API_510_for_Exam.pdf

API 510 - for Exam
Prepared by: DSc PhD Dževad Hadžihafizović (DEng)
Sarajevo 2023

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API 510
Current for 2014 Exam

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API 510 Examination Preparatory
Fast Track Technical
Table of Contents
Lesson 1 Service Restrictions/Joint Efficiencies/Radiography 5
Lesson 2 Shell and Head Calculations 29
Lesson 3 Maximum Allowable Working Pressure 43
Lesson 4 Hydrostatic Head Pressure 45
Lesson 5 Hydrostatic, Pneumatic Tests and Test Gauges 53
Lesson 6 Postweld Heat Treatment 57
Lesson 7 External Pressure Calculations 65
Lesson 8 Charpy Impact Testing 75
Lesson 9 Fillet Welds and Reinforcement Calculations 91
Lesson 10 Materials, Nameplates, and Data Reports 101
Lesson 11 Corrosion Calculations 105
Lesson 12 ASME Section IX Overview 115
Lesson 13 Writing a Welding Procedure Specification 143
Lesson 14 Welder Performance Qualification Test 175
Lesson 15 Review of WPS’s and PQR’s 195
Lesson 16 Section V NDE 217
Lesson 17 RP 577 Welding Inspections and Metallurgy 251
Lesson 18 RP 571 Damage Mechanisms 279
Lesson 19 API 510 Pressure Tests and Welded Repairs 295
Lesson 20 API 572 Inspection of Pressure Vessels 305

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Introduction
The API Exams are given the first Wednesday of December and June. There is now a trial examination
being given in September, which coincides with the API 653. This September exam may or may not
become permanent.
The Exam is given in two sections.
 The morning is the Open Book portion.
 The afternoon portion is Closed Book.
 4 hours are allotted to each.
The Open Book portion you are allowed to have all the Code books and the API Recommended
Practices to use as needed.
These books can be highlighted, tabbed and may have hand written notes in the margins of the pages.
The Closed Book portion you are not allowed to use the Code books or the API Recommended
Practices.
All questions are multiple choices; there are a total of 150 questions, each worth 2/3 point.
The Open Book half will have from 45 to 55 questions.
The balance of the questions will be on the second half Closed Book portion in the afternoon.
The examination is scored using a curve. The number of correct questions to pass varies from exam to
exam.
A passing score is based on the difficulty level of questions from one exam to another. Each question
when written is assigned a level of difficulty from 1 to 4, 4 being the hardest.
If a given exam has a high number of harder questions, it may only require 98 correct to pass, likewise
easier exams may require as many as 115 correct.

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ASME CODES OVERVIEW
Section VIII Div.1 Unfired Pressure Vessels.
Section VIII is divided into 3 Subsections.
Subsection A
General Rules which apply to all vessels
Subsection B Methods of Fabrication
Specific Rules based on method (s) used
Subsection C
Specific Rules Based on Material (s)
On the Exam you are responsible for:
Subsection A
Only Part of the UG rules that apply to all vessels no matter how they are made or what they are
made of.
Subsection B
Only Part of the UW rules for vessel fabricated by welding.
Subsection C
Only Part of the UCS rules for vessel made of Carbon or Low Alloys.
About the course
• The course will start with Section VIII Subsection B Methods of Fabrication Part UW, Welding.
• Next we will cover Part UG General Rules and calculations.
• Then we will cover the material specific rules for Carbon and Low Alloy Steels Part UCS.
• Upon completion of the Section VIII coverage we will commence Section IX Welding.
• After Section IX we will be covering Section V NDE.
• Then Selected Coverage from, RP 577 and RP 571 and API 510.
• There will be quizzes throughout the course.
• There will be homework reading assignments.
• There will be exercises to complete between classes.
• There will be a final exam.
• You will receive 3 Practice Exams each with 150 questions and the solutions sheets for self study.
• You need Yellow highlighters.
• You need tabs.
• A red pen or pencil (optional).
• The KIS (Keep It Simple) principle applies. You need a simple calculator it need not have any
math function higher than square root.
• You may want a backup calculator during the exam.

