Asme Code Calculations Cylindrical Components

Revised 03/06 to conform with the 2004 ASME Extract

Part A1 CHAPTER 1

ASME Code Calculations:
Cylindrical Components

Here is what you w i l l be able to do when you complete each objective:

1. Calculate the required minimum thickness or the maximum allowable working
pressure of piping, tubes, drums, and headers of ferrous tubing up to and
including 125 mm O.D.

2. Calculate the required minimum thickness or the maximum allowable working
pressure of ferrous piping, drums, and headers.

3. Calculate the required thickness or maximum allowable working pressure of a
seamless, unstayed dished head.

4. Calculate the minimum required thickness or maximum allowable working
pressure of unstayed flat heads, covers, and blind flanges.

5. Calculate the acceptability of openings in a cylindrical shell, header, or head.

6. Calculate the compensation required to reinforce an opening in a cylindrical
shell, header, or head.

Revised 03/06 to conform with the 2004 ASME Extract

1

Chapter 1 • ASME Code Calculations: Cylindrical Components 3

INTRODUCTION

As power engineers acquire their second and first class power engineering
certification, they find that their roles and areas of responsibility require them to
have a more detailed working knowledge of the key engineering codes and
standards with which their facility must comply. Power engineers often work on
teams or lead teams that are responsible for upgrades within their facilities
and/or for making changes to major pressure piping or equipment. Although
power engineers are not required to design a boiler or pressure vessel, they often
work as team members for equipment design, upgrade, process change,
commissioning, operation, or repair. These activities require work to be done in
accordance with applicable codes. As well, when you become chief engineer of a
facility, you may be called upon to lead teams and give approval for various
projects that must comply with specific engineering codes and standards.

In the early 1900’s, the American Society of Mechanical Engineers (ASME)
appointed various committees to draw up standards for the construction of
boilers and pressure vessels together with standards for welding and guidelines
for the care of boilers in service. These standards and guidelines have been
improved over the years with the improvement in materials and technology.

One important component of the standards for pressure vessels is the use of a
safety factor. The measured physical properties of a material, including ultimate
tensile strength, are divided by a defined safety factor to derive the maximum
allowable stress. In this way, allowance is made for limitations in the testing
technology, unusual stress concentrations, non-uniform materials, and material
flaws. Technological improvements, especially in materials testing, have allowed a
reduction in the safety factor to 3.5 in current editions of Section I; this is the
same factor used in Sections VIII-1 and VIII-2.

Pressures calculated or given in this module refer to gauge pressure unless
otherwise indicated.

Consult the latest ASME Codes (currently the 2004 Edition)—Section I and
Section VIII, Division 1—while studying this module. Figures referenced with a
Code section prefix, such as “Fig. PG-32” or “Fig. UG-34,” can be found in the
ASME Codes or the Academic Extract and are generally not reproduced here.

Note: Material and formulae used in this chapter refer to the 2004 edition of
the ASME Codes. Most relevant sections can be found in the 2004
ASME Academic Extract (visit www.powerengineering.ca for more info).

Note: Correct units of measure are very important to accurate calculations, and
students should be well versed in their use. However, due to the size and
complexity of Code calculations, it is common practice to omit the units

Conforms with the 2004 ASME Extract • Revised 03/06

4 Revised Second Class Course • Section A1 • SI Units

until the final answer is derived. This convention has been used
throughout this chapter.

Note: It should be noted that many US customary unit values presented
in the ASME codes do not convert directly into metric values in
the current ASME edition or the 2004 ASME Academic Extract (
i.e. 5 in. converts to 127mm, ASME shows 5 in. (125 mm); ¼ in.
coverts to 6.35 mm, ASME shows ¼ in. (6 mm)). You are required
to use the ASME values as presented and not to convert US
customary numbers to metric.

ASME SECTION I - POWER BOILERS

Paragraphs PG-1, PG-2: This Code covers rules for construction of power
boilers, electric boilers, miniature boilers, and high temperature water boilers.
The scope of jurisdiction of Section I applies to the boiler proper and the boiler
external piping. Superheaters, economizers, and other pressure parts connected
directly to the boiler, without intervening valves, are considered to be parts of
the boiler proper and their construction shall conform to Section I rules.

Materials

Paragraph PG-6 states that steel plates for any part of a boiler subject to
pressure, whether or not exposed to the fire or products of combustion, shall be
in accordance with specifications listed in paragraph PG-6.1. Paragraph PG-9
states that pipes, tubes, and pressure containing parts used in boilers shall
conform to one of the specifications listed in paragraph PG-9.1.

Design

Paragraph PG-16.3 states that the minimum thickness of any boiler plate under
pressure shall be 6 mm. The minimum thickness of plates to which stays may be
attached (in other than cylindrical outer shell plates) shall be 8 mm. When pipe
over 125 mm O.D. is used in lieu of plate for the shell of cylindrical components
under pressure, its minimum wall thickness shall be 6 mm.

Paragraph PG-16.4 states that plate material not more than 0.3 mm thinner than
the required thickness calculated by Code formula may be used provided the
manufacturing process is such that the plate will not be more than 0.3 mm
thinner than that specified in the order.

Paragraph PG-16.5 states that pipe or tube material shall not be ordered thinner
than the required thickness calculated by Code formula. Also, the ordered
thickness shall include provisions for manufacturing tolerance.



Paragraph PG-21 states that the term maximum allowable working pressure
(MAWP) refers to gauge pressure, except when noted otherwise in the
calculation formula of PG-27.2.

Paragraph PG-27 Cylindrical Components Under Internal Pressure

The formulae in this section are used to determine the minimum required
thickness of piping, tubes, drums, and headers, when the maximum allowable
working pressure is known. These formulae can be transposed to determine the
maximum allowable working pressure if the minimum required thickness is
given.

The symbols used in the formulae are found in paragraph PG-27.3 and are
defined as follows:
C = minimum allowance for threading and structural stability (mm) (see
PG-27.4, note 3)
D = outside diameter of cylinder (mm)
E = efficiency of longitudinal welded joints or of ligaments between
openings, whichever is lower (the values allowed for E are listed in
PG-27.4, note 1)
e = thickness factor for expanded tube ends (mm) (see PG-27.4, note 4)
P = maximum allowable working pressure (MPa). (see PG-21, refers to
gauge pressure)
R = inside radius of cylinder (mm)
S = maximum allowable stress value at the operating temperature of the
metal (Section II, Part D, Table 1A. See PG-27.4, note 2)
t = minimum required thickness (mm) (see PG-27.4, note 7)
y = temperature coefficient (see PG-27.4, note 6)

ASME SECTION VIII, DIVISION 1 - PRESSURE VESSELS

Foreword

The Boiler and Pressure Vessel Committee established rules for new
construction of pressure vessels that ensure safe and reliable performance. The
Code is not a handbook and cannot replace education, experience, and the use of
good engineering judgement. This can be seen in that Section VIII-1 applies to
small compressed-air receivers sold commercially to the general public as well as
to very large pressure vessels used by the petrochemical industry. The Code
contains mandatory requirements, specific prohibitions, and non-mandatory
guidance for pressure vessel construction activities.



Materials

Paragraph UG-4 states that materials subject to stress due to pressure are to
conform to the specifications given in Section II, except as otherwise permitted
in paragraphs UG-9, UG-10, UG-11, UG-15 and the Mandatory Appendices.
Paragraph UG-23 (a) lists the tables in Section II, D for various materials.

