmit a couple or bending moment (other than by friction which is ignored), deformation of the members are relatively small. Consequently, changes in angle at the joints are also small. [11]

2.8 STATICAL DETERMINANCY:

If the number of unknown reactions or internal forces in the structure or component is greater than the number of equilibrium equations available, then the problem is said to be statically indeterminate.

Additional equations have to be found by considering the displacement or deformation of the body.

Conditions:

There are three conditions;

Under-stiff: if there are more equilibrium equations than unknown forces or reactions the system is unstable and is not a structure but a mechanism

Just-stiff: this is the statically determinate case for which there are the same number of equilibrium equations as unknown forces. If any member is removed then a part of the whole of the frame will collapse

Over-stiff: this is the statically indeterminate case in which there are more unknown forces than available equilibrium equations. There is at least one member more than is required for the frame to be just stiff

2.9 STRESS STRAIN DIAGRAM:

During testing of a material sample, the stress–strain curve is a graphical representation of the relationship between stress, derived from measuring the load applied on the sample, and strain derived from measuring the deformation of the sample, i.e. elongation, compression, or distortion.

The nature of the curve varies from material to material.

The following diagrams illustrate the stress–strain behaviour of typical materials in terms of the engineering stress and engineering strain where the stress and strain are calculated based on the original dimensions of the sample and not the instantaneous values.

fig 2.29 Stress-strain diagram of ductile material:

Steel generally exhibits a very linear stress–strain relationship up to a well defined yield point . The linear portion of the curve is the elastic region and the slope is the modulus of elasticity or Young's Modulus. After the yield point, the curve typically decreases slightly because of dislocations. As deformation continues, the stress increases on account of strain hardening until it reaches the ultimate strength. Until this point, the cross-sectional area decreases uniformly because of Poisson contractions. The actual rupture point is in the same vertical line as the visual rupture point.

However, beyond this point a neck forms where the local cross-sectional area decreases more quickly than the rest of the sample resulting in an increase in the true stress. On an engineering stress–strain curve this is seen as a decrease in the stress. Conversely, if the curve is plotted in terms of true stress and true strain the stress will continue to rise until failure. Eventually the neck becomes unstable and the specimen ruptures.

2.9 (a) Offset method:

Less ductile materials such as aluminum and medium to high carbon steels do not have a well-defined yield point.

For these materials the yield strength is typically determined by the "offset yield method", by which a line is drawn parallel to the linear elastic portion of the curve and intersecting the abscissa at some arbitrary value (most commonly 0.2%). The intersection of this line and the stress–strain curve is reported as the yield point. Also the yield point is how much pressure and weight a piece of metal can hold before it gets to the elasticity point.

figure 2.30 stress strain diagram for a ductile material

2.9 (b) stress-diagram for brittle material:

Brittle materials such as concrete and carbon fiber do not have a yield point, and do not strain-harden which means that the ultimate strength and breaking strength are the same.. Typical brittle materials like glass do not show any plastic deformation but fail while the deformation is elastic. One of the characteristics of a brittle failure is that the two broken parts can be reassembled to produce the same shape as the original component as there will not be a neck formation like in the case of ductile materials. A typical stress strain curve for a brittle material will be linear.

Testing of several identical specimen, cast iron, or soil, tensile strength is negligible compared to the compressive strength and it is assumed zero for many engineering applications. Glass fibers have a tensile strength stronger than steel, but bulk glass usually does not. This is because of the Stress Intensity Factor associated with defects in the material. As the size of the sample gets larger, the size of defects also grows. In general, the tensile strength of a rope is always less than the tensile strength of its individual fibers.

figure 2.31 stress strain diagram for a brittle material

2.9 ( c ) LINEAR ELASTICITY:

Linear elasticity is the mathematical study of how solid objects deform and become internally stressed due to prescribed loading conditions..

The fundamental "linearizing" assumptions of linear elasticity are: infinitesimal strains or "small" deformations (or strains) and linear relationships between the components of stress and strain. In addition linear elasticity is only valid for stress states that do not produce yielding . These assumptions are reasonable for many engineering materials and engineering design scenarios. Linear elasticity is therefore used extensively in structural analysis and engineering design, often through the aid of finite element analysis.

2.9 (d) YIELDING:

The yield strength or yield point of a material is defined in engineering and materials science as the stress at which a material begins to deform plastically.

Prior to the yield point the material will deform elastically and will return to its original shape when the applied stress is removed. Once the yield point is

passed some fraction of the deformation will be permanent and non-reversible.In the three-dimensional space of the principal stresses (σ1,σ2,σ3), an infinite number of yield points form together a yield surface.

Knowledge of the yield point is vital when designing a component since it generally represents an upper limit to the load that can be applied. It is also important for the control of many materials production techniques such as forging, rolling, or pressing. In structural engineering, this is a soft failure mode which does not normally cause catastrophic failure or ultimate failure unless it accelerates buckling.

fig 2.32

fig 2.33

(e) RESIDUAL STRESSES :

Residual stresses are stresses that remain after the original cause of the stresses (external forces, heat gradient) has been removed.

They remain along a cross section of the component, even without the external cause. Residual stresses occur for a variety of reasons, including inelastic deformations and heat treatment.

Example:

Heat from welding may cause localized expansion, which is taken up during welding by either the molten metal or the placement of parts being welded. When the finished weldment cools, some areas cool and contract more than others, leaving residual stresses.

2.9 (f) STRESS CONCENTRATION:

“A stress concentration (often called stress raisers or stress risers) is a location in an object where stress is concentrated”

. An object is strongest when force is evenly distributed over its area, so a reduction in area, e.g. caused by a crack, results in a localized increase in stress. A material can fail, via a propagating crack, when a concentrated stress exceeds the material's theoretical cohesive strength. The real fracture strength of a material is always lower than the theoretical value because most materials contain small cracks that concentrate stress. Fatigue cracks always start at stress raisers, so removing such defects increases the fatigue strength.

figure 2.34 stress concenctration

2.9 (g) fatigue strength:

In materials science, fatigue is the progressive and localized structural damage that occurs when a material is subjected to cyclic loading. The maximum stress values are less than the ultimate tensile stress limit, and may be below the yield stress limit of the material.

2.9 (h)ultimate tensile strength:

Tensile strength (σUTS or SU ) is indicated by the maxima of a stress-strain curve and, in general, indicates when necking will occur. As it is an intensive property, its value does not depend on the size of the test specimen. It is, however, dependent on the preparation of the specimen and the temperature of the test environment and material.

Tensile strength, along with elastic modulus and corrosion resistance, is an important parameter of engineering materials used in structures and mechanical devices. It is specified for materials such as alloy, composite materials, ceramics, plastics and wood.

figure 2.35 figure showing ultimate tensile strength of a ductile material

2.9 (i) frature stress:

The true normal stress on the minimum • cross-sectional area at the beginning of fracture. In a tensile test, it is the load at fracture divided by the cross-sectional area of the specimen.

figure 2.36

2.9(j) necking:

Necking, in engineering or materials science, is a mode of tensile deformation where relatively large amounts of strain localize disproportionately in a small region of the material

The resulting prominent decrease in local cross-sectional area provides the basis for the name "neck". Because the local strains in the neck are large,

necking is often closely associated with yielding, a form of plastic deformation associated with ductile materials, often metals or polymers

figure 2.37 necking

figure 2.38 stress analysis of ductile material showing necking phenomenon

2.10 OCTAHEDRAL STRESSES:

Considering the principal directions as the coordinate axes, a plane whose normal vector makes equal angles with each of the principal axes is called an octahedral plane. There are a total of eight octahedral planes . The normal and shear components of the stress tensor on these planes are called octahedral normal stress and octahedral shear stress , respectively

Figure 2.39 octahedral stresses

2.11 VON MISSES CRITERION:

The von Mises yield criterion suggests that the yielding of materials begins when the second deviatoric stress invariant reaches a critical value . For this reason, it is sometimes called the -plasticity or flow theory. It is part of a

plasticity theory that applies best to ductile materials, such as metals. Prior to yield, material response is assumed to be elastic.

In material science and engineering the von Mises yield criterion can be also formulated in terms of the von Mises stress or equivalent tensile stress, , a scalar stress value that can be computed from the stress tensor. In this case, a material is said to start yielding when its von Mises stress reaches a critical value known as the yield strength, . The von Mises stress is used to predict yielding of materials under any loading condition from results of simple uniaxial tensile tests. The von Mises stress satisfies the property that two stress states with equal distortion energy have equal von Mises stress.

Figure 2.40 von-misses criterion yield envelope

2.12 PLANE STRESS:

A state of plane stress exist when one of the three principal , stresses is zero.

This usually occurs in structural elements where one dimension is very small compared to the other two, i.e. the element is flat or thin. In this case, the stresses are negligible with respect to the smaller dimension as they are not able to develop within the material and are small compared to the in-plane stresses. Therefore, the face of the element is not acted by loads and the

structural element can be analyzed as two-dimensional, e.g. thin-walled structures such as plates subject to in-plane loading or thin cylinders subject to pressure loading.

Figure 2.41

2.13 PLANE STRAIN:

If one dimension is very large compared to the others, the princpal strain in the direction of the longest dimension is constrained and can be assumed as zero, yielding a plane strain condition.

In this case, though all principal stresses are non-zero, the principal stress in the direction of the longest dimension can be disregarded for calculations. Thus, allowing a two dimensional analysis of stresses, e.g. a dam analyzed at a cross section loaded by the reservoir.

figure 2.42 plane stress and plane strain

2.14 MOHR STRESS CIRCLE:

The Mohr's circle, named after Otto mohr

is a two-dimensional graphical representation of the state of stress at a point.

The abscissa , , and ordinate , , of each point on the circle are the normal stress and shear stress components, respectively, acting on a particular cut plane with a unit vector with components . In other words, the circumference of the circle is the locus of points that represent state of stress on individual planes at all their orientations.

Mohr stress circle for plane stress and plane strain:

The circle represents all possible states of normal and shear stress on any plane through a stressed point in a material.

σ n – ½ ( σ x + σ y ) = 1/ 2 (σ x - σ y ) cos 2θ + τ xy sin 2θ

-τ s = ½ (σ x - σ y ) sin 2θ - τ xy cos 2θ

squaring both sides and adding the equations;

[σ n – ½ ( σ x + σ y ) ] ^2 + τ s ^ 2 = ¼ (σ x - σ y ) ^ 2 + τ xy ^ 2

this the equation of circle of radius

[ 1 /4(σ x - σ y ) ^ 2 + τ xy ^ 2] ^1/2

Figure 2.43

SIGN CONVENTIONS:

The sign conventions used on the circle will be, for normal stress, positive to right and negative to the left of the origin. Shear stresses which might be described as trying to cause a clockwise rotation of an element are plotted above the abscissa axis i.e, “positive” and shear stresses appearing as antclockwise rotation are plotted above the axis i.e, “negative”.Maximum shear stress:

τ smax = [1 /4(σ x - σ y )^2 + τ xy ^ 2 ] ^ 1/2 τ xy

plane on which maximum shear stress acts is calculated by the formula

tan 2θ = - (σ x - σ y / 2 τ xy)

principle stresses and planes:

σ1=( σ x +σ y)/2 + ½ [σ x - σ y )^2 +4τ xy ^ 2] ^1/2

σ 2 =( σ x +σ y)/2 – 1/2[σ x - σ y )^2 +4τ xy ^ 2] ^1/2planes:

θ = ½ tan -1 ( 2τxy/ σx-σy)

maximum shear stress in terms of principal stresses:

τ smax = ½ (σ1 - σ 2)

2.14 (b) mohr stress circle in three dimensional state of stress:

figure 2.44

to obtain the true maximum shear stress for use in design calculations it is necessary to consider all three principal planes. The three dimensional element subjected to the principal stresses is considered. The principal stress σ 3 is zero in this principal case because only plane stress condition is considered. Considering each of the three principal stresses to be labeled as 1, 2 and 3 it is possible to construct the Mohr’s diagram for each. Then the composite Mohr’s diagram is constructed by superimposing these diagrams then enables the maximum shear stress in the material to be determined.