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Lesson 1 Service Restrictions, Joint Efficiencies, and Radiography
Objectives
• Understand the service restrictions placed on weld joints based on service conditions.
• Identify weld joints by Categories (location in vessel).
• Identify welds by Types. (How made, double welded etc.).
• Determine the accept/reject values for weld imperfections located using radiography.
• Define the extent of radiography required by Code for a desired joint efficiency.
• Find weld joint efficiency (E) by using Table UW-12.
• Determine weld joint efficiencies based on RT markings.
• Determine the E to be used for calculating the required thickness or allowed pressure for
Seamless Shell sections and Seamless heads.
• Understand the rules for using welded pipe and tubing.

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ASME Section VIII
UW-2 Service Restrictions
(a) When vessels are to contain lethal substances footnote
1
, either liquid or gaseous, all butt welded
joints shall be fully radiographed, except under the provisions of UW-2(a)(2) and UW-2(a)(3) below, and
UW-11(a)(4).
When fabricated of carbon or low alloy steel, such vessels shall be postweld heat treated.
Footnote
1
When a vessel is to contain fluids of such a nature that a very small amount mixed or
unmixed with air is dangerous to life………..
If determined as lethal, …
(1) The joints of various categories (see UW-3) shall be as follows.
(a) Except under the provisions of (a)(2) or (a)(3) below, all joints of Category A shall be
Type No. (1) of Table UW-12.
(b) All joints of Categories B and C shall be Type No. (1) or No. (2) of Table UW-12.
These are the only two types which are considered acceptable for radiography by Section VIII Div.1
Type 1
Double Welded butt joint or equivalent. Backing if used must be removed.
Type 2
Single welded butt joint with backing which remains in place.

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UW-3 Welded Joint Category
The term “Category” as used herein defines the location of a joint in a vessel, but not the type of
joint.
(a) Category A. Longitudinal welded joints within the main shell, communicating chambers,
transitions in diameter, or nozzles; any welded joint within a sphere, within a formed or flat
head, or within the side plates of a flat-sided vessel; circumferential welded joints connecting
hemispherical heads to main shells, to transitions in diameters, to nozzles, or to
communicating chambers.
(b) Category B. Category B. Circumferential welded joints within the main shell, communicating
chambers,
2
nozzles, or transitions in diameter including joints between the transition and a
cylinder at either the large or small end; circumferential welded joints connecting formed
heads other than hemispherical to main shells, to transitions in diameter, to nozzles, or to
communicating chambers.
2
Circumferential welded joints are butt joints if the half-apex
angle, a, is equal to or less than 30 deg and the angle joints when a is greater than 30 deg.
(See Fig. UW-3.)
(c) Category C. Welded joints connecting flanges, Van Stone laps, tubesheets, or flat heads to
main shell, to formed heads, to transitions in diameter, to nozzles, or to communicating
chambers any welded joint connecting one side plate to another side plate of a flat sided
vessel.
(d) Category D. Welded joints connecting communicating chambers or nozzles to main shells, to
spheres, to transitions in diameter, to heads, or to flat-sided vessels, and those joints
connecting nozzles to communicating chambers (for nozzles at the small end of a transition
in diameter, see Category B).
An important note:
Hemispherical heads form a Category A joint between themselves and any other part, whether it is the
shell, another hemispherical head, etc. Hemispherical Heads are never considered Seamless by Code
rules. The Category A weld made by attaching the Hemispherical Head to shell is considered part of the
Head for calculation purposes. Later on in this lesson we begin our discussion of formed seamless Heads.
The formed heads on the exam that are considered seamless are Torispherical and Ellipsoidal,
Hemispherical is not seamless by Code.

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UW-51
Radiographic and Radioscopic Examination of Weld Joints
(a) All welded joints to be radiographed shall be examined in accordance with Article 2 of Section V except
as specified below.
(1) A complete set of radiographs and records, as described in Article 2 of Section V, for each vessel or
vessel part shall be retained by the Manufacturer, as follows:
(a) Films until the Manufacturer's Data Report has been signed by the Inspector;
(b) Records as required by this Division (10-13).
(2) A written radiographic examination procedure is not required. Demonstration of density and
penetrameter image requirements on production or technique radiographs shall be considered
satisfactory evidence of compliance with Article 2 of Section V.
(3) The requirements of T-285 of Article 2 of Section V are to be used only as a guide. Final acceptance of
radiographs shall be based on the ability to see the prescribed penetrameter image and the specified
hole or the designated wire of a wire penetrameter.