Design

ASME Boiler Code Section I, as well as Section VIII, Division 2 (VIII-2),
requires all major longitudinal and circumferential butt joints to be examined by
full radiograph. Section VIII-1 lists various levels of examination for these major
joints. A fully radiographed major longitudinal butt-welded joint in a cylindrical
shell would have a joint efficiency factor (E) of 1.0. This factor corresponds to
a safety factor (or material quality factor) of 3.5 in the parent metal. Non-
radiographed longitudinal butt-welded joints have a joint efficiency factor (E) of
0.7, which corresponds to a safety factor of 0.5 in plates. This results in an
increase of 43% in the thickness of the plates required.

Paragraph UG-20: Design temperature

With pressure vessels, the maximum temperature used in the design is important,
as is the minimum temperature.

The minimum temperature used in design shall be the lowest temperature that
the vessel will experience from any factor, including normal operation, upset
condition, or environmental conditions.

Paragraph UG-27: Thickness of shells under internal pressure

The formulae in this section are used to determine the minimum required
thickness of shells when the maximum allowable working pressure is known.
These formulae can be transposed to determine the maximum allowable working
pressure if the minimum required thickness is given.

The symbols used in the formulae are found in paragraph UG-27 (b) and are
defined as follows:
t =
minimum required thickness (mm)
P =
internal design pressure (MPa) (see UG-21. refers to gauge pressure)
R =
inside radius of shell course under consideration (mm)
S =
maximum allowable stress value (see UG-23 and the stress limitations
specified in UG-24)
E = joint efficiency for, or the efficiency of, appropriate joint in
cylindrical or spherical shells, or the efficiency of ligaments between
openings, whichever is less (use UW-12 for welded vessels. Use UW-
53 for ligaments between openings)



OBJECTIVE 3 1
OBJECTIVE
Calculate the required minimum thickness or the maximum allowable
working pressure of piping, tubes, drums and headers of ferrous
tubing up to and including 125 mm O.D.

SECTION I

The following formulae are found in ASME Section I, paragraph PG-27.2.1.

Formula for minimum required thickness

PD
t = + 0.005 D + e 1.1
2S + P

Formula for MAWP

⎡ 2t - 0.01D - 2e ⎤
P = S ⎢ ⎥ 1.2
⎢ D - ( t - 0.005D - e ) ⎥
⎣ ⎦

Example 1: boiler tube
Calculate the minimum required wall thickness of a watertube boiler tube 70 mm
O.D. that is strength welded into place in a boiler. The tube is located in the
furnace area of the boiler and has an average wall temperature of 350°C. The
maximum allowable working pressure is 4000 kPa gauge. The tube material is
carbon steel SA-192.

Note: Check PG-6 for plate materials and PG-9 for boiler tube materials
before starting calculations; the information will direct you to the
correct stress table in ASME Section II, Part D by indicating if the
metal is carbon steel or an alloy steel.

Solution
For tubing up to and including 125 mm O.D. use equation 1.1.
(See paragraph PG-27.2.1 )

PD
t = + 0.005D + e
2S + P



Where
P = 4000 kPa = 4.0 MPa
D = 70 mm
e = 0 (see PG-27.4, note 4, strength welded)
S = 87.8 MPa (see Section II, Part D, Table 1A,
SA-192 at 350°C)

4 × 70
t = + 0.005(70) + 0
2(87.8) + 4
280
= + 0.35
179.6
= 1.56 + 0.35
= 1.9 mm (Ans.)

Note: This value is exclusive of the manufacturer’s tolerance allowance (see
PG-16.5). The manufacturing process does not produce absolutely
uniform wall thickness; add an allowance of approximately 12.5% to
the minimum thickness calculated.

The formula for minimum thickness may be transposed to solve for the
maximum allowable working pressure if the tube size and thickness are known.

Example 2: superheater tube
Calculate the maximum allowable working pressure, in kPa, for a 75 mm O.D.
and 4.75 mm minimum thickness superheater tube connected to a header by
strength welding. The average tube temperature is 400°C. The tube material is
SA-213-T11.

Note: Check PG-9 for boiler tube materials before starting calculations; the
information will direct you to the correct stress table in ASME Section
II, Part D. SA-213-T11 is alloy steel.

Solution
For tubing up to and including 125 mm O.D. Use equation 1.2.
(See paragraph PG-27.2.1.)

⎡ 2t − 0.01D − 2e ⎤
P = S⎢ ⎥
⎢ D − ( t − 0.005D − e ) ⎥
⎣ ⎦

Where
t = 4.75 mm
D = 75 mm
e = 0 (see PG-27.4, note 4, strength welded.)
S = 102 MPa (Section II, Part D, Table 1A,
SA-213-T11 at 400°C)



⎡ ( 2 × 4.75 ) - ( 0.01 × 75 ) - ( 2 × 0 ) ⎤
P = 102 × ⎢ ⎥
⎢ 75 - ( 4.75 - ( 0.005 × 75 ) - 0 )
⎣ ⎥
⎦
⎡ 9.5 - 0.75 ⎤
= 102 × ⎢ ⎥
⎢ 75 - ( 4.75 - 0.375 ) ⎥
⎣ ⎦
8.75
= 102 ×
70.625
= 12.64 MPa = 12 640 kPa (Ans.)

The tubes were strength welded in Example 1 and Example 2. For calculations
involving tubes expanded into place, the appropriate value of e is found in
paragraph PG-27.4, note 4.

SECTION VIII

The following formulae (found in ASME Section VIII-1, paragraph UG-27(c))
are used for calculating wall thickness and design pressure. Paragraph UG-31(a)
states that these calculations are used for tubes and pipes under internal pressure.

Thin Cylindrical Shells

(1) Circumferential stress (longitudinal joints)

PR
t = 1.3
(SE - 0.6P )
Or
SEt
P = 1.4
(R + 0.6t )

When t < 0.5R or P < 0.385SE

(2) Longitudinal stress (circumferential joints)
.
PR
t = 1.5
(2SE + 0.4P)
Or
2SEt
P = 1.6
(R - 0.4t )

When t < 0.5 R or P < 1.25SE



Thick Cylindrical Shells

As internal pressures increase higher than 20.6 MPa, special considerations must
be given to the construction of the vessel as specified in paragraph U-1 (d). As
the ratio of t/R increases beyond 0.5, a more accurate equation is required to
determine the thickness. The formulae for thick walled vessels are listed in
Appendix 1, Supplementary Design Formulas 1.1 to 1.3.

SE =
(
P R02 + R 2 )
(R2
0 - R2 )
Where R0 and R are outside and inside radii, respectively.
By substituting R0 = R + t

⎛ 1 ⎞
t = R ⎜ Z 2 - 1⎟ Where Z =
( SE + P ) 1.7
⎝ ⎠ ( SE - P )
Where t > 0.5 R or P > 0.385SE

And

⎡ ( Z - 1) ⎤ ⎡ (R + t)⎤
2

P = SE ⎢ ⎥ Where Z = ⎢ ⎥ 1.8
⎢ ( Z + 1) ⎥
⎣ ⎦ ⎣ R ⎦

For longitudinal stress with t > 0.5R or P > 1.25SE

⎛ 1 ⎞ ⎛ P ⎞
t = R ⎜ Z 2 - 1⎟ Where Z =⎜ ⎟ +1 1.9
⎝ ⎠ ⎝ SE ⎠

And

⎡(R + t)⎤
2

P = SE ( Z - 1) Where Z =⎢ ⎥ 1.10
⎣ R ⎦

Note: Formulae 1.3 to 1.10 are for internal pressure only.