figure 2.45 [12]

2.15

CYLINDER UNDER PRESSURE:

A

equation A;

CASE STUDY : STRESS ANALYSIS OF THICK WALLED CYLINDER:

Figure 7. Mohr's circle for a three-dimensional state of stress

CASE STUDY: THIN WALLED THEROY APPLIED TO CYLINDRICAL ANALYSIS

COMBINED STRESSES IN PRESSURE VESSELS:

THIN-WALLED PRESSURE VESSEL:

thin-walled pressure vessel

A

Cylindrical thin-walled pressure vessel showing co-ordinate axes And cutting planes ( a, b and c)

free body diagram of segment of cylindrical pressure vessel showing pressure and internal hoop stresses

free body diagram of end section of cylindrical thin-walled pressure vessel showing pressure and internal axial stress

SPHERICAL PRESSURE VESSEL:

Spherical pressure vessel can be analyzed in the similar way as the cylindrical pressure vessel. The axial stresses results from the pressure acting on the projected area of the sphere such that;

C

free body diagram of end section of spherical thin-walled pressure vessel showing pressure and internal hoop and axial stresses

Analysis of equation A and C shows that element either cylindrical or spherical is subjected to biaxial stresses ( a normal stress acting in two direction)

THICK WALLED PRESSURE VESSEL:

stress distribution of radial and hoop stresses

elasticity method is used for thick walled pressure vessel it is very difficult only results are displayed

[13]

CHAPTER # 4

DESIGN OF PRESSURE VESSEL

CONTENTS: Summary Diagram Material Properties Shell & Plate design Head design N-1 4`` sch-160 N-2 4`` sch-160 on head N-1&N-2 Flanges N-3 1``NPT6000# H.cplg N-4&N-5 4`` process conn M-1 12``x16`` MWY on shell M-2 12``x16`` MWY on Head Vessel Weight & Volume Lifting Lugs

Pressure Vessel Design Summary:

CustomerVesselPart NumberDrawingJob

Outside Diameter [inch]straight Shell (not including straight flange on heads) Volume [cuft]Fluid (value from Material Properties) Weight Empty [lbs.]Weight FullWeight Under Test

Maximum Allowed Working Pressure

Maximum Design Metal Temperature

Hydrostatic Test

Seismic ZoneFoundation Factor

Primary Material of ConstructionAllowable StressMinimum allowed thickness per UG-16(b) Material NormalizedMaterial Impact Tested (not required per UG-20(f)) Radiography requiredCorrosion Allowance

Code Cases Required

UG-22 Loadings Considered(a) Internal pressure(a) External pressure(b) Vessel weight full, empty and at hydro test (c) Weight of attached equipment and piping (d)(1) Attachment of internals(d)(2) Attachment of vessel supports(d) Cyclic or dynamic reactions(f) Wind(f) Snow(f) Seismic(g) Fluid impact shock reactions(h) Temperature gradients(h) Differential thermal expansion(i) Abnormal pressures like deflagration

YesNoYesNoNoYesNoNoNoYesNoNoNoNo

ASME VIII-1 Code Edition Addenda Materials

2007-

IIDNone

SA-516 Gr.7020,000

0.09375NoNo

None0.125

0.750.3

Maximum Internal pressure, psi

150Maximum External Pressure, psi

0At Temperature, ºF

120Maximum Temperature, ºF

120Minimum Temperature, ºF

-20At Pressure, psi

150Test Pressure, psi

195At a Minimum Temperature of: ºF

AmbientFor a Minimum Duration of:

1/2 hr

60"120"213

Non-lethal6000

1900019000

PVE Sample VesselsSample Vertical Vessel

Sample 4Sample 4Sample 4

Material Properties ver

2.01

1

2

<- Vessel34

Design Pressure UG-

22(a)5

<- P, internal operating pressure at top of vessel (psig)<- mPa, external operation pressure<- Operating Fluid<- h, fluid height (ft)<- rho, fluid density (1.0 for water)

6

7

8

9

10

Design Pressure = P + 0.4331*rho*h

Hydro Test (UG-99(b))

Test Press = P * 1.3 * MR

= 150 + 0.4331 * 1 * 12 mDp =1112

pressure measured at top of vessel, rounded up13

= 150 * 1.3 * 1 mTp =1415

Material Properties

(ASME IID)16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

120 <- mTemp, design temp ºF Test at ambient temp

Material Where Used AmbientStrength

DesignStrength

StrengthRatio

Max ºF ExtGraph

SA-516 70 Plate Shell, Heads 20000 20000 1.000 1000 CS-2

SA-106 B Seamless Pipe Nozzles, Manway 17100 17100 1.000 1000 CS-2

SA-240 316 Plate Flange Pads 20000 20000 1.000 1500 HA-2

SA-105 Forging Flanges, Couplings 20000 20000 1.000 1000 CS-2

SA/CSA-G40.21 44W (38W Stresses used) Legs 17100 17100 1.000 650 CS-2

SA-193 B7 Bolts <= 2.5" 25000 25000 1.000 1000

Min Ratio (MR) = 1.000

195

155.2

150.00.0

Water. fresh12.000

1.000

Sample Vertical Vessel

Pipe and Shell :1

Description2Do

Options:3

4 ip? - Calculate interior pressureep? - Calculate exterior pressurepr? - Pipe or rolled platept? - Type of piperelief? - Stress Relief Calculations Required

5

6

7

t

8

Dimensions:9

10 Do [in] - outside diametert [in] - nominal wall thicknesstminUG16b [in] - minimum wall per UG-16(b)L [in] - length for volume and weightCorr [in] - corrosion allowance

11

12

13

14

Material and Conditions:

15

16 MaterialS [psi] - allowable stress levelEl - longitudinal efficiency (circ. stress)Ec - circ. connecting efficiency (longitudinal stress)UTP [%] - undertolerance allowance UTI [in] - undertolerance allowance P [psi] - interior pressure

17

18

19

20

21

22

Stress Classification:NOTE: Both validity checks need to be "Acceptable" in order to use this sheetIf not, refer to sheet "Thick Cylindrical Shell"

23

24

25

26

27

ckValidity1 =ckValidity2 =

tmin < 0.5*(Do/2)P< 0.385*S*El

0.331 < 0.5*(60/2) =155.2< 0.385*20000*0.7 =

Variables:28

29

30

Td = 0.000UT [in] = t*UTP+UTInt [in] = t-Corr-UT-TdRi [in] = Do/2-nt

Volume [cuft] = ((Do/2-t)^2)* *L/1728Weight [lb] = (Do-t)* *L*t*40.84/144

0 =0.5*0+0 =

0.5-0.125-0-0 =60/2-0.375 =

((60/2-0.5)^2)*3.1416*120/1728 =

0.0000.0000.37529.625

31

32

33

(60-0.5)*3.1416*120*0.5*40.84/144 =34

Interior Pressure: VIII-1 UG-27(c)(1,2)

ta [in] = P*Ri/(S*El-0.6*P)tb [in] = P*Ri/(2*S*Ec+0.4*P)

tmin [in] = MAX(ta,tb,tminUG16b)tr1 [in] = P*Ri/(S*1-0.6*P)

Checkt = tmin <= nt

PMaxA [psi] = (S*El*nt)/(Ri+0.6*nt)PMaxB [psi] = (2*S*Ec*nt)/(Ri-0.4*nt)

PMax [psi] = Min(PMaxA,PMaxB) CheckP = PMax >= P

35

155.2*29.625/(20000*0.7-0.6*155.2) =155.2*29.625/(2*20000*0.7+0.4*155.2) =

MAX(0.331,0.164,0.094) =155.2*29.625/(20000*1-0.6*155.2) =

0.331 <= 0.375 =

(20000*0.7*0.375)/(29.625+0.6*0.375) = (2*20000*0.7*0.375)/(29.625-0.4*0.375) =

MIN(176,356) =176 >= 155.2 =

0.3310.164

36

37

38

0.231Acceptable

176356

39

40

41

42

43

44 Acceptable

Lo

ng

Se

am

Le

ng

th

176

0.331

189.859

3180.84

AcceptableAcceptable

SA-516 7020,000

0.700.70

0.000%0.000

155.20

60.0000.5000

0.094120.000

0.125

InteriorNo Exterior

Rolled PlateNon-Threaded

No

Rolled Plate Shell

Heads : Torispherical39

40

<- Vessel<- Desc

22

4243

Dimensions:44

<- Do, outside diameter<- L, inside crown radius<- IKR, inside knuckle radius<- tb, thickness before forming<- tf, thickness after forming<- tminUG16(b) - Min.t. Per UG-16(b)<- Corr, corrosion allowance<- Skirt, straight skirt length

45

48

49

51

53

54

55

5657

Material and Conditions:

58

<- material<- S, allowable stress level (psi)<- E, efficiency<- P, interior pressure<- Pa, exterior pressure

59

60

61

65

6669

Calculated Properties:70

<- Approx. blank dia (inch)<- Approx. weight (lbs, steel)

<- Volume (cuft, includes skirt)<- Spherical Limit<- Depth of Head

74

75

7678

Variables:115

D =t =

L /r = M =

Ro =

Do-2*ttf-corrL/IKR0.25*(3+sqrt(L/ikr)) L + tb

= 60-2*0.55= 0.675-0.125= 60/3.6= 0.25*(3+sqrt(60/3.6))= 60 + 0.75

D =t =

L /r = M =

Ro =

58.900.5516.6671.77160.750

116

123

125

126

128131

Interior Pressure App 1-4(a), App 1-4(d):134

App. 1-4(a) check: 0.0005 =< tf/L < 0.002 = 0.0005=<0.675/60<0.002 tf/L =137

App. 1-4(f) calculation not requiredIF(tf/L<0.002,IF(tf/L>=0.0005,"Calculation required","Error"),"Calculation not required")138

TMinI ==

TMin = PMax =

=

(P*L*M)/(2*S*E - 0.2*P) <= t(155.197*60*1.771)/(2*20000*0.85 - 0.2*155.197) <= 0.55

TminI =141

142

Max(Tminl,tminUG16(b))<=tf-corr(2*S*E*t)/(L*M + 0.2*t) >= P (2*20000*0.85*0.55)/(60*1.771 + 0.2*0.55) >= 155.197

AcceptableAcceptable

TMin =PMax =

146

149

150153

Interior Pressure for Nozzles App 1-4(a), App 1-4(d), UG-37(a)(1):

157

TMinE1 ==

TSp ==

(P*L*M)/(2*S*1 - 0.2*P) <= t (Nozzle in Knuckle) TMinE1 =

158

(155.197*60*1.771)/(2*20000*1 - 0.2*155.197) <= 0.55(P*L*1)/(2*S*1 - 0.2*P) (155.197*60*1)/(2*20000*1 - 0.2*155.197)

159

(Nozzle in Crown) Tsp =

165

166167

Head stress relief UCS-79(d), UNF-79(d), UHA-44(d)

177

Rf = IKR+tb/2% elong = ((75*tb)/Rf)*(1-Rf/Ro)

= 3.6+0.75/2= ((75*0.75)/3.975)*(1)

Rf = 3.975180

% elongation =181

<- Max Elongation<- Cold Formed

184

14.2% <- Elongation185

<- Vessel carries lethal substances(Yes/no)<- Impact testing is required (Yes/no)<- Formed between 250 and 900 Degrees F

no <- Greater than 10% reduction in thicknessYes <- Head is greater than 5/8" thick before forming