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(b) Indications shown on the radiographs of welds and characterized as imperfections are unacceptable
under the following conditions and shall be repaired as provided in UW-38, and the repair radiographed
to UW-51 or, at the option of the Manufacturer, ultrasonically examined in accordance with the method
described in Appendix 12….
(1) Any indication characterized as a crack or zone of incomplete fusion or penetration ;
(2) Any other elongated indication on the radiograph which has length greater than:
(a) 1/4 in. for t up to 3/4 in.
(b)1/3t for t from 3/4 in. to 2-1/4 in.
(c) 3/4 in. for t over 2-1/4 in.
Where;
t = the thickness of the weld excluding any allowable reinforcement.
For a butt weld joining two members having different thicknesses at the weld, t is the thinner of these two
thicknesses. Since the value of t must be the lesser thickness this decreases the size of the maximum
acceptable indication.

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(3) Any group of aligned indications that have an aggregate (total) length greater than t in a length
of 12t,
Example: t = 1” total length (L) cannot exceed 1” in 12”
Also individual lengths cannot exceed the following:
(b) 1/3t for t from 3/4 in. to 2-1/4 in. * In this example none of the individual indications can
exceed 1/3 x 1” = 1/3” (.333”)
(3) Except when the distance between the successive imperfections exceeds 6L where L is the
length of the longest imperfection in the group; * This means that if the two groups are
isolated from each other, they can be evaluated separately within a length of 12t.
(4) Rounded indications in excess of that specified by the acceptance standards given in
Appendix 4.
Example from Appendix 4: More on this during the Section V Coverage.

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UW-52
Spot Examinations of Weld Joints
(a). Butt welded joints that are to be spot radiographed shall be examined locally as provided
herein.
(b) Minimum Extent of Spot Radiographic Examination
(1) One spot shall be examined on each vessel for each 50 ft increment of weld or fraction thereof
for which a joint efficiency from column (b) of Table UW-12 is selected. However, for identical
vessels, each with less than 50 ft of weld for which a joint efficiency from column (b) of Table
UW-12 is selected, 50 ft increments of weld may be represented by one spot
examination.
* The idea of this rule is that each 50’ increment is to be a hold point for approval; the next increment is
not to be started until the previous one has been accepted. The drawing below is the simplest case;
you will not see this often.
* This rule also addresses smaller, often machine welded vessels such as small air receivers. One is
picked at random for spot radiography. If it passes, all are approved.
18' fraction
18' fraction 18' fraction
50'
increment
50'
increment
25' fraction

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(2) For each increment of weld to be examined, a sufficient number of spot radiographs shall be
taken to examine the welding of each welder or welding operator. Under conditions where two
or more welders or welding operators make weld layers in a joint, or on the two sides of a
double-welded butt joint, one spot may represent the work of all welders or welding operators.
* Every welder in a given 50’ increment must have his work radiographed. It can be a individual photo
(radiograph) or a group picture. Here welder A was radiographed alone and B & C’s work was
examined on the same radiograph.
(3) Each spot examination shall be made as soon as practicable…... The location of the spot shall
be chosen by the Inspector,… except that when the Inspector cannot be present or otherwise
make the selection, the fabricator may exercise his own judgment in selecting the spots.
(4) Radiographs required at specific locations to satisfy the rules of other paragraphs, such as
UW-9(d), UW-11(a)(5)(b), and UW-14(b), shall not be used to satisfy the requirements for spot
radiography.
Note: UW-11(a)(5)(b), will be covered in depth later in this lesson.
UW-9(d)
(d) Except when the longitudinal joints are radiographed 4 in. each side of each circumferential
welded intersection, vessels made up of two or more courses shall have the centers of the
welded longitudinal joints of adjacent courses staggered or separated by a distance of at least
five times the thickness of the thicker plate.
* Longitudinal Welds Aligned must be radiographed for at least 4 inches on each side of the joint.
UW-14(b) Single openings meeting the requirements given in UG-36(c)(3) may be located in head-to-shell
or Category B or C butt welded joints, provided the weld meets the radiographic requirements
in UW-51 for a length equal to three times the diameter of the opening with the center of the
hole at mid-length. Defects that are completely removed in cutting the hole shall not be
considered in judging the acceptability of the weld. ** UW-51, not 52 to grade film.
50' increment Welder B & C on opposite sides
of the weld.
Welders
B & C
Welder A
Alone

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3.5" x 3 = 10.5"
* UG-36 (c)(3) addresses small opening which do not require reinforcement calculations.
Summary
The special radiography requirements given in UW-9 (d), UW-11(a)(5)(b) and UW-14 (b) cannot be
substituted for any of the spot radiography required by UW-52.
* We will see why this is significant when we commence our studies of “Joint Efficiencies” later.