Example 3: thin shell thickness
A vertical boiler is constructed of SA-515-60 material in accordance with the
requirements of Section VIII-1. It has an inside diameter of 2440 mm and an
internal design pressure of 690 kPa at 230°C. The corrosion allowance is 3 mm,
and joint efficiency is 0.85. Calculate the required thickness of the shell if the
allowable stress is 138 MPa.



Solution
The quantity 0.385SE = 45.16 MPa; since this is greater than the design pressure
P = 690 kPa, use equation 1.3. (See Section VIII-1, UG-27.)
Note R must be in the fully corroded state to determine the minimum thickness.

PR
t = + corrosion allowance
( SE - 0.6 P )
0.69 × (1220 + 3)
= + 3
(138 × 0.85) - ( 0.6 × 0.69 )
843.87
= + 3
116.886
= 7.22 + 3
= 10.22 mm (Ans.)

The calculated thickness is less than 0.5R; therefore, equation 1.3 is acceptable.

Example 4: thick shell thickness
Calculate the required shell thickness of an accumulator with P = 69 MPa,
R = 45.7 cm, S = 138 MPa, and E = 1.0.
Assume a corrosion allowance of 6 mm.

Solution
The quantity 0.385SE = 53.13 MPa; since this is less than the design pressure
P = 69 MPa, use equation 1.7.

⎛ 1 ⎞ SE + P
t = R ⎜ Z 2 - 1⎟ Where Z =
⎝ ⎠ SE - P
(138 × 1) + 69
Z =
(138 × 1) − 69
207
=
69
= 3
⎛ 1 ⎞
t = ( 457
+ 6 ) ⎜ 3 2 - 1⎟
⎝ ⎠
= 463 × 0.732
= 338.92 mm

Total including corrosion allowance
t = 338.92 + 6
= 344.92 mm (Ans.)



Example 5
Calculate the required shell thickness of an accumulator with P = 52.75 MPa,
R = 45.7 cm, S = 138 MPa, and E = 1.0. Assume corrosion allowance = 0.

Solution
The quantity 0.385SE = 53.13 MPa; since this is greater than the design pressure
P = 52.75 MPa, use equation 1.3.

PR
SE - 0.6P
52.75 × 457
= +0
(138 × 1) - 0.6 ( 52.75 )
24106.75
=
106.35
= 226.67 mm (Ans.)

This example used equation 1.3; compare the answer using equation 1.7
⎛ 1 ⎞ SE + P
t = R ⎜ Z 2 - 1⎟ Where Z =
⎝ ⎠ SE - P
(138 × 1) + 52.75
Z =
(138 × 1) - 52.75
190.75
=
85.25
= 2.2375
⎛ 1
⎞
t = 457 ⎜ 2.2375 2 - 1⎟
⎝ ⎠
= 457 × 0.4958
= 226.59 mm (Ans.)

This shows that the 'simple to use' equation (1.3) is accurate over a wide range of
R/t ratios.



OBJECTIVE 3 2
OBJECTIVE
Calculate the required minimum thickness or the maximum allowable
working pressure of ferrous piping, drums, and headers.

In cylindrical vessels, the stress set up by the pressure on the longitudinal joints is
equal to twice the stress on the circumferential joints.

SECTION I

The following formulae are found in ASME Section I, paragraph PG-27.2.2.

The information for piping, drums, or headers may be given with either the
inside (R) or outside (D) measurement.

Using the outside diameter

PD
t = + C 2.1
2SE + 2 yP

2SE ( t - C )
P = 2.2
D - ( 2 y )( t - C )

Using the inside radius
PR
t = + C 2.3
SE - (1 - y ) P

SE ( t - C )
P = 2.4
R + (1 - y )( t − C )

Example 6: steam piping
Calculate the required minimum thickness of seamless steam piping which carries
steam at a pressure of 6200 kPa gauge and a temperature of 375°C. The piping is
plain end, 273.1 mm O.D. (nominal pipe size of 10 inches) and the material is
SA-335-P11. Allow a manufacturer's tolerance allowance of 12.5%.

Note: Check PG-6 and PG-9 for materials before starting calculations; the
information will direct you to the correct stress table in ASME Section II,
Part D. The material SA-335-P11 is alloy steel.


Note: Plain-end pipe does not have its wall thickness reduced when joining to
another pipe. For example, lengths of pipe welded together, rather than
being joined by threading, are classed as plain-end pipes.

Solution
Use equation 2.1 (See PG-27.2.2.)

PD
t = + C
2 SE + 2 yP

Where
P =
6200 kPa = 6.2 MPa
D =
273.1 mm
C =
0 (see PG-27.4, note 3, 4-inch nominal and larger)
S =
104 MPa (see Section II, Part D, Table 1A,
SA-335-P11 at 375°C)
E = 1.0 (see PG-27.4, note 1, seamless pipe as per PG-9.1)
y = 0.4 (see PG-27.4, note 6, ferritic steel less than 475°C)

6.2 × 273.1
t = + 0
2 (104 × 1) + 2 ( 0.4 × 6.2 )
1693.22
=
208 + 4.96
1693.22
=
212.96
= 7.95 mm

This value does not include a manufacturer's tolerance allowance of 12.5%.

Therefore

7.95 × 1.125 = 8.94 mm (Ans.)

Example 7: steam piping using outside diameter
Calculate the maximum allowable working pressure in kPa for a seamless steel
pipe of material SA-209-T1. The nominal pipe size is 323.9 mm (~12 in. pipe)
with a wall thickness of 11.85 mm. The operating temperature is 450°C. The pipe
is plain ended. Assume that the material is austenitic steel.

Part D. The material SA-209-T1 is alloy steel.



Solution
Use equation 2.2. (See PG-27.2.)

2SE ( t - C )
P =
D - ( 2 y )( t - C )

Where
D = 323.9 mm (see 2005 Academic Supplement, Formulae and
Physical Constants, "Table of Actual Pipe Dimensions.")
t = 11.85 mm
C = 0 (see PG-27.4, note 3, 4-inch (100 mm) nominal and
larger)
S = 101 MPa (Section II, Part D, Table 1A, SA-209-T1 at
450°C)
E = 1.0 (see PG-27.4, note 1, seamless pipe as per PG-9.1)
y = 0.4 (see PG-27.4, note 6, austenitic steel at 450°C)

2 (101 × 1) × (11.85 - 0 )
P =
323.9 - ( 2 × 0.4 ) × (11.85 - 0 )
202 × 11.85
=
323.9 - 9.48
2393.7
=
314.42
= 7.613 MPa
= 7613 kPa (Ans.)

Example 8: drum using inside radius
A welded watertube boiler drum of SA-515-60 material is fabricated to an inside
radius of 475 mm on the tubesheet and 500 mm on the drum. The plate
thickness of the tubesheet and drum are 59.5 mm and 38 mm respectively. The
longitudinal joint efficiency is 100%, and the ligament efficiencies are 56%
horizontal and 30% circumferential. The operating temperature is not to exceed
300°C. Determine the maximum allowable working pressure based on:

(a) the drum
(b) the tubesheet



FIGURE 1
Welded Watertube
Boiler Drum
DRUM

TUBESHEET

Note: This is a common example of a watertube drum fabricated from two
plates of different thickness. Greater material thickness is required where
the boiler tubes enter the drum than is required for a plain drum. For
economy, the drum is designed to meet the pressure requirements for
each situation.

Part D. The material SA-515-60 is carbon steel plate.