186

187

188

189

190

Stress Relieve ?191

Required Yes ?no nono nono nono noYes ? YES

YES

5.0%Yes

nonono

14.2

0.233

0.413

0.485175.8

0.485

0.0113

11.6654.574

9.770153

68.7789.5

SA-516 7020,000

0.85155.2

0.0

60.00060.000

3.6000.7500.6750.0940.1251.500

Sample Vertical VesselF & D Heads

Nozzle Reinforcement UW16(c) <- SavedDesign

30

Automatic dh - not hillside Automatic Limit Diameter

Curved Shell or Head Section

31

<- Vessel<- Description

22

33

Shell:34

<- Shell Material<- Sv, shell allowable stress level, PSI<- E1, efficiency of shell at nozzle<- Ds, Shell ID<- Vt, shell wall thick, uncorroded, UT removed<- tr, required shell wall thickness int. press.(E=1)<- trE, required shell wall thickness ext. press.(E=1)<- sca, shell corrosion allowance<- tmin16b, Min allowed wall per UG-16(b)

35

Do36

37

Nt38

39

40

41

42

43

Nozzle:44

<- Nozzle Material<- Sn, allowable stress level (Sn)

45

Vt46

<- B, from A =<- E, nozzle efficiency

47

48

<- P, internal design pressure<- Pa, external design pressure<- Do, outside diameter<- Nt, wall thick, uncorroded<- UTp, undertolerance (%)<- nca, nozzle corrosion allowance<- L, exterior Projection

49

UW-16.1 (c)50

51

54

55

57

58

Reinforcing:61

<- Leg41, size of weld fillet<- F

71

74

Variables:87

UT = Rn =

t =tn = d =

fr1 = fr2 =

tcLeg41 = F =

Nt*UTpDo/2 - (Nt-nca) + UT Vt-scaNt-nca Do-2*tn MIN(Sn/Sv,1) MIN(Sn/Sv,1)Min(0.25,0.7*Min(0.75,tn,t)) Min(Fenterered, 1)

= 0.531 * 0.125= 4.5/2 - (0.531-0.125) + 0.066= 0.5 - 0.125= 0.531-0.125= 4.5 - 2*0.406= MIN(17100/20000, 1)= MIN(17100/20000, 1)= Min(0.25,0.7*Min(0.75,0.406,0.375))

UT = Rn =

t =tn = d =

fr1 = fr2 =

tc41 = F =

0.0661.9100.3750.4063.6880.8550.8550.2501.000

Undertolerance

Effective Radius

Effective Shell Thickness

Avail. Nozzle Thick. No UT

Opening Dia.

88

90

95

101

102

108

111

127

133

Pipe Required Wall Thickness - trn from internal, trnE from external pressure

141

LDo = L/Do LDo = 1.333 Dot == (155.2*1.91)/(17100*1 - 0.6*155.2)= (155.2*1.91)/(17100*1 - 0.6*155.2)= (3*4.5*0)/(4*17600)

Do/trnE Dot = trn =

trnR = trnE =

0.000142

trn = (P*Rn)/(Sn*E - 0.6*P) <= tn-UT

trnR = (P*Rn)/(Sn*1 - 0.6*P) trnE = (3*Do*Pa)/(4*B) <= tn-ut

Geometry Constraints:

Acceptable143

E=1145

Acceptable146

148

0.7*Leg41 >= tc41Appendix 1-7 Necessary Check

0.7*0.375 >= 0.25 0.263 >= 0.250 Acceptable149

180

when Ds>60,if(2*Rn<=Ds/3,if(2*Rn<=40, "App. 1-7 calculations not required","App. 1-7 calculations required"),"App. 1-7 calculations required")when Ds<=60,if(2*Rn<Ds/2,if(2*Rn<20,"App. 1-7 calculations not required","App. 1-7 calculations required"),"App. 1-7 calculations required")

App. 1-7 calculations not required

181

182

183

Area Replacement: Fig UG-37.1

Pressure From: Internal External207

A ==

Ae = A1 =

= A1e =

= A2 =

= A2e =

= A41 =

1.0*d*tr*F + 2*tn*tr*F*(1-fr1)1.0*3.688*0.231*1 + 2*0.406*0.231*1*(1-0.855)

A Required (internal) =

208

209

0.5*(d*trE*1 + 2*tn*trE*1*(1-fr1))

= 0.5*(3.688*0*1 + 2*0.406*0*1*(1-0.855))

A Requi212

max(d, 2*(t+tn)) * (E1*t-F*tr)-2*tn*(E1*t-F*tr)*(1-fr1)max(3.688,2*(0.375+0.406))* (1*0.375-1*0.231)-2*0.406*(1*0.375-1*0.231)*(1-0.855)max(d, 2*(t+tn)) * (E1*t-F*trE)-2*tn*(E1*t-F*trE)*(1-fr1) max(3.688,2*(0.375+0.406))* (1*0.375-1*0)-2*0.406*(1*0.375-1*0)*(1-0.855) min((tn-trnR)*fr2*Min(5*t,2*L) , (tn-trnR)*fr2*Min(5*tn,2*L))min((0.406-0.017)*0.855*Min(5*0.375,2*6) , (0.406-0.017)*0.855*Min(5*0.406,2*6))min((tn-trnE)*fr2*Min(5*t,2*L) , (tn-trnE)*fr2*Min(5*tn,2*L))min((0.406-0)*0.855*Min(5*0.375,2*6) , (0.406-0)*0.855*Min(5*0.406,2*6))

A1 =215

216

A1e =219

220

A2 =225

226

A2e =230

231

Leg41^2*fr2 = 0.375^2*0.855

A41 = Actual Area =

Actual-Required =

Tstd =

240

249

Acceptable250

Tstd = Swre = Nact =

Tt =

Standard pipe wall thickness from chart

0.2370.0000.4650.000Acceptable

331

tr * Pa / PNt * (1-UTp)0.8/Nth

= 0.231 * 0 / 155.197= 0.531 * (1-0.125)= 0.8/0

Swre = Nact =

Tt =

Req. Exterior pressure

Actual Wall Thick.

Ug-31(c)(2) threads

332

333

334

UG-45335

UG45 = UG45a = UG45b =

UG45b1 = UG45b2 = UG45b3 = UG45b4 =

Max(UG45a, UG45b) <= Nact Max(trn,trnE) + Nca + Tt Min(UG45b3,UG45b4)Max(tr + Sca, tmin16b + Sca) Max(Swre + Sca,tmin16b + Sca) Max(UG45b1,UG45b2) Tstd*0.875 + Nca

= Max(0.142, 0.332) <= 0.465= Max(0.017,0) + 0.125 + 0= Min(0.356, 0.332)= Max(0.231 + 0.125, 0.094 + 0.125)= Max(0 + 0.125,0.094 + 0.125)= Max(0.356,)= 0.237*0.875 + 0.125

UG45 = UG45a = UG45b = UG45b1 = UG45b2 = UG45b3 = UG45b4 =

336

337

338

0.356339

340

0.356341

0.332342

Nozzle

0.3320.1420.332

red (external) = 0.0000.514

0.623

0.120

1.339

0.651

0.1201.257 2.1100.378 2.110

0.879

0.0170.0170.000

0.3751.000

0.09590

Leg41

Leg41 SA-106B

17,10017,600

1.00155.20

0.04.5000.531

12.5%0.1256.000

t

SA-516 7020,000

1.0059.000.5000.2310.0000.1250.094

Sample Vertical VesselN1 - 4" SCH 160 Pipe

Nozzle Reinforcement 30

Automatic dh - not hillside Automatic Limit Diameter

Curved Shell or Head Section

31

<- Vessel<- Description

22

33

Shell:34

<- Shell Material<- Sv, shell allowable stress level, PSI<- E1, efficiency of shell at nozzle<- Vt, shell wall thick, uncorroded, UT removed<- tr, required shell wall thickness int. press.(E=1)<- trE, required shell wall thickness ext. press.(E=1)<- sca, shell corrosion allowance<- tmin16b, Min allowed wall per UG-16(b)

35

Do36

37

Nt39

40

41

42

43

Nozzle:44

<- Nozzle Material<- Sn, allowable stress level (Sn)

45

46

Vt<- B, from A =<- E, nozzle efficiency

47

48

<- P, internal design pressure<- Pa, external design pressure<- Do, outside diameter<- Nt, wall thick, uncorroded<- UTp, undertolerance (%)<- nca, nozzle corrosion allowance<- L, exterior Projection

49

50

UW-16.1 (c)51

54

55

57

58

Reinforcing:61

<- Leg41, size of weld fillet<- F

71

74

Variables:87

UT = Rn =

t =tn = d =

fr1 = fr2 =

tcLeg41 = F =

Nt*UTpDo/2 - (Nt-nca) + UT Vt-scaNt-nca Do-2*tn MIN(Sn/Sv,1) MIN(Sn/Sv,1)Min(0.25,0.7*Min(0.75,tn,t)) Min(Fenterered, 1)

= 0.531 * 0.125= 4.5/2 - (0.531-0.125) + 0.066= 0.5 - 0.125= 0.531-0.125= 4.5 - 2*0.406= MIN(17100/20000, 1)= MIN(17100/20000, 1)= Min(0.25,0.7*Min(0.75,0.406,0.375))

UT = Rn =

t =tn = d =

fr1 = fr2 =

tc41 = F =

0.0661.9100.3750.4063.6880.8550.8550.2501.000

Undertolerance

Effective Radius

Effective Shell Thickness

Avail. Nozzle Thick. No UT

Opening Dia.

88

90

95

101

102

108

111

127

133

Pipe Required Wall Thickness - trn from internal, trnE from external pressureLDo = 1.333

141

LDo = trn =

trnR = trnE =

L/Do Dot = Do/trnE Dot = trn =

trnR = trnE =

0.000142

= (155.2*1.91)/(17100*1 - 0.6*155.2)= (155.2*1.91)/(17100*1 - 0.6*155.2)= (3*4.5*0)/(4*17600)

Acceptable(P*Rn)/(Sn*E - 0.6*P) <= tn-UT

(P*Rn)/(Sn*1 - 0.6*P) (3*Do*Pa)/(4*B) <= tn-ut

143

E=1145

Acceptable146

Geometry Constraints:0.7*Leg41 >= tc41

Area Replacement: Fig UG-37.1

148

0.7*0.375 >= 0.25 0.263 >= 0.250Internal

AcceptableExternal

149

Pressure From:A Required (internal) =

207

A ==

Ae = A1 =

= A1e =

= A2 =

= A2e =

= A41 =

1.0*d*tr*F + 2*tn*tr*F*(1-fr1)1.0*3.688*0.233*1 + 2*0.406*0.233*1*(1-0.855)

208

209

0.5*(d*trE*1 + 2*tn*trE*1*(1-fr1)) = 0.5*(3.688*0*1 + 2*0.406*0*1*(1-0.855)) A Requi

A1 =212

max(d, 2*(t+tn)) * (E1*t-F*tr)-2*tn*(E1*t-F*tr)*(1-fr1)max(3.688,2*(0.375+0.406))* (1*0.375-1*0.233)-2*0.406*(1*0.375-1*0.233)*(1-0.855)max(d, 2*(t+tn)) * (E1*t-F*trE)-2*tn*(E1*t-F*trE)*(1-fr1) max(3.688,2*(0.375+0.406))* (1*0.375-1*0)-2*0.406*(1*0.375-1*0)*(1-0.855) min((tn-trnR)*fr2*Min(5*t,2*L) , (tn-trnR)*fr2*Min(5*tn,2*L))min((0.406-0.017)*0.855*Min(5*0.375,2*6) , (0.406-0.017)*0.855*Min(5*0.406,2*6))min((tn-trnE)*fr2*Min(5*t,2*L) , (tn-trnE)*fr2*Min(5*tn,2*L))min((0.406-0)*0.855*Min(5*0.375,2*6) , (0.406-0)*0.855*Min(5*0.406,2*6))

215

216

A1e =219

220

A2 =225

226

A2e =230

231

Leg41^2*fr2 = 0.375^2*0.855 A41 = Actual Area =

Actual-Required =

Tstd =

240

249

Acceptable250

Tstd = Swre = Nact =

Tt =

Standard pipe wall thickness from chart 0.2370.0000.4650.000Acceptable

331

tr * Pa / PNt * (1-UTp)0.8/Nth

= 0.233 * 0 / 155.197= 0.531 * (1-0.125)= 0.8/0

Swre = Nact =

Tt =

Req. Exterior pressure

Actual Wall Thick.