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UW-52
Spot Examinations of Weld Joints
(c) Standards for Spot Radiographic Examination.
Spot examination by radiography shall be made in accordance with the technique prescribed in UW-
51(a). The minimum length of spot radiograph shall be 6 in.
(c)(3) Rounded indications are not a factor in the acceptability of welds that are not required to be fully
radiographed.
(d) Evaluations and Retests
When a spot, radiographed as required in (b)(1) or (b)(2) above has been examined and the
radiograph discloses welding which does not comply……….The locations shall be determined by the
Inspector… if the two additional pass, repair the failed spot, if either of the two additional spots fail
the entire rejected weld shall be removed and the joint re-welded or the entire increment
completely radiographed and all defects corrected.
50' increment Second Additional Radiograph
Orig. Rejected
First Additional Radiograph
50' increment
Second Additional Radiograph
Acceptable
Repair and
radiograph
Acceptable

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…, if either of the two additional spots fail the entire rejected weld shall be removed and the joint
rewelded or the entire increment completely radiographed and all defects corrected.
50' increment
Failed.
Original
Rejected OrSecond Additional
Radiograph Failed.
50' increment
Repair all defects in the 50' or
remove all 50' then weld and
apply spot radiography again.
Radiograph
Entire 50'

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UW-11
Radiographic and Ultrasonic Examinations of Weld Joints
(a) Full Radiography. The following welded joints shall be examined radiographically for their full length
….
(1) All butt welds in the shell and heads of vessels used to contain lethal substances [see UW-2(a)];
* Remember, UW-2(a) demands that in lethal service the welds be of Type 1 for Category A and must be
of either Type1 or 2 for Categories B and C.
Type 1
Type 2
Lethal Service
Full Radiography

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….
(2) All butt welds in vessels in which the nominal thickness [ see (g) below] at the welded joint
exceeds 1-1/2 in. (38mm), or exceeds the lesser thicknesses prescribed in UCS-57…. * This
paragraph is on the examination.
(g) For radiographic and ultrasonic examination of butt welds, the definition of nominal thickness at the
welded joint under consideration shall be the nominal thickness of the thinner of the two parts
joined. Nominal thickness is defined in 3-2.
(c) Nominal Thickness – …….For plate material, the nominal thickness shall be, at the Manufacturer’s
option, either the thickness shown on the Material Test Report {or material Certificate of Compliance
[UG-93(a)(1)]} before forming, or the measured thickness of the plate at the joint or location under
consideration.
* Information only this is not on the exam.
….
(2) All butt welds in vessels in which the nominal thickness [see (g) below] at the welded joint
exceeds 1-1/2 in. (38 mm), or exceeds the lesser thicknesses prescribed in UCS-57, UNF-57,
UHA-33, UCL-35, or UCL-36 for the materials covered therein, or as otherwise prescribed in
UHT-57, ULW-51, ULW-52(d), ULW-54, or ULT-57; however, except as required by UHT-57(a),
Categories B and C butt welds in nozzles and communicating chambers that neither exceed
NPS10 nor 1-1/8 in. (29 mm) wall thickness do not require any radiographic examination;
* If none of the rules in the paragraphs above apply then use the default thickness of 1-1/2”.
This means;
If the material of construction is not one of those referenced UW-11(a)(2) then the default value for
the thinner thickness exceeded becomes 1-1/2”. Since the API 510 examination is restricted to UCS
materials (carbon and low alloy steels) this rule will be demonstrated using a Carbon Steel that is
classified as a P-Number 1.