Solution
This example has two parts:
a) The drum - consider the drum to be plain with no penetrations.
b) The tubesheet - consider the drum to have penetrations for boiler tubes.

(a) Use equation 2.4 (inside radius R). (See PG-27.2.2.)

SE (t - C )
Drum P =
R + (1 - y )(t - C )

Where
S = 115 MPa (see Section II, Part D, Table 1A,
SA-515-60 at 300°C)
E = 1 (see PG-27.4, note 1)
t = 38 mm
C = 0 (see PG-27.4, note 3, 4-inch (100 mm) nominal
and larger)
R = 500 mm (for the drum)



Drum P =
(115 × 1) (38 - 0)
500 + (1 - 0.4)(38 - 0)
4370
=
500 + 22.8
= 8.36 MPa (Ans.)

Note: In cylindrical vessels, the stress set up by the pressure on the longitudinal
joints is equal to twice the stress on the circumferential joints.

(b) Use equation 2.4 (inside radius R). (See PG-27.2.2.)

SE (t - C )
Tubesheet P =
R + (1 - y )(t - C )

Where
S = 115 MPa (see Section II, Part D, Table A1, SA-515-60 at
300°C)
E = 0.56 (circumferential stress = 30% and longitudinal stress
= 56%; therefore, 0.56 < 2 x 0.30)
T = 59.5 mm
C = 0 (see PG-27.4, note 3, 4-inch (100 mm) nominal and
larger)
R = 475 mm (for the tubesheet)

Tubesheet P =
(115
× 0.56 )( 59.5 - 0 )
475 + (1 - 0.4)(59.5 - 0)
3831.8
=
475 + 35.7
= 7.5 MPa (Ans.)

Note: The maximum allowable working pressure is based on the lowest number.

SECTION VIII-1

Section VIII-1 does not contain separate formulae for small and large bore
cylinders.

The formulae given in paragraph UG-27 are used as set out in Objective 1.



OBJECTIVE 3
Calculate the required thickness or maximum allowable working
pressure of a seamless, unstayed dished head.

Section I: DISHED HEAD CALCULATIONS

The paragraphs from PG-29 must be considered when performing calculations
on dished heads.

Paragraph PG-29.1 states that the thickness of a blank, unstayed dished head
with the pressure on the concave side, when it is a segment of a sphere, shall be
calculated by the following formula:

5PL
t = 3.1
4.8S

Where:
t = minimum thickness of head (mm).
P = maximum allowable working pressure (MPa).
L = radius (mm) to which the head is dished, measured on the
concave side
S = maximum allowable working stress (MPa) (see ASME
Section II, Part D, Table 1A).

Paragraph PG-29.2 states: "The radius to which the head is dished shall be not
greater than the outside diameter of the flanged portion of the head. Where two
radii are used, the longer shall be taken as the value of L in the formula.”

Example 9: the segment of a spherical dished head
Calculate the thickness of a seamless, blank unstayed dished head having
pressure on the concave side. The head has a diameter of 1085 mm and is a
segment of a sphere with a dish radius of 918 mm. The maximum allowable
working pressure is 2500 kPa and the material is SA-285 A. The metal
temperature does not exceed 250°C. State if this thickness meets Code.



Solution
Use equation 3.1. (See paragraph PG-29.1 for segment of a spherical dished
head.)

5PL
t =
4.8S

Where
P = 2.5 MPa
L = 918 mm
S = 88.9 MPa (see ASME Section II, Part D, Table 1A,
SA-285 A at 250°C)

5 ( 2.5 × 918 )
t=
4.8 × 88.9
= 26.89 mm (Ans.)

Note: PG-29.6 states “No head, except a full-hemispherical head, shall be of a
lesser thickness than that required for a seamless shell of the same diameter."

Therefore, to determine if this head thickness meets Code, the thickness of the
shell must be calculated.

Use equation 2.1 (See paragraph PG-27.2.2.)

PD
t = + C
2 SE + 2 yP

Where
D = 1085 mm
E = 1 (welded)

2.5 × 1085
t =
2 ( 88.9 × 1) + 2 ( 0.4 × 2.5 )
2712.5
=
177.8 + 2
= 15.086 mm

Therefore, the head thickness of 26.89 mm meets Code requirements.



Paragraph PG-29.3 states

When a head, dished to a segment of a sphere, has a
flanged-in manhole or access opening that exceeds 150
mm in any dimension, the thickness shall be increased by
15% of the required thickness for a blank head computed
by the above formula, but in no case less than 3.0 mm
additional thickness over a blank head. Where such a
dished head has a flanged opening supported by an
attached flue, an increase in thickness over that for a
blank head is not required. If more than one manhole is
inserted in a head, the thickness of which is calculated by
this rule, the minimum distance between the openings
shall be not less than one-fourth of the outside diameter
of the head.

Note: This applies to the manhole found on the end of a boiler drum.

Example 10: the segment of a spherical dished head with a flanged-in
manhole
Calculate the thickness of a seamless, unstayed dished head with pressure on the
concave side, having a flanged-in manhole 154 mm by 406 mm. The head has a
diameter of 1206.5 mm and is a segment of a sphere with a dish radius of 1143
mm. The maximum allowable working pressure is 1550 kPa, the material is SA-
285-C, and the metal temperature does not exceed 220oC.

Note: Check paragraph PG-44, "Inspection Openings" to see if this manhole
size is acceptable.

Solution
First thing to check: is the radius of the dish at least 80% of the diameter of the
shell? (per paragraph PG-29.5)

dish radius 1143
=
shell diameter 1206.5
= 0.9473
0.9473 > 0.8

Therefore, the radius of this dish meets the criteria.

Use equation 3.1. (See paragraph PG-29.1.)

5PL
t =
4.8S



Where
P = 1.55 MPa
L = 1143 mm
S = 108 MPa (see ASME Section II, Part D, Table 1A: use
250°C since 220°C is not listed; therefore, use the next
higher temperature)

5 (1.55 × 1143)
t =
4.8 (108 )
= 17.088 mm

This thickness is for a blank head. PG-29.3 requires this thickness to be
increased by 15% or 3.0 mm, whichever is greater.

Therefore

17.088 × 0.15 = 2.56 mm

This is less than 3.0 mm, so the thickness must be increased by 3.0 mm

Therefore

Required head thickness = 17.088 + 3.0
= 20.088 mm (Ans.)

Semi-ellipsoidal head

Paragraph PG-29.7 A blank head of a semi-ellipsoidal form in which half the
minor axis or the depth of the head is at least equal to one-quarter of the inside
diameter of the head shall be made at least as thick as the required thickness of a
seamless shell of the same diameter as provided in PG-27.2.2. If a flanged-in
manhole that meets the Code requirements is placed in an ellipsoidal head, the
thickness of the head shall be the same as for a head dished to a segment of a
sphere (see PG-29.1 and PG-29.5) with a dish radius equal to eight-tenths the
diameter of the shell and with the added thickness for the manhole as specified
in PG-29.3.

This rule combines two rules:
1. blank head rule
2. flanged-in manhole rule

A semi-ellipsoidal head is shown in Fig. 2.



FIGURE 2
h = 1/4 D 1/2 r = 1/4 D
Semi-ellipsoidal Head

h
r
D

L

Full-hemispherical head

The following rule applies to drums or headers with a full-hemispherical end.

Paragraph PG-29.11: The thickness of a blank, unstayed, full-hemispherical head
with the pressure on the concave side shall be calculated by the following
formula:

PL
t = 3.2
2 S - 0.2 P

Where
t = minimum thickness of head (mm).
P = maximum allowable working pressure (MPa).
L = radius to which the head was formed (mm)
(measured on the concave side of the head).
S = maximum allowable working stress (MPa)
(Table A1, Section II, Part D).