Ug-31(c)(2) threads

332

333

334

UG-45335

UG45 = UG45a = UG45b =

UG45b1 = UG45b2 = UG45b3 = UG45b4 =

Max(UG45a, UG45b) <= Nact Max(trn,trnE) + Nca + Tt Min(UG45b3,UG45b4)Max(tr + Sca, tmin16b + Sca) Max(Swre + Sca,tmin16b + Sca) Max(UG45b1,UG45b2) Tstd*0.875 + Nca

= Max(0.142, 0.332) <= 0.465= Max(0.017,0) + 0.125 + 0= Min(0.358, 0.332)= Max(0.233 + 0.125, 0.094 + 0.125)= Max(0 + 0.125,0.094 + 0.125)= Max(0.358,)= 0.237*0.875 + 0.125

UG45 = UG45a = UG45b = UG45b1 = UG45b2 = UG45b3 = UG45b4 =

336

337

338

0.358339

340

0.358341

0.332342

No

zzle

0.3320.1420.332

red (external) = 0.0000.507

0.623

0.120

1.339

0.651

0.1201.250 2.1100.364 2.110

0.887

0.0170.0170.000

0.3751.000

0.09590

Leg41

Leg41 SA-106B17,10017,600

1.00155.20

0.04.5000.531

12.5%0.1256.000

t

SA-516 7020,000

1.000.5000.2330.0000.1250.094

Sample Vertical VesselN2 - 4" SCH 160 Pipe on Bot. Head

18 B16.5/16.47 Flange 1920

SlipOn

<- Vessel<- Description

21

22

23

24 Select Flange<- Category<- Material Type<- Material<- Pressure Class<- Nominal Size

25

26

27

28

29

31

Nominal -Table -

Max Temp ºF - Pod, pipe OD -

C-Si2-1.110004.500

32

33

34

35

36 Nozzle<- tn, Nozzle Wall Thickness (inch)<- tnr, Required Nozzle Wall Thickness (inch)

37

38

39

40 Operating Conditions Acceptable<- T, temperature ºF<- P, pressure, psig<- Corr, corrosion allowance

Max press @ 100 ºF [p1]Max press @ 120 ºF [p2]

41

42

43

44

45 Flange Welds:<- F1, pipe fillet size<- F2, flange fillet size F2<- Sp, allowable stress, pipe<- Sf, allowable stress, flange

47

48

49

50

53

54 Geometry constraint: VIII UW-21 (b)wtmin = 0.7*tn

wt = 0.7*MIN(F1,F2)= 0.7*MIN(0.531,0.531)

= 0.7*0.531 wtmin = 0.372Req. weld throat

Actual weld throat

61

wt =63

Acceptable6467

68 Weld Strength:Min Sa =

Max Weld Stress = Weld Load = Weld Area =

= Weld Stress =

MIN(Sp,Sf)Sa * 0.49POD^2*pi*P/4Pod*pi*(F1-corr + F2)

= MIN(17100,20000)= 17100 * 0.49= 4.5^2*pi*155.197/4

Min Sa =Max S =

Load = Area =

17,10069

70

2,46813.247

72

74

4.5*pi*(0.531-0.125 + 0.531)75

Load/Area = 2468.305/13.247 Stress =78

Acceptable79

81

82

83

186

8,379

0.372

0.5310.531

1710020000

285280

120155.20.125

0.5310.017

SAForgedSA 105

1504.00

Sample Vertical VesselN1 & N2 - 4" Class 150 RFSO

Coupling UW16.1Z1M15

16

<- Vessel<- Description

22

18

CODShell:20

FULL PEN.

<- t, Shell Wall Thick (inch)<- tMin, Min Required Wall at E=1 (inch)<- D, Shell Opening Diameter (inch)<- P,design Pressure (psi)

POD

23

24

26 Outside28

29 tt1Coupling:30

Inside Vessel<- Coupling

<- Coupling Material<- Sn, Allowable Stress Level (Sn)<- F1, Weld Size<- tmin16b, Min allowed wall per UG-16(b)<- Corrc, Coupling Corrosion Allowance (inch)

31

32D

33

35

UW-16.1 (Z-1) (Modified)

37

38

2.2501.315

11.5000.358

12.5%

<- COD - Coupling OD<- POD - Pipe OD<- n, Threads Per Inch<- pt, Corresponding sch XXS Wall Thickness (inch)<- UT, Under Tolerence (%)

39

40

42

44

46

47

Geometry Restrictions Fig. UW-16.149

tcp =Tmin = tcmin =

t1 =t1 > =

(COD-POD)/2-CORRCMin(0.75,tcp,t) Min(0.25,0.7*Tmin)0.7*F1 tcMin

= (2.25-1.315)/2-0.125= Min(0.75,0.343,0.675)= Min(0.25,0.7*0.343)= 0.7*0.375= 0.263 >= 0.24

Tcp =Tmin = tcmin =

t1 =

0.3430.3430.2400.263Acceptable

50

51

53

56

64

74

Required Coupling Wall Thickness UG-44(c), B16.11 - 2.1.1 and UG-31(c)(2)= 1.315/2-0.8/11.5= (1-0.125)*0.358-0.125-0.8/11.5= 155*0.588/(20000*1+0.4*155. Acceptable

75

Ro =tp =

Min Thick =

POD/2-0.8/n(1-UT)*pt-Corrc-0.8/nP*Ro/(Sn*1+0.4*P)

Ro =tp =

trn =

0.5880.119

76

77

78

79

Pressure Weld Stress UW-18(d) - Pressure Load only UW-16(f)(3)(a)(3)(b)80

Load =Weld Area =

= Max Stress =

Weld Stress =

COD^2*(PI()/4)*P = 2.25^2*(PI()/4)*155.197 Load =Weld Area =

6171.436

81

pi()*((COD+F1)^2-COD^2)/4pi()*((2.25+0.375)^2-2.25^2)/4

82

83

Min(Sn,Sv) * 0.55Load / Area

= Min(20000,0) * 0.55= 617 / 1.436

Max Stress =Weld Stress =

1100088

89

Acceptable90

UG-4595

Tstd =Nact =

Tt = UG45 =

= UG45a =

Standard pipe wall thickness from chartPt * (1-UT)0.8/nMax(UG45a, UG45b) <= Nact Max(0.199, 0.241) <= 0.313 trn + corrc + Tt0.005 + 0.125 + 0.07Min(UG45b1, UG45b4) Min(0.358, 0.241)Max(tmin+ CORRC, Tmin16b + CORRC) Max(0.233 + 0.125, 0.094 + 0.125)

Tstd =Nact =

Tt =UG45 =

0.1330.3130.070

96

Actual Wall Thick.

Ug-31(c)(2) threads

97

98

99

Acceptable100

UG45a =101

102

UB45b ==

UG45b1 ==

UG45b4 =

UB45b =103

104

UG45b1 = 0.358105

106

Tstd*0.875 + corrc = 0.133*0.875 + 0.125 UG45b4 = 0.241107

108

109

0.241

0.199

0.241

430

0.005

1 inch 6000#SA-10520,0000.3750

0.0940.125

tF1

0.6750.2331.875155.2

Sample Vertical VesselN3 - 1" Class 6000 NPT Half Coupling

Nozzle Reinforcement UG40(a-2) <- SavedDesign 30

Automatic dh - not hillside Automatic Limit Diameter

Curved Shell or Head Section

31

<- Vessel<- Description

22

33

Shell:34

<- Shell Material<- Sv, shell allowable stress level, PSI<- E1, efficiency of shell at nozzle<- Vt, shell wall thick, uncorroded, UT removed<- tr, required shell wall thickness int. press.(E=1)<- trE, required shell wall thickness ext. press.(E=1)

35

Full36 Penn.37

39

40

41

<- sca, shell corrosion allowance<- P, internal design pressure<- nca, nozzle corrosion allowance

Vtdi42

49

57

UG-40 (a-2)Flange Pad:62

<- Flange Pad Material<- Sp, allowable stress level<- Dp, outside diameter<- di, inside (uncorroded)<- tp, pad thick<- Leg42, size of weld fillet<- F<- GOD - gasket OD<- GID - gasket ID<- m - gasket factor<- gy - gasket factor y<- varC - bolt circle dia<- BoltOD, bolt size<- Nbolt, number of bolts<- DepthT, depth of bolt holes<- Sb - allowable bolt stress at DESIGN temp<- Sba - allowable bolt stress at ASSEMBLY temp

64

65

66

67

70

72

74

77

78

79

80

81

82

83

84

85

86

Variables:87

Dp = t =

te = d =

fr2 = fr4 =

varN = b0 =

varb = varG =

= Ro =

tcLeg42 = F =

Min(2*d,DpEntered) Vt-scatp-Vt di+2*nca MIN(Sp/Sv,1) MIN(Sp/Sv,1) (GOD-GID)/2 varN / 2min(Sqrt(b0)/2,b0)

= Min(2*4.75,9)= 0.5 - 0.125= 1.5-0.5= 4.5 - 2 * 0.125= MIN(20000/20000, 1)= MIN(20000/20000,1)= (5.5-4.5)/2= 0.5 / 2= min(Sqrt(0.25)/2,0.25)

Dp = t = te = d =

fr2 = fr4 =

varN = b0 =

varb = varG =

9.0000.3751.0004.7501.0001.0000.5000.2500.2505.000

Effective Reinforcing

Effective Shell Thickness

Effective Reinf. Thick.

Finished Opening Dia.

91

95

98

104

112

115

Gasket Width in Contact

gasket seating width

eff seating width

gasket load reaction diameter

119

120

121

max(GOD-2*varb,(GOD-GID)/2 + GID)max(5.5-2*0.25,(5.5-4.5)/2 + 4.5)

122

123

Dp/2Min(0.25,0.7*MIN(0.75,te,t)) Min(Fenterered, 1)

= 9/2= Min(0.25,0.7*MIN(0.75,1,0.375))

Ro = tc42 =

F =

4.5000.2501.000

124

130

133

Geometry Constraints:148

0.7*Leg42 >=Bolt Loads:

H = HP = HD = HT =

Wm1 = Wm2 =

Am = Ab =

tc42 0.7*0.5 >= 0.25 0.350 >= 0.250 Acceptable156

185

0.785*varG^2*P2*varb*3.14*varG*m*Ppi/4 * di^2 * P H - HDH + HP pi*varb*varG*gy max(Wm1/Sb, Wm2/Sa) Root*Nbolt >= Am

= 0.785*5^2*155.197= 2*0.25*3.14*5*3*155.197= pi/4 * 4.5^2 * 155.197= 3046 - 2468= 3046 + 3655= pi*0.25*5*1800= max(6701/25000, 7069/25000)= 0.207*8

H = HP = HD = HT =

Wm1 = Wm2 =

Am = Ab =

Internal

30463655246857767017069

end load

contact load

end load

face load

bolt load

seating load

req bolt area

Acceptable Pressure From:

A Required (internal) =A Required (external) =

A1 = A1e = A5 =

A42 =Actual Area =

199

200

201

202

203

204

205

1.656External

206

Area Replacement: Fig UG-37.1207

A = Ae = A1 =

A1e = A5 =

A42 =

1.0*d*tr*F0.5*d*trE*1 (d)* (E1*t-F*tr)(d) * (Eone*t-F*trE)((Dp - d)te-BoltOD*DepthT*2)*fr4Leg42^2*fr2

= 1.0*4.75*0.231*1= 0.5*4.75*0*1= (4.75) * (1*0.38-1*0.23)

= (4.75) * (1*0.375-1*0)

= ((9-4.75)*1-0.625*1*2)*1= 0.5^2*1

210

213

217

221

237

245

249

Acceptable Actual-Required =250

1.0970.000

0.684

3.0000.250

1.7813.0000.250

3.934 5.0312.895 5.031

0.283

SA-240 31620,000

9.0004.5001.5000.5001.0005.5004.5003.0001,8007.5000.625

81.000

25,00025,000

Dp

F

Leg42 Shelltp

tLeg42

SA-516 7020,000

1.000.5000.2310.0000.125

155.200.125

Sample Vertical VesselN4 & N5 - 4" Double Sided Flange Pad

Flange ASME VIII Div I Appendix 2

1

2

3 DescriptionhGDimensions:4

5

6

Wfd? - Select a flange designA [in] - flange ODBn [in] - ID, uncorrodedt [in] - flange thicknesstn [in] - nozzle wall thickness

B7

8

9

Gasket: G10

11 GOD [in] - gasket ODGID [in] - gasket IDm - gasket factorgy - gasket factor y

12

13

14

A

Fig 2-13.2 Modified

Bolting:15

16

17

varC [in] - bolt circle diaBoltOD [in] - bolt size Nbolt - number of bolts DepthT [in] - thread depth Leg1 [in]

18

19

20

Operating Conditions:21

22

23

Corr [in] - corrosion allowanceP [psi] - internal operating pressure

Material Properties:24

25

26

CastMaterial? - Cast Or NonCastSf [psi] - allowable flange stress at DESIGN temp.Sfa [psi] - Allowable Flange Stress at ASSEMBLY temp.Efo [psi] -Operating Flange ModulusEfs [psi] - Seating Flange ModulusSb [psi] - allowable bolt stress at DESIGN tempSba [psi] - allowable bolt stress at ASSEMBLY temp

27

28

29

30

31

Geometry Constraints:tmin = min(0.75,tn,t)

tc = max(0.25,0.7*tmin)ThroatLeg1 = 0.7*Leg1

ChTL1 = ThroatLeg1 >= tc

Calculated Dimensions:B = Bn+2*Corr

32

MIN(0.75,0.5,1.5) =MAX(0.25,0.7*0.5) =

0.7*0.5 =0.35 >= 0.35 =

0.50033

34

35

Acceptable36

37

38 4.5+2*0.125 =(5.5-4.5)/2 =

0.5 / 2 =

4.7500.5000.250

varN = (GOD-GID)/2 Gasket width in contact39

40 b0 = varN / 2 Gasket seating width

varb = IF(b0>0.25,Sqrt(b0)/2,b0) Effective seating width

IF(0.25>0.25,SQRT(0.25)/2,0.25) =41

42

43

0.250varG = IF(b0>0.25,GOD-2*varb,(GOD-GID)/2 + GID)

IF(0.25>0.25,5.5-2*0.25,(5.5-4.5)/2 + 4.5) = 5.00044

45

46

ThreadMin = 0.75*Sf/Sb UG-43(g)

0.75*20000/20000 =0.75 <= 1 =CheckTrdMin = ThreadMin <=

DepthT

Bolt Loads: (VIII App 2-5)

Acceptable47

48

49 H = 0.785*varG^2*P 0.785*5^2*155.2 =2*0.25*3.14*5*3*155.2 =

PI()/4 * 9^2 * 155.2 =3046 - 9873 =

3,0463,6559,873-6,827

end load

HP = 2*varb*3.14*varG*m*P contact load50

51

52

HD = pi()/4 * A^2 * P end load

HT = H - HD face load

0.750

0.3500.350

NonCast20,00020,000

27,900,00027,900,000

20,00020,000

0.125155.2

7.5000.625

8.01.0000.500

5.5004.5003.00

1,800

HG

t

tn Shell hT

HD

hD

HT

C

Fig2-13.2modified9.0004.5001.5000.500

N4 & N5 - 4" Process Connections

Flange :Wm1 = H + HP 3046 + 3655 =

PI()*0.25*5*1800 =6,7017,069

bolt load1

2 Wm2 = pi()*varb*varG*gy seating load

Am = Max(Wm1/Sb, Wm2/Sba)

Bolt area required

MAX(6701/20000, 7069/20000) =3

4

RootArea [sq. in] = PVELookup("BoltSizing","Lookup","Root Area",BoltOD)Ab = RootArea*Nbolt

CheckExcess = Ab>=Am

Flange Loads: (App 2-5)

0.2085

6

7

0.208*8 =1.664>=0.353 = Acceptable

8

9

10

W [lb] = (Am + Ab)*Sba/2 (0.353 + 1.664)*20000/2 =6701 - 3046 =

(20174+6701)/8 =

20,1743,6553,359

seating conditions

HG [lb] = Wm1 - H operating conditions

TBoltLoad [lb] = (W+Wm1)/Nbolt

Flange Moment Arms: (Table App 2-6 - loose flanges)

mhD [in] = (varC-A)/2mhT [in] = (varC-(A+varG)/2)/2mhG [in] = (varC-varG)/2

Flange Moments: (App 2-6)

11

12

13 (7.5-9)/2 =(7.5-(9+5)/2)/2 =

(7.5-5)/2 =

-0.7500.2501.250

14

15

16

MD [in-lb] = HD * mhDMT [in-lb] = HT * mhTMG [in-lb] = HG * mhG

9873 * -0.75 =-6827 * 0.25 =3655 * 1.25 =

-7405+-1707+4569 =20174*(7.5-5)/2 =

-7,405-1,7074,569-4,54325,218

end pressure

face pressure

gasket load

17

18

19

Mo1 [in-lb] = MD+MT+MGMo2 [in-lb] = W*(varC-varG)/2

Graph: App 2-7.1 Value of Y

total operating

total seating

20

21

22

K = A/B 9/4.75 = 1.8953.205

23

24 Y = PVELookup("Y","FlangeFactorK",K)

Flange Seating Stress: (App 2-7,8)

STs = Y*ABS(Mo2) / (t^2*B)CheckSTs = ABS(STs) <= Sfa

Flange Operating Stress: (App 2-7,8)

STo = Y*ABS(Mo1) / (t^2*B)CheckSTo = STo <= Sf

Flange Flexibility: (App 2-14)

Jseating = (109.4*Mo2) / (Efs*t^3*ln(K)*0.2)

25

3.205*ABS(25218) / (1.5^2*4.75) =ABS(7563) <= 20000 =

26

27 Acceptable

28

3.205*ABS(-4543) / (1.5^2*4.75) =1362 <= 20000 =

29

30 Acceptable

31

32

33

34

(109.4*25218) / (27900000*1.5^3*LN(1.895)*0.2) =ABS(0.229) <= 1 =CheckJSt = ABS(Jseating) <=

1Acceptable

Joperating = (109.4*Mo1) / (Efo*t^3*ln(K)*0.2)(109.4*-4543) / (27900000*1.5^3*LN(1.895)*0.2) =

35

36

37 CheckJOp = ABS(Joperating) <= 1

ABS(-0.041) <= 1 = Acceptable-0.041

0.229

1,362

7,563

1.664

0.353

Nozzle Reinforcement : UW16(h) <- SavedDesign 30

Automatic dh - not hillside Automatic Limit Diameter

Curved Shell or Head Section

31

<- Vessel<- Description

22

33

Shell:34

<- Shell Material<- Sv, shell allowable stress level, PSI<- E1, efficiency of shell at nozzle<- Vt, shell wall thick, uncorroded, UT removed<- tr, required shell wall thickness int. press.(E=1)<- trE, required shell wall thickness ext. press.(E=1)<- sca, shell corrosion allowance<- tmin16b, Min allowed wall per UG-16(b)

35

Do36

Nt37

39

40

41

42

43

Nozzle:44

<- Nozzle Material<- Sn, allowable stress level (Sn)

g45 Vt46

<- B, from A =<- E, nozzle efficiency

47

Dp48

<- P, internal design pressure<- Pa, external design pressure<- Do, outside diameter<- Nt, wall thick, uncorroded<- UTp, undertolerance (%)<- nca, nozzle corrosion allowance<- L, exterior Projection<- Ip, interior projection

49

Weld to connect to reinforcing pad50

51

54

UW-16.1 (h)55

57

58

59

Reinforcing:61

<- Reinforcing plate material<- Sp, allowable stress level<- Dp, outside diameter<- te, reinforcement thick<- Leg41, size of weld fillet<- Leg42, size of weld fillet<- Leg43, size of weld fillet<- LegG, depth of groove

At least one telltale hole (max. size NPS 1/4 tap) in repad required63

65

66

69

71

72

73

75

Variables:87

UT = Rn = Dp =

t =ti =

te = tn = d =

fr1 = fr2 = fr3 = fr4 =

h =tcLeg43 =

F =

Nt*UTpDo/2 - (Nt-nca) + UT Min(2*d,DpEntered) Vt-scaNt-2*nca teEntered Nt-ncaDo-2*tn MIN(Sn/Sv,1) MIN(Sn/Sv,1) MIN(Sn/Sv,Sp/Sv,1) MIN(Sp/Sv,1)MIN(Ip-sca,2.5*t,2.5*ti) Min(0.25,0.7*Min(0.75,t,tn))1.000

= 0.75 * 0= 17.5/2 - (0.75-0) + 0= Min(2*16,21.5)= 0.675 - 0.125= 0.75 - 2 * 0

UT = Rn = Dp =

t = ti = te = tn = d =

fr1 = fr2 = fr3 = fr4 =

h = tc43 =

F =

0.0008.00021.5000.5500.7500.5000.75016.0000.8550.8550.8551.0000.6250.2501.000

Undertolerance

Effective Radius

Effective Reinforcing

Effective Shell Thickness

Nom Thick of Int. Proj.

Effective Reinf. Thick.

Avail. Nozzle Thick. No UT

Opening Dia.

88

90

91

95

96

97

= 0.75-0= 17.5 - 2*0.75= MIN(17100/20000, 1)= MIN(17100/20000, 1)= MIN(17100/20000, 20000/20000,1)= MIN(20000/20000,1)= MIN(0.75-0.125,2.5*0.55,2.5*0.75)= Min(0.25,0.7*Min(0.75,0.55,0.75))

101

102

108

111

114

115

126

131

132

Pipe Required Wall Thickness - trn from internal, trnE from external pressure141

trn = trnR = trnE =

= (155.2*8)/(17100*1 - 0.6*155.2)= (155.2*8)/(17100*1 - 0.6*155.2)= (3*17.5*0)/(4*17600)

trn = trnR = trnE =

Acceptable(P*Rn)/(Sn*E - 0.6*P) <= tn-UT

(P*Rn)/(Sn*1 - 0.6*P) (3*Do*Pa)/(4*B) <= tn-ut

143

E=1145

Acceptable146

Geometry Constraints:0.7*Leg41 >= 0.7*min(0.75,te,tn)0.7*Leg42 >= 0.5*Min(0.75,te,t)0.7*Leg43-nca >= tc43

148

0.7*0.5 >=0.7*0.375 >=

0.7*0.375-0 >=

0.7*Min(0.75,0.5,0.75)0.5*Min(0.75,0.5,0.55)0.25

0.3500.2630.263

>=>=>=

0.3500.2500.250

Acceptable Acceptable Acceptable

151

155

158171

Nozzle

0.0730.0730.000

SA-516 7020,00021.500

0.5000.5000.3750.3750.675

t

Leg43

0.09590

SA-106B17,10017,600

1.00155.20

0.017.500

0.7500.0%0.0001.5000.750

Ring te

Shell

Proj

Leg42

Leg41

SA-516 7020,000

1.000.6750.4130.0000.1250.094

Sample Vertical VesselM1 - 12" x 16" Manway on Head c/w 3" x 3/4" Ring

M1 - 12" x 16" Manway on Head c/w 3" x 3/4" RingArea Replacement: Fig UG-37.1

Pressure From:A Required (internal) =

175

Internal External207

A ==

Ae = A1 =

= A1e =

= A2 =

= A2e =

= A3 =

= A5 =

A41 = A42 = A43 =

1.0*d*tr*F + 2*tn*tr*F*(1-fr1)1.0*16*0.413*1 + 2*0.75*0.413*1*(1-0.855)