The above formula shall not be used when the required thickness of the head
given by the formula exceeds 35.6% of the inside radius. Instead, use the
following formula:

⎛ 1 ⎞ 2(S + P)
t = L ⎜ Y 3 - 1⎟ where Y = 3.3
⎝ ⎠ 2S - P

Example 11: full-hemispherical head
Calculate the minimum required thickness (mm) for a blank, unstayed, full-
hemispherical head with the pressure on the concave side. The radius to which
the head is dished is 190.5 mm. Maximum allowable working pressure is 6205
kPa, and the head material is SA-285-C. The average temperature of the header is
300oC.



Solution
Use equation 3.2. (See PG-29.11.)

PL
t =
2S - 0.2 P

Where
P = 6.205 MPa
L = 190.5 mm
S = 107 MPa (see ASME Section II, Part D, Table 1A, SA-285-C at 300oC)

6.205 × 190.5
t =
2 (107 ) - 0.2 ( 6.205 )
1182.05
=
214 - 1.241
1182.05
=
212.759
= 5.56 mm (Ans.)

Check if this thickness exceeds 35.6% of the inside radius:

190.5 × 0.356 = 67.8 mm

It does not exceed 35.6%, therefore

The thickness of the head meets Code requirements.

Paragraph PG-29.12: If a flanged-in manhole that meets the Code requirements
(see PG-44) is placed in a full-hemispherical head, the thickness of the head shall
be the same as for a head dished to a segment of a sphere (see PG-29.1 and
PG-29.5), with a dish radius equal to eight-tenths the diameter of the shell and
with the added thickness for the manhole as specified in PG-29.3.

SECTION VIII-1: DISHED HEAD CALCULATIONS

Sections VIII-1 and VIII-2 each contain rules for the design of spherical shells,
heads, and transition sections. There are significant differences in the equations
due to the different design approaches used. This chapter uses only Section VIII-
1 equations.

Section VIII-1 has rules for head configurations including spherical,
hemispherical, ellipsoidal, and torispherical shapes.



Spherical Shells and Hemispherical Heads

Paragraph UG-27 (d) gives the required thickness of a thin spherical shell due to
internal pressure.

PR
t = 3.4
2SE - 0.2 P

or

2SEt
P = 3.5
R + 0.2t

Where t < 0.356R or P < 0.665SE

For thick shells, where t >0.356R or P > 0.665SE, use Mandatory Appendix 1
sections 1-3.

As the ratio t/R increases beyond 0.356, use the following equations

⎛ 1 ⎞ 2 ( SE + P )
t = R ⎜ Y 3 -1⎟ where Y = 3.6
⎝ ⎠ 2 SE - P

or

⎛ R+t⎞
3
⎛ Y -1 ⎞
P = 2SE ⎜ ⎟ where Y = ⎜ ⎟ 3.7
⎝Y + 2⎠ ⎝ R ⎠

Where t > 0.356R or P > 0.665SE

Example 12: hemispherical head
A pressure vessel is built of SA-516-70 material and has an inside diameter of
2440 mm. The internal design pressure is 690 kPa at 232°C. The corrosion
allowance is 3 mm, and the joint efficiency is 0.85. What is the required thickness
of the hemispherical heads if the allowable stress is 138 MPa?



Solution
The quantity 0.665SE = 78 MPa; since this is greater than the design pressure of
690 kPa, use equation 3.4. (See paragraph UG-32 (f).)
The inside radius in a corroded condition is equal to
R = 1220 + 3 (corrosion allowance)
= 1223 mm
PR
2 SE - 0.2 P
0.69 × 1223
= +3
2 (138 × 0.85) - 0.2 ( 0.69 )
843.87
= +3
234.46
= 3.6 + 3
= 6.6 mm (Ans.)

The calculated thickness is less than 0.356R; therefore, equation 3.3 is acceptable.

Example 13: spherical head
A spherical pressure vessel with an internal diameter of 3048 mm has a head
thickness of 25.4 mm. Determine the design pressure if the allowable stress is
113 MPa. Assume joint efficiency E = 0.85.

Solution
As no corrosion allowance is stated the design pressure will act on the given
internal diameter.
Use equation 3.5 since t is less than 0.356R.

2 SEt
P =
R + 0.2t
2 (113 × 0.85 × 25.4 )
=
1524 + 0.2 ( 25.4 )
4879.34
=
1529.08
= 3.191 MPa (Ans.)

The calculated pressure is less than 0.665SE; therefore, equation 3.4 is
acceptable.



Example 14: thick hemispherical head
Calculate the required hemispherical head thickness of an accumulator with
P = 69 MPa, R = 460 mm, S = 103 MPa, and E = 1.0.
Assume a corrosion allowance of 6 mm.

Solution
The quantity 0.665SE = 68.495 MPa; since this is less than the design pressure of
69 MPa, use equation 3.6.

⎛ 1 ⎞ 2 ( SE + P )
t = R ⎜ Y 3 - 1⎟ where Y =
⎝ ⎠ 2 SE - P

2 (103 × 1 + 69 )
Y =
2 (103 × 1) - 69
344
=
137
= 2.51
⎛ 1 ⎞
t = R ⎜ Y 3 - 1⎟
⎝ ⎠
⎛ 1
⎞
= 460 + 6 ⎜ 2.513 - 1⎟
⎝ ⎠
= 466 ( 0.359 )
= 167.3 mm
This is the minimum thickness i.e. fully corroded state.
Total head thickness is 167.3 + 6 mm (corrosion allowance) = 173.3 mm (Ans.).

Connecting this head to the accumulator shell would require special treatment,
which is outside of the scope of this module.

Ellipsoidal Heads

The commonly used ellipsoidal head has a ratio of base radius to depth of 2:1
(shown in Fig. 3a). The actual shape can be approximated by a spherical radius of
0.9D and a knuckle radius of 0.17D (shown in Fig. 3b.) The required thickness
of 2:1 heads with pressure on the concave side is given in paragraph UG-32
(d).

PD
t = 3.8
2SE - 0.2 P
or
2 SEt
P = 3.9
D + 0.2t


Where
D = inside base diameter
E = joint efficiency factor
P = pressure on the concave side of the head
S = allowable stress for the material
t = thickness of the head

FIGURE 3
Ellipsoidal Head

(a)

(b)

Section VIII-1 does not give any P/S limitations or rules for ellipsoidal heads
when the ratio of P/S is large.

Torispherical Heads

Shallow heads, commonly referred to as flanged and dished heads (F&D heads),
can be built according to paragraph UG-32 (e). A spherical radius L of 1.0D and
a knuckle radius r of 0.06D, as shown in Fig. 4, approximates the most common
F&D heads.



FIGURE 4
Torispherical Head

The required thickness of an F&D head is

0.885PL
t = 3.10
SE - 0.1P
or
SEt
P = 3.11
0.885L + 0.1t

Where
E = joint efficiency factor
L = inside spherical radius
P = pressure on the concave side of the head
S = allowable stress
t = thickness of the head

Shallow heads with internal pressure are subjected to a stress reversal at the
knuckle. This stress reversal could cause buckling of the shallow head as the ratio
D/t increases.



OBJECTIVE 4
Calculate the minimum required thickness or maximum allowable
working pressure of unstayed flat heads, covers, and blind flanges.