208

209

0.5*(d*trE*1 + 2*tn*trE*1*(1-fr1)) = 0.5*(16*0*1 + 2*0.75*0*1*(1-0.855)) A Requi

A1 =212

max(d, 2*(t+tn)) * (E1*t-F*tr)-2*tn*(E1*t-F*tr)*(1-fr1)max(16,2*(0.55+0.75))* (1*0.55-1*0.413)-2*0.75*(1*0.55-1*0.413)*(1-0.855) max(d, 2*(t+tn)) * (E1*t-F*trE)-2*tn*(E1*t-F*trE)*(1-fr1) max(16,2*(0.55+0.75))* (1*0.55-1*0)-2*0.75*(1*0.55-1*0)*(1-0.855)min((tn-trnR)*fr2*min(5*t,2*L) , (tn-trnR)*(Min(2.5*tn+te,L)*fr2*2)min((0.75-0.073)*0.855*min(5*0.55,2*1.5) , (0.75-0.073)*(Min(2.5*0.75+0.5,2*1.5)*0.855*2)min((tn-trnE)*fr2*Min(5*t,2*L) , 2*(tn-trnE)*Min(2.5*tn+te,L)*fr2)min((0.75-0)*0.855*Min(5*0.55,2*1.5) , 2*(0.75-0)*Min(2.5*0.75+0.5,1.5)*0.855) Min(5*t*ti*fr2, 5*ti*ti*fr2, 2*h*ti*fr2)Min(5*0.55*0.75*0.855, 5*0.75*0.75*0.855, 2*0.625*0.75*0.855)

215

216

A1e =219

220

A2 =223

224

A2e =228

229

A3 =233

234

(Dp - d - 2tn)te*fr4Leg41^2*fr3Leg42^2*fr4 (Leg43-nca)^2*fr2

=(21.5 - 16 - 2*0.75)*0.5*1 A5 = A41 = A42 = A43 =

Actual Area = Actual-Required =

236

A41 = A42 =

0.5^2*0.8550.375^2*1= (0.375-0)^2*0.855

241

244

247

249

Acceptable250

Internal Weld Load: (UG-41)WmaxI = (A - A1 + 2*Tn*Fr1*(E1*t-F*tr))*Sv, min0

W1-1 = MIN((A2 + A5 + A41 + A42)*Sv,WmaxI)

256

WmaxI =

W1-1 = W2-2 = W3-3 =

= (6.69 - 2.17 + 2*0.75*0.855*(1*0.55-1*0.413))*20000257260

= MIN((1.592 + 2 + 0.214 + 0.141)*20000,93927)261

W2-2 = Min((A2 + A3 + A41 + A43 + 2*Tn*t*frone)*Sv,WmaxI)

W3-3 = Min((A2 + A3 + A5 + A41 + A42 + A43 + 2*Tn*t*fr1)*Sv,WmaxI)= Min((1.592 + 0.802 + 2 + 0.214 + 0.141 + 0.12 + 2*0.75*0.55*0.855)*20000,93927)

External Weld Load: (UG-41)

= Min((1.592 + 0.802 + 0.214 + 0.12 + 2*0

Weld load262

266

267271

272

WmaxE =

W1-1 ==

WmaxE =

W1-1e =

(Ae - A1e + 2*Tn*Fr1*(E1*t-F*tr))*Sv, min0 = (0 - 8.68 + 2*0.75*0.855*(1*0.55-1*0.413))*20000

MIN((A2e + A5 + A41 + A42)*Sv,WmaxE) MIN((1.763 + 2 + 0.214 + 0.141)*20000,0)

273276

Weld load277

278

W2-2 = Min((A2e + A3 + A41 + A43 + 2*Tn*t*frone)*Sv,WmaxE)

W3-3 = Min((A2e + A3 + A5 + A41 + A42 + A43 + 2*Tn*t*fr1)*Sv,WmaxE)= Min((1.763 + 0.802 + 2 + 0.214 + 0.141 + 0.12 + 2*0.75*0.55*0.855)*20000,0)

Component Strength (UG-45(c), UW-15(c))

W2-2e = W3-3e =

= Min((1.763 + 0.802 + 0.214 + 0.12 + 2*0

Weld load279

283

284288

294

A2 shear = g tension =

A41 shear = A42 shear =

PI()/2*(Do-tn)*tn*Sn*0.7PI()/2*Do*LegG*Min(Sv,Sn)*0.74PI()/2*Do*Leg41*Min(Sn,Sp)*0.49PI()/2*DP*Leg42*Min(Sv,Sp)*0.49

= PI()/2*(17.5-0.75)*0.75*17100*0.7= PI()/2*17.5*0.675*Min(20000,17100)*0.74= PI()/2*17.5*0.5*Min(17100,20000)*0.49= PI()/2*21.5*0.375*Min(20000,20000)*0.49

A2s = gt =

A41s = A42s =

236,206234,795115,165124,113

295

296

297

301308

Failure mode along strength path (Greater than Weld Load, see App L-7) S1-1 = A42s + A2s >= W1-1

= 124113 + 236206 >= 78923S2-2 = A41s + gt >= W2-2

= 115165 + 234795 >= 68654S3-3 = gt + A42s >= W3-3

= 234795 + 124113 >= 93927

309

Acceptable S1-1 =312

313

Acceptable S2-2 =320

321

S3-3 =Acceptable326

327

Tstd = Swre = Nact =

Tt =

Standard pipe wall thickness from chart Tstd = Swre = Nact =

Tt =

0.3750.0000.7500.000Acceptable

331

tr * Pa / PNt * (1-UTp)0.8/Nth

= 0.413 * 0 / 155.197= 0.75 * (1-0)= 0.8/0

Req. Exterior pressure

Actual Wall Thick.

Ug-31(c)(2) threads

332

333

334

UG-45335

UG45 = UG45a = UG45b =

UG45b1 = UG45b2 = UG45b3 = UG45b4 =

Max(UG45a, UG45b) <= Nact Max(trn,trnE) + Nca + Tt Min(UG45b3,UG45b4)Max(tr + Sca, tmin16b + Sca) Max(Swre + Sca,tmin16b + Sca) Max(UG45b1,UG45b2) Tstd*0.875 + Nca

= Max(0.073, 0.328) <= 0.75= Max(0.073,0) + 0 + 0= Min(0.538, 0.328)= Max(0.413 + 0.125, 0.094 + 0.125)= Max(0 + 0.125,0.094 + 0.125)= Max(0.538,)= 0.375*0.875 + 0

UG45 = UG45a = UG45b = UG45b1 = UG45b2 = UG45b3 = UG45b4 =

336

337

338

0.538339

340

0.538341

0.328342

0.3280.0730.328

358,908

349,960

360,318

00

0

0

93,927

78,92368,65493,927

red (external) = 0.0002.170

1.592

0.802

2.0000.2140.1410.120

8.680

1.763

0.802

2.0000.2140.1410.120

7.038 13.7200.348 13.720

6.690

Nozzle Reinforcement UW16(c)mod <- SavedDesign Page 16 of 2230

Automatic dh - not hillside Automatic Limit Diameter

Curved Shell or Head Section

31

<- Vessel<- Description

22

33

Shell:34

<- Shell Material<- Sv, shell allowable stress level, PSI<- E1, efficiency of shell at nozzle<- Ds, Shell ID<- Vt, shell wall thick, uncorroded, UT removed<- tr, required shell wall thickness int. press.(E=1)<- trE, required shell wall thickness ext. press.(E=1)<- sca, shell corrosion allowance<- tmin16b, Min allowed wall per UG-16(b)

35

OD Nozzle36

37

Nt Leg4338

Full39

enn.40

41

42

43

Nozzle:44

<- Nozzle Material<- Sn, allowable stress level (Sn)

45

Vt46

<- B, from A =<- E, nozzle efficiency

47

48

<- P, internal design pressure<- Pa, external design pressure<- Do, outside diameter<- Nt, wall thick, uncorroded<- UTp, undertolerance (%)<- nca, nozzle corrosion allowance<- L, exterior Projection<- Ip, interior projection

49

50

51

54 UW-16.1 (c) modified55

57

58

59

Reinforcing:61

<- Leg41, size of weld fillet<- Leg43, size of weld fillet<- F

71

73

74

Variables:87

UT = Nt*UTpRn = Do/2 - (Nt-nca) + UT

t = Vt-scati = Nt-2*nca

tn = Nt-ncad = Do-2*tn

fr1 = MIN(Sn/Sv,1)fr2 = MIN(Sn/Sv,1)

h = MIN(Ip-sca,2.5*t,2.5*ti) tcLeg41 = Min(0.25,0.7*Min(0.75,tn,t)) tcLeg43 = Min(0.25,0.7*Min(0.75,t,tn))

F = Min(Fenterered, 1)

= 0.75 * 0= 17.5/2 - (0.75-0.125) + 0= 0.5 - 0.125= 0.75 - 2 * 0.125= 0.75-0.125= 17.5 - 2*0.625= MIN(17100/20000, 1)= MIN(17100/20000, 1)= MIN(0.875-0.125,2.5*0.375,2.5*0.5)= Min(0.25,0.7*Min(0.75,0.625,0.375))= Min(0.25,0.7*Min(0.75,0.375,0.625))

UT = Rn =

t =ti =

0.0008.1250.3750.500

Undertolerance

Effective Radius

Effective Shell Thickness

Nom Thick of Int. Proj.

Avail. Nozzle Thick. No UT

Opening Dia.

88

90

95

96

tn = 0.625d = 16.250

fr1 = 0.855 fr2 = 0.855

h = 0.750 tc41 = 0.250 tc43 = 0.250

F = 1.000

101

102

108

111

126

127

131

133

Pipe Required Wall Thickness - trn from internal, trnE from external pressure141

LDo = L/Do LDo = 0.150 Dot = Do/trnE= (155.2*8.125)/(17100*1 - 0.6*155.2)

Dot = 0.000142

trn = (P*Rn)/(Sn*E - 0.6*P) <= tn-UT

trnR = (P*Rn)/(Sn*1 - 0.6*P) trnE = (3*Do*Pa)/(4*B) <= tn-ut

Geometry Constraints:

trn = trnR = trnE =

Acceptable143

= (155.2*8.125)/(17100*1 - 0.6*155.2)= (3*17.5*0)/(4*17600)

E=1145

Acceptable146

148

0.7*Leg41 >= tc410.7*Leg43-nca >= tc43

0.7*0.375 >= 0.250.7*0.625-0.125 >= 0.25

0.2630.313

>=>=

0.2500.250

AcceptableAcceptable

150

158171

Appendix 1-7 Necessary Checkwhen Ds>60,if(2*Rn<=Ds/3,if(2*Rn<=40, "App. 1-7 calculations not required","App. 1-7 calculations required"),"App. 1-7 calculations required")when Ds<=60,if(2*Rn<Ds/2,if(2*Rn<20,"App. 1-7 calculations not required","App. 1-7 calculations required"),"App. 1-7 calculations required")

App. 1-7 calculations not required

180

181

182

183

Area Replacement: Fig UG-37.1A = 1.0*d*tr*F + 2*tn*tr*F*(1-fr1)

= 1.0*16.25*0.231*1 + 2*0.625*0.231*1*(1-0.855)

Pressure From: Internal External207

A Required (internal) =208

209

Ae = 0.5*(d*trE*1 + 2*tn*trE*1*(1-fr1)) = 0.5*(16.25*0*1 + 2*0.625*0*1*(1-0.855)) A Required (external) =212

A1 = max(d, 2*(t+tn)) * (E1*t-F*tr)-2*tn*(E1*t-F*tr)*(1-fr1)= max(16.25,2*(0.375+0.625))* (1*0.375-1*0.231)-2*0.625*(1*0.375-1*0.231)*(1-0.855)

A1e = max(d, 2*(t+tn)) * (E1*t-F*trE)-2*tn*(E1*t-F*trE)*(1-fr1)= max(16.25,2*(0.375+0.625))* (1*0.375-1*0)-2*0.625*(1*0.375-1*0)*(1-0.855)

A2 = min((tn-trnR)*fr2*Min(5*t,2*L) , (tn-trnR)*fr2*Min(5*tn,2*L))= min((0.625-0.074)*0.855*Min(5*0.375,2*2.625) , (0.625-0.074)*0.855*Min(5*0.625,2*2.625))

A2e = min((tn-trnE)*fr2*Min(5*t,2*L) , (tn-trnE)*fr2*Min(5*tn,2*L))= min((0.625-0)*0.855*Min(5*0.375,2*2.625) , (0.625-0)*0.855*Min(5*0.625,2*2.625))

A3 = Min(5*t*ti*fr2, 5*ti*ti*fr2, 2*h*ti*fr2)= Min(5*0.375*0.5*0.855, 5*0.5*0.5*0.855, 2*0.75*0.5*0.855)

A1 =215

216

A1e =219

220

A2 =225

226

A2e =230

231

A3 =233

234

A41 = Leg41^2*fr2A43 = (Leg43-nca)^2*fr2

= 0.375^2*0.855= (0.625-0.125)^2*0.855

A41 = A43 =

Actual Area = Actual-Required =

240

247

249

Acceptable250

Tstd = Standard pipe wall thickness from chart Tstd = 0.375Swre = 0.000Nact = 0.750

Tt = 0.000Acceptable

331

Swre = tr * Pa / P Nact = Nt * (1-UTp)

Tt = 0.8/Nth

= 0.231 * 0 / 155.197= 0.75 * (1-0)= 0.8/0

Req. Exterior pressure

Actual Wall Thick.