UNSTAYED FLAT HEADS, COVERS, AND BLIND
FLANGES

Flat plates, covers, and flanges are used extensively in boilers and pressure
vessels. When a flat plate or cover is used as an end closure or head of a pressure
vessel, it may be designed as an integral part of the vessel (having been formed
with the cylindrical shell) or welded to it. Alternately, it may be a separate
component that is attached by bolts or some quick-opening mechanism utilizing
a gasket joint attached to a companion flange on the end of the shell. Bolted
flanges are not covered in the scope of this module.

The concepts of unstayed flat heads, covers, and especially blind flanges are
often misunderstood and can be challenging to anyone learning and working on
this type of equipment. It is very important for power engineers to have good
working knowledge of thickness requirements as this allows them to work safely
and provide sound and safe advice.

SECTION 1

Paragraph PG-31.1 states that the minimum thickness of unstayed flat heads,
cover plates, and blind flanges shall conform to the requirements. Paragraph PG-
31.2 defines the notations used in this paragraph and in Fig. PG-31. Paragraph
PG-31.3 states formulae for calculating the minimum thickness of flat, unstayed
circular heads, covers, and blind flanges.

When the circular head, cover, or blind flange is attached by welding

CP
t = d 4.1
S

When the circular head, cover, or blind flange is attached by bolts
(Fig. PG-31 (j), (k))



CP 1.9Whg
t = d + 4.2
S Sd 3

Note: W = the total bolt loading and hg = the gasket moment arm. The gasket
moment arm is the radial distance from the centre line of the bolts to the
line of the gasket reaction force (Fig. PG-31 (j), (k)).

When using equation 4.2, the thickness t shall be calculated for both design
conditions (flange sketches j and k) and the greater value used.

Note: The formulae used to determine thickness may be transposed to solve for
P and find the maximum allowable working pressure for a flat head or
cover of known thickness.

Paragraph PG-31.3.3 states two formulae for the required thickness of flat
unstayed heads, covers, or blind flanges that are square, rectangular, elliptical,
obround, or segmental in design and attached by welding.

ZCP
t = d 4.3
S

Where Z is a factor from the ratio of the short and long spans
2.4d
Z = 3.4 - to a maximum of 2.5
D

When the non-circular head, cover, or blind flange is attached by bolts (Fig. PG-
31. (j), (k))

ZCP 6Whg
t = d + 4.4
S SLd 2

Paragraph PG-31.4 lists the values for C to be used in the formulae 4.1, 4.2, 4.3,
and 4.4.



Example 15: circular flat head welded to a shell
(Illustrated by Fig. PG-31 (e) and Fig. 5.)

FIGURE 5
Circular Flat Head

Calculate the minimum thickness for the circular head and the depth of the fillet
welds required. The material for head and shell is SA-285-A. The shell is
seamless. The thickness t is 19 mm. Maximum allowable working pressure is
2500 kPa. Shell’s inside diameter d is 762 mm. Head joint welding meets Code
requirements.

Solution
Use equation 4.1

CP
t = d
S
Where
P = 2.5 MPa
d = 762 mm
S = 88.9 MPa (ASME Section II, Part D, Table 1A)
As no temperature is given, the saturation temperature of
steam (224°C at 2500 kPa) may be used; therefore, use
the value for 250°C.
C = 0.33 m (see PG-31.4, Fig PG-31 sketch (e), where m is
defined as the ratio of tr/ts from paragraph PG-31.2)
tr = required minimum thickness of the shell
ts = actual thickness of the shell as given
Use equation 2.3 to find the value of tr (see paragraph PG-27.2.2).

PR
t = + C
SE - (1 - y ) P



Where
R = d/2 = 381 mm
E = 1 (see PG-27.4, note 1)
y = 0.4 (see PG-27.4, note 6)
C = 0 (see PG-27.4, note 3)
2.5 × 381
tr = + 0
(88.9 × 1) - (1 - 0.4 ) × 2.5
952.5
=
87.4
= 10.898 mm

Therefore
tr
m =
ts
10.898
=
19
= 0.574

C = 0.33 m (from PG - 31.4)
= 0.33 × 0.574
= 0.19

As this value is less than 0.2, use 0.2 in the formula from PG-31.4 or in
equation 4.1.

CP
t = d
S
0.20 × 2.5
= 762
88.9
= 762 × 0.0750
= 57.15 mm (Ans.)

For a welded circular flat head (Fig PG-31 (e)), a minimum thickness of 57.15
mm is required.

The depth of each weld would be 0.7 ts (see Fig PG-31 (e)).

ts = 19 mm (given)
= 0.7 × 19
= 13.3 mm



It is interesting to note that the required minimum shell thickness is 10.898 mm,
yet the required minimum thickness of the blank head is approximately 5.2 times
thicker at 57.15 mm.

Example 16: circular flat head maximum allowable working pressure
Calculate the maximum allowable working pressure for a circular flat head with
the following specifications. Head design to Fig. PG-31, sketch (d). Shell and
head thickness of 30.5 mm. Material is SA-285-B. Head joint weld meets Code
requirements. Shell diameter is 610 mm. Operating temperature not to exceed
300°C.

Solution
t = 30.5 mm
S = 96.6 MPa (see ASME Section II, Part D, Table 1A)
d = 610 mm
C = 0.13 (see Fig. PG-31 (d))

Use equation 4.6. (See PG-32.3.2.)

t 2S
P = 4.6
d 2C

30.52 × 96.6
P =
6102 × 0.13
= 1.858 MPa (Ans.)

The maximum allowable working pressure for this flat, unstayed head is
1858 kPa.

SECTION VIII-1

The equations for the design of unstayed plates and covers are found in
paragraph UG-34.

CP
t = d 4.7
SE

Where
d = effective diameter of the flat plate (mm)
C = coefficient between 0.1 and 0.33 (depending on the
corner details as shown in Fig. UG-34)
P = design pressure
S = allowable stress at design temperature



E = butt-welded joint efficiency of the joint within the flat
plate
t = minimum required thickness of the flat plate

The value of E depends on the degree of non-destructive examination
performed. E is not a weld efficiency value of the head to shell corner joint.

Example 17: integral flat plate
Using the rules of paragraph UG-34, determine the minimum required thickness
of an integral flat plate with an internal pressure P = 17 MPa, an allowable stress
S = 120 MPa, and a plate diameter d = 610 mm. There are no butt weld joints
within the head. There is a corrosion allowance of 4 mm. The corner detail
conforms to Fig. UG-34 (b-2) (assume that m = 1).

Solution
Use equation 4.7. (See Fig UG-34 (b-2))

Where
C = 0.33(m) = 0.33(1) = 0.33
d = 610 + 4 = 614 mm (fully corroded state)

CP
t = d + corrosion allowance
SE
0.33×17
= 614 × +4
120 ×1
= 136.76 mm (Ans.)



OBJECTIVE 5
Calculate the acceptability of openings in a cylindrical shell, header,
or head.

Openings through the pressure boundary of a vessel require extra care to keep
loading and stresses at acceptable levels. An examination of the pressure
boundary may indicate that extra material is needed near the opening to keep
stresses at acceptable levels. This extra material may be provided by increasing
the wall thickness of the shell or nozzle or by adding a reinforcement plate
around the opening.

The stress analysis basis used in the ASME Codes to analyze nozzle
reinforcement is called Beams on Elastic Foundation (Hetenyi, 1946). Although the
methods used are a simplified application of the elastic foundation theory,
experience has shown that they are acceptable.

ASME Codes Section I and Section VIII give two methods for examining the
acceptability of openings in the pressure boundary for pressure loads only. The
first method, called the reinforced opening or area replacement method is
used when nearby substitute areas replace the area removed by the opening. The
second method is the ligament efficiency method. This method determines the
effectiveness of the material between adjacent openings to carry the stress
compared with the area of metal that was there before the openings existed.
Curves have been developed to simplify this examination. For single openings,
only the area replacement method is used. For multiple openings, either method
may be used.

Since stress is related to load and cross-sectional area, areas are substituted when
making calculations. Placement and location of the replacement area are very
important. Equations have been developed to set the limits for reinforcement.

Reinforcement limits are developed parallel and perpendicular to the shell
surface from the opening.



Figure 6
Reinforcement Limits Rn
tn
t rn
A B
smaller of
2.5t or WL1
2.5tn + t e tr

t

smaller of
d
2.5t or
2.5tn ABCD = Limits of reinforcement

C d or d or D
Rn + t n + t Rn + t n + t
Use larger value Use larger value

When an opening is cut into a vessel wall for the attachment of a nozzle with
diameter d (as in Fig. 6), the vessel wall thickness t is usually thicker than the
minimum thickness required tr. The area (tr x d) is the cross-sectional area that is
removed and has to be compensated for. ASME Section I, paragraph PG-36
(ASME Section VIII, paragraph UG-40) gives the rules for the “Limits of Metal
Available for Compensation." The limit is shown by box ABCD in Fig. 6 above.

If greater than the cross-sectional area removed, the additional material in the
shell wall and the additional material in the nozzle wall (the hatched cross-
sectional area shown in Fig. 6 within the limit of compensation boundary) may
provide adequate compensation.

SECTION I

ASME Section I, paragraph PG-32 "Openings in Shells, Headers and Heads"
contains rules to be applied to maintain the vessel pressure boundary. Paragraph
PG-32.1.1 states that paragraphs PG-32 to PG-39 shall apply to all openings
(except for flanged-in manholes as stated in paragraph PG-29) and to tube holes
in a definite pattern that are designed according to paragraph PG-52. Paragraph
PG-32.1.2 provides the rules for openings that do not require reinforcement
calculations, providing the diameter of the opening does not exceed that
permitted by the chart in Fig. PG-32.



To determine if reinforcement is required, the value K is calculated from the
formula

PD
K = 5.1
1.82St

Using the chart in Fig. PG-32, the value for the x-axis is calculated from the shell
diameter times the shell thickness. The point where the x-axis value meets the K
value curve is read to the y-axis and gives the maximum diameter of the opening,
allowed without reinforcement.

Example 18: reinforcement of nozzle abutting vessel
Determine if reinforcement is required for a 100 mm I.D. nozzle located in a
cylindrical boiler shell. The nozzle abuts the vessel wall and is attached by a full-
penetration weld. The O.D. of the shell is 1000 mm. The thickness of the shell
wall is 25.4 mm. The thickness of the nozzle wall is 10 mm. The shell material is
SA-515-60 and the nozzle material is SA-192. The maximum allowable working
pressure is 4500 kPa, and the design temperature is not to exceed 200°C. All
joint efficiencies E = 1.0.

Solution
As this is a boiler shell, ASME Section I rules apply. (See PG-32.1.2.)
Use equation 5.1 to calculate the K value.

PD
K =
1.82St

Where
P = 4.5 MPa
D = 1000 mm
S = 118 MPa
t = 25.4 mm

PD
K =
1.82St
4.5 × 1000
=
1.82 (118 × 25.4 )
= 0.825

Using Fig. PG-32, calculate the x-axis value.

Shell diameter × shell thickness = 1000 × 25.4
= 25400 mm 2



The intersection of the x-axis value (2540) and the K value curve (0.825) give a y-
axis value of approximately 134 mm.

Therefore, no additional reinforcement is required (Ans.) for an opening of
100 mm diameter.

SECTION VIII-1

Section VIII-1 requires all openings in pressure vessels, not subjected to rapid
fluctuations, to use reinforcement calculations in paragraph UG-37, unless
certain dimensional requirements are met as listed in paragraph UG-36(c)(3).

Determine the reinforcement requirements for a 60 mm I.D. nozzle located in a
cylindrical shell. The nozzle abuts the vessel wall and is attached by a full-
wall is 25.4 mm, and the thickness of the nozzle wall is 10 mm. The shell material
is SA-516-60 and the nozzle is SA-192. The maximum allowable working
pressure is 4500 kPa, and the design temperature is not to exceed 200°C. All
joint efficiencies E = 1.0

Solution
As this is a not a boiler shell, ASME Section VIII-1 rules apply.
(See UG-36(c)(3).)

UG-36(c)(3) states that reinforcement is not required if
(a)the opening is not larger than 89 mm diameter and the shell is 10 mm thick or
less; or (b) the opening is not larger than 60 mm diameter and the shell thickness
is greater than 10 mm.

In this example, the nozzle diameter is 60 mm

This falls within the second condition, i.e. not larger than 60 mm in a shell that is
thicker than 10 mm.

Therefore, no reinforcement is required (Ans.).



OBJECTIVE 6
Calculate the compensation required to reinforce an opening in a
cylindrical shell, header, or head.

SECTION I

ASME Section I, paragraph PG-33, "Compensation required for openings in
shells and formed heads", states the rules for compensation. Paragraph PG-33.2
states that the total cross-sectional area of compensation required in any given
plane for a vessel under internal pressure shall not be less than A as defined in
PG-33.1, shown in Fig. 7.

For an opening in a shell with a nozzle abutting the shell wall (such as an
opening for a safety valve), the requirements are illustrated in Fig. 7.
FIGURE 7
Nozzle Wall Abutting
Vessel Wall
Dp

tn Rn
tr n
rn
A B
WL1 te
smaller of WL2
2.5t or
2.5tn + t e tr

t

smaller of d
2.5t or
2.5tn ABCD = Limits of reinforcement

C d or d or D
Rn + t n + t Rn + t n + t
Use larger value Use larger value
NOZZLE WALL ABUTTING VESSEL WALL



Where

(a) The area to be replaced A (shown as the cross-hatched area)
= dtrF
where F is taken from the chart Fig. PG-33

(b) The area in the shell wall thickness available to be used as
compensation A1
(shown as the forward sloped hatched areas on either side of the opening)
= the larger of d(t – Ftr) or 2(t + tn)(t – Ftr)

(c) The area in the nozzle wall thickness available to be used as
compensation A2
(shown as the backward sloped hatched area on either side of the nozzle)
= the smaller of 2(tn – trn)(2.5tfr1) or 2(tn – trn)(2.5tn + te)fr1
where fr1 is the ratio of Snozzle/Sshell

(d) The area available from the nozzle to the reinforcement plate welds A41
= (WL1)2 × fr1
where fr1 is the ratio of the lesser of Snozzle or Splate / Sshell

(e) The area available from the reinforcement plate to shell weld A42
= (WL2)2fr3

(f) The area available in the reinforcement plate (shown as herring-bone
brick hatch) A5
= (Dp – d – 2tn)te/fr3
`Where fr3 is Splate/Sshell

If A1 + A2 + A41 > A
The opening is adequately reinforced.

If A1 + A2 + A41 < A
The opening is not adequately reinforced, and reinforcing elements
(reinforcement plate and welds) must be added and/or the thickness must be
increased.

Therefore, if A1 + A2 + A41 +A42 + A5 > A
The opening is adequately reinforced.