Ug-31(c)(2) threads

332

333

334

UG-45335

UG45 = Max(UG45a, UG45b) <= Nact UG45a = Max(trn,trnE) + Nca + Tt UG45b = Min(UG45b3,UG45b4)

UG45b1 = Max(tr + Sca, tmin16b + Sca) UG45b2 = Max(Swre + Sca,tmin16b + Sca) UG45b3 = Max(UG45b1,UG45b2)UG45b4 = Tstd*0.875 + Nca

= Max(0.199, 0.356) <= 0.75= Max(0.074,0) + 0.125 + 0= Min(0.356, 0.453)= Max(0.231 + 0.125, 0.094 + 0.125)= Max(0 + 0.125,0.094 + 0.125)= Max(0.356,)= 0.375*0.875 + 0.125

UG45 = UG45a = UG45b =

336

337

338

UG45b1 = 0.356339

UG45b2 =340

UG45b3 = 0.356341

UG45b4 = 0.453342

No

zzle

0.3560.1990.356

0.0002.315

0.883

0.641

0.1200.214

6.026

1.002

0.641

0.1200.214

4.173 8.0030.378 8.003

3.795

0.0740.0740.000

0.3750.6251.000

0.09590

P

Leg41 Shell

Leg43Proj

SA-106B17,10017,600

1.00155.20

0.017.500

0.7500.0%0.1252.6250.875

t

Leg41

SA-516 7020,000

1.0059.000.5000.2310.0000.1250.094

Sample Vertical VesselM2 - 12" x 16" Manway c/w 4 x 3/4" Ring on Shell

Vessel Weight and Volume 1

Description2

Volume:3

4 nhead - Number of heads?SG - Fluid Specific GravityVE [ft3] - Volume of Each HeadVS [ft3] - Volume of Shell

5

6

7

Construction:8

9

10

Wh [lb] - Weight of Each HeadWs [lb] - Weight of ShellWm [lb] - Misc Weight11

Calculations:V [ft3] = VE*nhead + VS

V2 [Imp. Gallons] = V*6.229V3 [US Gallons] = V*7.4805

Wf [lb] = 62.37*SG*VWC [lb] = Wh*nhead + Ws + WmWT [lb] = WC + Wf

12

13 total volume 12*2 + 190 =214*6.229 =

214*7.4805 =62.37*1*214 =

789*2 + 3181 + 650 =5409.83 + 13347.18 =

14

15

16 fluid weightconstruction weight

total weight17

18

214.001,333.011,600.8313,347.185,409.8318,757.01

7893181650

21.00

12.00190.00

Sample Vertical Vessel

Lifting Lugs 1

Description Load Case 1

2

Dimensions:3

4 Load [lb] - vessel weight emptyW [in] - widthThick [in] - lug thicknessH [in] - hole heightDia [in] - hole diameter OR [in] - outside radius Weld [in] - leg size

MaterialSA [psi] - allowed stress in tension

5

6

7

Load Case 2

8

9

10 H

11

12

13

14

15

16

17

18

All of load assumed carried by one lug All load cases analyzed independently Never load lug perpendicular to face Contour lug to fit vesselDo not move or support vessel with this lug when full or pressurized

eld

W

SB = UG-34(b) Max Bending Stress, SS = IID Tbl 1A(d) Max Shear Stress, SSw = UW-15(c) UW-15 Max Weld Shear

SB [psi] = SA * 1.5SS [psi] = SA * 0.8

SSw [psi] = SA * 0.49

Tensile Stress (case 1):A1 [in

2] = Thick*(OR-Dia/2)

A [in2] = A1 * 2

Stress [psi] = Load / ACheckTenStr = Stress <= SA

Pin Bearing Stress (case 1 and 2):Area [in

2] = Dia * Thick PinStress

[psi] = Load / Area CheckPinStr = PinStress <= (1.6 * SA)

Bending Stress (case 2):Moment [in-lb] = Load * H

I [in4] = Thick * W^3 / 12

c [in] = W/2BendStress [psi] = Moment*c/ICheckBendStr = BendStress <= SB

Shear Stress (case 2):ShrArea [in

2] = W*Thick

ShrStress [psi] = Load/ShrArea CheckShStr = ShrStress <= SS

Weld Stress (case 1):Circ [in] = W*2+Thick*2+Weld*4

WeldArea [in2] = Circ * Weld

WeldStress [psi] = Load / WeldAreaCheckWldStr = WeldStress <= SSw

Weld Stress (case 2):Moment2 [in-lb] = Load * H

I2 [in4] = (Thick +2*Weld)* (W+2*Weld)^3 / 12 -

I

20000 * 1.5 =20000 * 0.8 =

20000 * 0.49 =

30,00016,0009,800

19

20

21

22

23

24

0.5*(2.5-1.5/2) =0.875 * 2 =

6000 / 1.75 =3429 <= 20000 =

0.8751.750

25

26 Acceptable

27

1.5 * 0.5 =6000 / 0.75 =

8000 <= (1.6 * 20000) =

0.75028

29

Acceptable30

31

32 6000 * 2.5 =0.5 * 8^3 / 12 =

8/2 =15000*4/21.333 =2813 <= 30000 =

15,00021.3334.000

33

34

35

Acceptable36

37

38 8*0.5 =6000/4 =

1500 <= 16000 =

4.00039

40 Acceptable

41

8*2+0.5*2+0.25*4 =18 * 0.25 =

6000 / 4.5 =1333 <= 9800 =

18.0004.500

42

43

44

Acceptable45

46

47 6000 * 2.5 = 15,000

(0.5 +2*0.25)* (8+2*0.25)^3 / 12 - 21.333 =8/2 + 0.25 =

15000*4.25/29.844 =2136 <= 9800 =

29.8444.250

48

49 c2 [in] = W/2 + WeldWldStress2 [psi] = Moment2*c2/I2CheckWldStr2 = WldStress2 <= SSw

50

51 Acceptable2,136

1,333

1,500

2,813

8,000

3,429

W

SA-516 7020,000

6,0008.0000.5002.5001.5002.5000.250

OR

Dia

Load C

Sample Vessel 4 Liftng Lugs

Vessel On Beams Ver 2.24 7-Jan-09IBC-2000

15

<- Vessel1718

Vessel Dimensions (Inch and Lbs):19

<- H, height<- L, center of gravity<- ls, leg free length<- Do, shell outside diameter<- ds, leg pitch diameter<- t, shell corroded thickness<- ws - leg weld size<- lw - length of leg to shell weld<- lwf - length of weld on foot<- W, Weight lbs<- Pr, Pressure

20

21

22

23

24

25

26

27

28

29

3031

Site Specific Seismic Information per IBC-2000:32

ion importance factor35

38

<- Ss, Acceleration at Short Periodsration at a period of one second

39

40

<- Fa, Site Coefficient<- Fv, Site Coefficient

41

42

<- R, Response Modification Factor4351

Leg Supports:52

<- Structural Description<- n, number of legs<- Ix, for one leg<- Iy, for one leg<- fFactor, Least radius of Gyration<- A, Leg Cross Sectional Area<- 2cx, Beam Depth<- 2cy, Beam Width<- K1, Leg Anchor Factor

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55

56

57

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59

60

6162

Material Properties:63

<- maximum leg bending stress (Sb)<- maximum shell stress (Sa)

64

6566

Attachment Dimensions:67

<- 2C1, Width of rectangular loading<- 2C2, Length of rectangular loading

68

69

Static DeflectionE = 30,000,000

bc = 12.0

71

72

leg boundary condition based on fixed or loose leg73

y = (2*W*ls^3)/(bc*n*E*(Ix + Iy))= (2*19000*27^3)/(12*4*30000000*(29.1 + 9.32))

Period of Vibrationg = 386

y = 0.01474

7576

77

78

T = 2*pi*sqrt(y/g) =2 * 3.14 * sqrt(0.01/386) T = 0.0377980

Base ShearSms = Fa*SsSm1 = Fv*S1Sds = 2/3*SmsSd1 = 2/3*Sm1Cs = Sds/(R/I)

CsMAX = Sd1/(T*R/I) CsMIN = 0.044*Sds*I

84

= 1.2*0.75= 2.8*0.3= 2/3*0.9= 2/3*0.84= 0.6/(3/1)= 0.56/(0.037*3/1)= 0.044*0.6*1

Sms = 0.9Sm1 = 0.84Sds = 0.600Sd1 = 0.560

Cs = 0.200CsMAX = 5.020CsMIN = 0.026Csfinal = 0.200

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95

96

97

98

99

100

Csfinal = if(cs<=csmax, if(cs>csmin, cs, csmin), csmax)101

V = Csfinal*W = 0.2*19000 V =102112

3800

6.0006.000

17,10020,000

W6x154

29.1009.3201.4604.4306.0006.0000.800

1.000 <- I, occupat<- Site Class<- Ss, Acc

E0.7500.300

1.2002.800

3.000 <- R, Res

162.50090.00027.00060.00060.5000.5000.250

18.00035.00019,000155.2

Sample Vertical Vessel

115 Sample Vertical Vessel Vessel On Beams117 Horizontal Seismic Force at Top of Vessel

7-Jan-09

Ftmax = 0.25*VFtp = 0.07 * T * V

= 0.25 * 3800= 0.07 * 0.037 * 3800

Ftmax = 950Ftp = 9.89

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119

Ft = if (T < 0.7, 0, min(0.07*T*V, Ftmax)) Ft =120121

122 Horizontal Seismic Force at cgFh = V - Ft = 3800 - 0 Fh =123

124

125 Vertical force at cgFv = W Fv =126

127

128 Overturning Moment at BaseMb = L*Fh + H*Ft = 90 * 3800 + 162.5 * 0 Mb = 342,000129

130

131 Overturning Moment at Bottom Tangent LineMt = (L-ls)*Fh + (H-ls)*Ft = (90 - 27) * 3800 + (162.5 - 27) * 0 Mt = 239,400132

133

134 Maximum eccentric loadf1 = Fv/n + 4*Mto/(n*Do) = 19000/4 + 4*239400/(4 * 60) f1 = 8,740135

136

137 Axial Load138139

140 Leg Loads141

f2 = Fv/n + 4*Mb/(n*ds) = 19000/4 + 4*342000/(4 * 60.5) f2 = 10,403

f3x = 0.5*V*Ix/(Ix+Iy)

f3y = 0.5*V*Iy/(Ix+Iy)

=0.5* 3800*29.1 /( 29.1+9.32)