Determine the reinforcement requirements for a 100 mm I.D. nozzle located in a
cylindrical boiler shell. The nozzle abuts the vessel wall and is attached by a full-
wall is 22.5 mm and the thickness of the nozzle wall is 8 mm. The nozzle fillet
welds are 5 mm wide. The shell material is SA-516-60 and the nozzle is SA-192.
The maximum allowable working pressure is 4900 kPa, and the design



temperature is not to exceed 200°C. All joint efficiencies E = 1.0. The
reinforcement plate (if required) shall be of SA-192 material and 18 mm thick.

Solution
As this is a boiler shell, ASME Section I rules apply. Use equation 5.1.
(See PG-32.1.2.)

PD
K =
1.82St

Where
P = 4.9 MPa
D = 1000 mm
S = 118 MPa
t = 22.5 mm
4.9 × 1000
K =
1.82 (118 × 22.5 )
= 1.014

ASME Section I, Fig. PG-32, "General Notes," states that K is limited to a value
of 0.99. Therefore PG-32.1.2 cannot be used.

Allowable tensile stress for SA-516-60 is 118 MPa and for SA-192 is 92.4 MPa.

Therefore:
118
f r1 =
92.4
= 1.28

Use equation 2.3 to determine the minimum required shell thickness (additional
thickness may be used towards reinforcement requirements). (See PG-27.2.2)

Where
P = 4.5 MPa
R = 500 – 22.5 = 477.5 mm
S = 118 MPa
E = 1
y = 0.4 (see PG-27.4, note 6)
C = 0 (see PG-27.4, note 3)



PR
tr = +C
SE - (1 - y ) P
4.5 × 477.5
= +0
(118 × 1) - (1 - 0.4)4.5
= 20.408 mm
Therefore
tr = 20.408 mm and t = 22.5 mm

Use equation 1.1 to determine the minimum required nozzle thickness.
(See PG-27.2.1)

Where
P = 4.5 MPa
D = 100 + (2 x 8) = 116 mm
S = 92.4 MPa
e = 0 (see PG-27.4, note 4)

PD
t = + 0.005D + e
2S + P
4.5 × 116
= + 0.005 (116 ) + 0
2 ( 92.4 ) + 4.5
522
= + 0.58
189.3
= 3.3375 mm

Therefore
tr n = 3.3375 mm and tn = 8 mm

Limit of compensation parallel to shell surface
X = d or X = (0.5d + tn + t), whichever is larger
X = 100 or X = (0.5 ×100 + 8 + 22.5) = 80.5
Therefore
X = 100 mm

Limit perpendicular to the shell surface
Y = 2.5t or Y = (2.5tn + te), whichever is smaller
Y = 2.5 × 22.5 = 56.25 or Y = (2.5 × 8 + 18) = 38
Therefore
Y = 38 mm

(a) Reinforcement area required A (according to Fig. PG-33.1)
A = dtrF (where F is taken from the chart Fig. PG-33.3, F=1)
Ar = 100 ×20.408 ×1 = 2040.8 mm2



(b) Reinforcement area available in the shell
(X replaces d in the equation)
A1 = X(t – Ftr)
A1 = 100(22.5 – 1 x 20.408) = 209.2
Therefore
A1 =209.2 mm2

(c) Reinforcement area available in the nozzle
Y replaces (2.5tn + te) in the equation
A2 = 2(tn - trn)(Y)fr1
A2 = 2(8 - 3.3375)(38) × 1.17 = 414.59
Therefore
A2 = 414.59 mm2

(d) Reinforcement area available in the nozzle weld
A41 = (WL1)2fr2 where fr2 = Sn/Ss
A41 = (5)2 × 92.4/118 = 19.58
Therefore
A41 = 19.58 mm2

Total area available from shell, nozzle, and nozzle weld
Ar = A1 + A2 + A41
Ar = 209.2 + 414.59 + 19.58 = 643.37 mm2

(e) Area provided by the reinforcement plate weld
A42 = (WL2)2Fr3
A42 = (5)2 × 92.4/118 = 19.58
Therefore
A42 = 19.58 mm2

Area required by reinforcement pad
A5 = A – (Ar + A42)
A5 = 2040.8 - (643.37 + 19.58) = 1377.85
Therefore
A5 = 1377.85 mm2

(f) Diameter of the reinforcement pad
A5 = ( D p - d - 2tn )te f r 3
92.4
1377.85 = ( D p - 100 - 2 × 8) × 18 ×
118
1377.85
( D p - 100 - 16) =
14.095
Dp = 97.75 + 100 + 16
= 213.75 mm ( Ans.)



Thus, a reinforcing pad 213.75 mm diameter and 18 mm thick is required to
carry the tensile stress and maintain the vessel pressure boundary. This pad size
falls within the limits of compensation.

SECTION VIII-1

The limits of compensation stated in paragraph UG-40 (b) and (c) are the same
used in Section I, except that the vessel shell and nozzle must be treated as being
in a corroded condition.

Therefore, the limit of compensation parallel to the shell surface

X = diameter of the finished opening in corroded condition
Or
X = radius of the finished opening in corroded condition + shell
wall thickness+ nozzle wall thickness

Whichever is larger

The limit of compensation normal to the shell surface

Y = 2.5 × nominal shell thickness less the corrosion allowance
Or
Y = 2.5 × nozzle wall thickness + the thickness of the
reinforcing plate (te)

Whichever is smaller



CHAPTER QUESTIONS

The following questions provide candidates with experience using the ASME
Codes.

1. Calculate the minimum required wall thickness of a watertube boiler tube 75
mm O.D. that is strength welded in place in a boiler drum. The tube will be
in the furnace area of the boiler and have an average wall temperature of
350°C. The maximum allowable working pressure is 3500 kPa. The tube
material is SA-192.

2. Calculate the required shell thickness for a hydraulic cylinder with a design
pressure of 62 000 kPa. The cylinder has an internal diameter of 36 cm,
S = 142 MPa, and E = 1.0. Assume no corrosion allowance for this cylinder.

3. Calculate the thickness of a boiler steam header designed with a seamless,
unstayed, full hemispherical head, with pressure on the concave side. The
inside radius of the header and the radius to which the head is dished is 304
mm, MAWP is 6205 kPa, and the header and head material is SA-204-A. The
average temperature of the header is 400°C. The header has a flanged-in
circular inspection opening 100 mm diameter.

4. An air receiver pressure vessel is constructed from SA-204-A with an inside
diameter of 1830 mm. The design pressure is 1034 kPa at 200°C. The
corrosion allowance is 4 mm, and the joint efficiency is 0.85. What is the
required thickness of the hemispherical heads if the allowable stress is 147.5
MPa?

5. Using the rules in Section VIII-1, determine the minimum required thickness
of the flat end plate of a rectangular box header 200 mm by 400 mm with an
internal pressure of 2500 kPa. The material used has a stress value of 103
MPa. The plate is integrally welded into place as per Fig UG-34(h). There is
no corrosion allowance and no butt-welded joints in the plate.

6. Using the rules in Section I, calculate the reinforcement requirements for a
150 mm I.D. nozzle located in a cylindrical boiler drum. The nozzle abuts the
vessel wall and is attached by a full-penetration weld. The I.D. of the drum is
780 mm. The thickness of the drum is 28.575 mm. The nozzle wall thickness
is 35 mm. The drum material is SA-516-60, and the nozzle material is SA-
209-T1. The maximum allowable working pressure is 6000 kPa, and the
design temperature is 250°C. All joint efficiencies E = 1. The reinforcement
plate material (if required) is of SA-515-55 and 10 mm thick.


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