=0.5* 3800*9.32 /( 29.1+9.32)

f3x = 1,439

f3y = 461142143

144 Leg Bending Momentse = (ds-Do)/2

Mx = f1*e + f3x*lsMy = f1*e + f3y*ls

=(60.5-60)/2=8740*0.25 + 1439*27=8740*0.25 + 461*27

e = 0.25Mx = 41,041My = 14,629

145

146

147148

149 Leg Bending StressSbmax = Sb * 1.25

fx = Mx*cx/Ix fy = My*cy/Iy

=17100 * 1.25=41041 * 3 / 29.1

=14629 * 3 / 9.32

Sbmax = 21,375150

AcceptableAcceptable

fx =fy =

151

152153

154 Leg axial stress155 K1*ls/fFactor = =0.8 * 27 / 1.46 K1*ls/fFactor = 14.795

Fa max = 25,675Fa max = AISC code lookup based on K1*ls/r156

fa = f2/A =10403 / 4.43 Acceptable fa =157158

159 Maximum Euler StressFe = 12*pi^2*E/(23*(K1*L/r)^2)

= 12*pi^2*30000000/(23*14.795^2)160

Fe = 705,785161162

163 Combined StressFc1 = fa/Famax + 0.85*fx/((1-fa/Fe)*Sbmax)

= 2348/25675 + 0.85*4231/((1-2348/705785)*21375) Fc2 = fa/Famax + 0.85*fy/((1-fa/Fe)*Sbmax)

= 2348/25675 + 0.85*4709/((1-2348/705785)*21375)

Acceptable164

Fc1 =165

Acceptable166

Fc2 =167168

0.28

0.26

2,348

4,2314,709

19,000

3,800

0

171 Sample Vertical Vessel Vessel On Bea172

173 Beam to Shell Attachment Stresses174

175 Beam Dimensions

7-Jan-09

cx = 2cx/2 cy = 2cy/2

cx = 3.000cy = 3.000

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177

178

179180

181 C dimensions for weld stressweld area = ws*lw

wcx = lw/2wcz = cy + ws

wa = 4.500 wcx = 9.000 wcz = 3.250

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183

184

wcy = sqrt(wcx^2 + wcy^2) = sqrt(9^2 + 9.569^2) wcy = 9.569185186

187 Shear Force DistributionVx = (V*Ix)/((n/2)*(Ix+Iy)) Vy = (V*Iy)/((n/2)*(Ix+Iy)) Vg = W/n

Vx = 1,439Vy = 461Vg = 4,750

= (3800*29.1)/((4/2)*(29.1+9.32))

= (3800*9.32)/((4/2)*(29.1+9.32))

188

189

gravity190191

192 Weld Moments of InertiasIwx = (ws*lw^3/12)*2Iwz = (lw*ws^3/12 + wa*(cy+ws/2)^2)*2

= (18*0.25^3/12 + 4.5*(3+0.25/2)^2)*2Iwy = Iwx + Iwz

= (0.25*18^3/12)*2 Iwx = 243.0Iwz = 87.9

193

194

195

= 243 + 88 Iwy = 330.9196197

198 Weld MomentsMx = Vx*(ls+lw/2) + Vg*(ds-Do)/2

= 1439*(27+18/2) + 4750*(60.5-60)/2My1 = Vy*(ls+lw/2)

Mz = Vy*(ds-Do)/2

Mx = 52,995199

200

= 461*(27+18/2)= 461*(60.5-60)/2

My1 = 16,593Mz = 115

201

202203

204 Weld StressesSx = Mx*wcx/Iwx Sy = My1*wcy/Iwy Sz = Mz*wcz/Iwz Sg = Vg/(wa*2)

Bending Twisting Torision Gravity

= 41041*9/243= 16593*9.569/330.9= 115*3.25/87.9= 4750/(4.5*2)

Sx = 1,520Sy = 480Sz = 4Sg = 528

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206

207

208209

210 Stress Limits and RatiosSlim = min(Sb,Sa)*0.49

SxR = Sx/Slim SyR = Sy/Slim SzR = Sz/Slim SgR = Sg/Slim

= min(17100,20000)*0.49

= 1520/8379= 480/8379= 4/8379= 528/8379

Acceptable

Slim = 8,379211212

SxR = SyR = SzR = SgR =

total (<1)

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214

215

216

217

218 Foot Plate Attachment Stresseswaf = ws*lwfVv = V/n

Sv = Vv/wafSgf = Vg/waf

SvRf = Sv/SlimSgRf = Sgf/Slim

weld area in foot = 0.25*35= 3800/4

= 950/8.75= 4750/8.75

= 109/8379= 542.857/8379

waf = 8.750Vv = 950

Sv = 109Sgf = 543

219

220221

222

223224

SvRf = SgRf =

total (<1)

225

226

Acceptable227

0.0130.0650.078

0.1810.0570.0010.0630.302

7-Jan-09231

232 WRC 107 - shell local stress at support233

234 Loads (psi and lb)<- P, Axial Load (=vx)<- VL, Longitudinal load(=f2)<- Vc, Circumferential load<- ML, Moment (=My)<- Mc, Moment<- MT, Torisional

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236

237

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239

240

241

242 Parameters243 MaxSPm = Sa244 MaxSPmb = 1.5*Sa245 MaxSPmbQ = 1.5*Sa

Pm - primary membrance stressPb - primary bending stressQ - secondary stress

MaxSPm = 20,000MaxSPmb = 30,000

MaxSPmbQ = 30,000Ri = 29.5

Rm = 29.75

for Pm stresses

for Pm + Pb stresses

for Pm + Pb + Q stressesRi = (Do-2*T)/2

Rm = (Do-T)/2 r = Rm/T

Beta1 = 2C1/2/RmBeta2 = 2C2/2/Rm

SL = (Ri-0.4*T)*Pr/(2*T) Sc = (Ri+0.6*T)*Pr/T

246

247

= 29.75/0.5= 6/2/29.75= 6/2/29.75= (29.5-0.4*0.5)*155.197/(2*0.5)= (29.5+0.6*0.5)*155.197/0.5

Kb = 1

r =248

Beta1 = 0.101Beta2 = 0.101

SL = 4,547Sc = 9,250Kn = 1

249

250

280

281

282 Stress concentration factors283 Shell Combined Stresses:284

285

286

287

288

289

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293

294

295

296

297

298

299

300

301

302

303

304

305

306

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311

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313

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315

316

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321

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323

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325

Lookup A Curve A Value A Value Equation Cat Au AL Bu BL Cu CL Du DLPressure Stress VIII-1 Code 4C 3C SC Pm 9250 9250 9250 9250 9250 9250 9250 9250

No/(P/Rm) 3C or 4C 9.80981 7.99937 Kn*A*P/(Rm*T) Pm -949 -949 -949 -949 -774 -774 -774 -774Mo/P 1C or 2C-1 0.10736 0.07214 Kb*A*6*P/T^2 Pb -2492 2492 -2492 2492 -3708 3708 -3708 3708

No/(Mc/(Rm^2*beta))

3A 2.00349 Kn*A*Mc/(Rm^2*beta*T) Pm 0 0 0 0Mo/(Mc/(Rm*beta)) 1A 0.08976 Kb*A*6*Mc/(Rm*beta*T^2) Q 0 0 0 0

No/(ML/(Rm^2*beta))

3B 6.62238 Kn*A*ML/(Rm^2*beta*T) Pm -2171 -2171 2171 2171Mo/(ML/(Rm*beta)) 1B or 1B-1 0.04054 Kb*A*6*ML/(Rm*beta*T^2) Q -4661 4661 4661 -4661

Pm So 6130 6130 10472 10472 8476 8476 8476 8476Pm+Pb So 3638 8621 7980 12963 4768 12184 4768 12184

Pm+Pb+Q So -1023 13282 12641 8302 4768 12184 4768 12184Pressure Stress VIII-1 Code SL Pm 4547 4547 4547 4547 4547 4547 4547 4547

Nx/(P/Rm) 3C or 4C 9.80981 7.99937 Kn*A*P/(Rm*T) Pm -774 -774 -774 -774 -949 -949 -949 -949Mx/P 1C-1 or 2C 0.07151 0.10957 Kb*A*6*P/T^2 Pb -3784 3784 -3784 3784 -2470 2470 -2470 2470

Nx/(Mc/(Rm^2*beta))

4A 3.02513 Kn*A*Mc/(Rm^2*beta*T) Pm 0 0 0 0Mx/(Mc/(Rm*beta)) 2A 0.04374 Kb*A*6*Mc/(Rm*beta*T^2) Q 0 0 0 0

Nx/(ML/(Rm^2*beta))

4B 1.96743 Kn*A*ML/(Rm^2*beta*T) Pm -645 -645 645 645Mx/(ML/(Rm*beta)) 2B or 2B-1 0.05817 Kb*A*6*ML/(Rm*beta*T^2) Q -6486 6486 6486 -6486

Pm Sx 3128 3128 4418 4418 3598 3598 3598 3598Pm+Pb Sx -656 6913 634 8203 1128 6068 1128 6068

Pm+Pb+Q Sx -7142 13399 7120 1717 1128 6068 1128 6068Shear VL VL/(Pi*sqrt(c1*c2)*T) -2208 -2208 2208 2208Shear VC VC/(Pi*sqrt(c1*c2)*T) 0 0 0 0

Total Shear Sum of shears Txo 0 0 0 0 -2208 -2208 2208 2208S1m ((Sx+So)/2)+SQRT(((Sx-So)/2)^2+Txo^2) 6,130 6,130 10,472 10,472 9,327 9,327 9,327 9,327S2m ((Sx+So)/2)-SQRT(((Sx-So)/2)^2+Txo^2) 3,128 3,128 4,418 4,418 2,747 2,747 2,747 2,747S12 abs(S1m - S2m) 3,001 3,001 6,053 6,053 6,579 6,579 6,579 6,579S23 abs(S2m-0) 3,128 3,128 4,418 4,418 2,747 2,747 2,747 2,747S31 abs(0-S1m) 6,130 6,130 10,472 10,472 9,327 9,327 9,327 9,327

Sm<= MaxSPmb max(S12,S23,S31)<=30000 Acceptable 6,130 6,130 10,472 10,472 9,327 9,327 9,327 9,327S1m+b ((Sx+So)/2)+SQRT(((Sx-So)/2)^2+Txo^2) 3,638 8,621 7,980 12,963 5,809 12,897 5,809 12,897S2m+b ((Sx+So)/2)-SQRT(((Sx-So)/2)^2+Txo^2) -656 6,913 634 8,203 87 5,354 87 5,354

S12 abs(S1m - S2m) 4,294 1,709 7,346 4,761 5,722 7,543 5,722 7,543S23 abs(S2m-0) 656 6,913 634 8,203 87 5,354 87 5,354S31 abs(0-S1m) 3,638 8,621 7,980 12,963 5,809 12,897 5,809 12,897

Smb<= MaxSPmb max(S12,S23,S31)<=30000 Acceptable 4,294 8,621 7,980 12,963 5,809 12,897 5,809 12,897S1m+b+Q ((Sx+So)/2)+SQRT(((Sx-So)/2)^2+Txo^2) -1,023 13,399 12,641 8,302 5,809 12,897 5,809 12,897S2m+b+Q ((Sx+So)/2)-SQRT(((Sx-So)/2)^2+Txo^2) -7,142 13,282 7,120 1,717 87 5,354 87 5,354

S12 abs(S1m - S2m) 6,119 116 5,521 6,585 5,722 7,543 5,722 7,543S23 abs(S2m-0) 7,142 13,282 7,120 1,717 87 5,354 87 5,354S31 abs(0-S1m) 1,023 13,399 12,641 8,302 5,809 12,897 5,809 12,897

Smb<= MaxSPmbQ max(S12,S23,S31)<=60000 Acceptable 7,142 13,399 12,641 8,302 5,809 12,897 5,809 12,897

59.50

1,439.110,402.9

0.014,629.5

0.00.0

123

124