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UCS-57
From paragraph UCS-57:
In addition to the requirements of UW-11, complete radiographic examination is required for each butt
welded joint at which the thinner of the plate or vessel wall thicknesses at the welded joint exceeds
the thickness limit above which full radiography is required in Table UCS-57.
Further Explained: For P No.1 materials the thinner of the two must exceed 1.25”.
Therefore the girth weld at the 1.25 to 1.5” joint and all above it are exempt.
Full RT
Carbon Steel P No.1
when the thinner of
the two exceeds1-¼"
t = 1.5"
t = 1.25
t = 1.75"
t = 1.25
t = 1.5"

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UW-11 Continued
(3) All butt welds in the shell and heads of unfired steam boilers ………Steam Boilers are NOT on
the Exam.
(4) All butt welds in nozzles, communicating chambers, etc., attached to vessel sections or heads
that are required to be fully radiographed under (1) or (3) above; however, .....Categories B and C
butt welds in nozzles and communicating chambers that **neither exceed NPS 10 (DNS 250) nor
1-1/8 in. (29mm) wall thickness do not require any radiographic examination;
** This only applies to circumferential welds in small (NPS 10 / 1-1/8” thick.) nozzles and chambers.
Longitudinal seams are not exempted by this rule.
Full RT for
Lethal
or
Thickness
Weld Neck Flange
Cat. C .75" thick NPS
6" No RT
Cat. A long
seam RT req.
1-1/8" thick
NPS 10" Weld
Neck Flange
Cat. C No RT
NPS 12" Nozzle and Weld Neck
Flange. Cat. C RT. Req.
Now for the hardest rule to understand!
(5) All Category A and D butt welds in vessel sections and heads where the design of the joint or
part is based on a joint efficiency permitted by UW-12(a), in which case:
(a) Category A and B welds connecting the vessel sections or heads shall be of Type No. (1) or
Type No. (2) of Table UW-12; * Just means they must be radiographable.
(b) Category B or C butt welds [but not including those in nozzles or communicating chambers
except as required in (2) above] which intersect the Category A butt welds in vessel sections
or heads or connect seamless vessel sections or heads shall, as a minimum, meet the
requirements for spot radiography in accordance with UW-52.

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* This paragraph is only mandatory when it is desired by the designer to use the highest joint efficiency
possible for calculations of thickness required or pressure allowed.
It is a choice the designer makes when there are no mandatory requirements based on service or
material as found in UW-11 (a) (1)*Lethal Service, (2)*Thickness exceeded
(6) All butt welds joined by… electrogas welding is not on the exam.
(7) Ultrasonic examination in accordance with UW- 53 may be substituted for radiography for the
final closure seam of a pressure vessel if the construction of the vessel does not permit
interpretable radiographs in accordance with Code requirements. The absence of suitable
radiographic equipment shall not be justification for such substitution.
(8) Exemptions from radiographic examination for certain welds in nozzles and communicating
chambers as described in (2), (4), and (5) above take precedence over the radiographic
requirements of Subsection C of this Division.
Note: This means that even though P-No. 5 for example requires RT in all thicknesses the
small/thin nozzles are exempt.
(b) Spot Radiography. Except as required in (a)(5)(b) above, butt welded joints made in
accordance with Type No. (1) or (2) of Table UW-12 which are not required to be fully
radiographed by (a) above, may be examined by spot radiography. Spot radiography shall be
in accordance with UW-52.
* If full RT is not mandatory Spot Radiography is done because the designers chose it.
If spot radiography is specified for the entire vessel, radiographic examination is not required of
Category B and C butt welds in nozzles and communicating chambers that exceed neither NPS
10 nor 1-1/8 in. wall thickness
(c) No Radiography. Except as required in (a) above, no radiographic examination of welded
joints is required when the vessel or vessel part is designed for external pressure only, or
when the joint design complies with UW-12(c).
* The designer can choose not to do RT if there is no mandatory requirement such as lethal, thickness, or
the desire for a higher joint E.
Before starting shell and head calculations let’s have a look at the types of welds and the weld joint
efficiencies that apply based on the amount of radiography applied.
These E values are found on Table UW-12 of Section VIII Division 1.
 The following is a simplification for the API 510 Exam; it does not reflect all of the possible
combinations of radiography, weld types and the resulting joint efficiencies

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UW-12
Joint Efficiencies
Table UW-12 gives the joint efficiencies E to be used in the formulas of this Division for joints completed
by an arc or gas welding process. Except as required by UW-11(a)(5), a joint efficiency depends only on
the type of joint and on the degree of examination of the joint and does not depend on the degree of
examination of any other joint.
(a) Value of E not greater than that given in column (a)* of Table UW-12 shall be used in the design
calculations for fully radiographed butt joints [seeUW-11(a)], except that when the requirements
of UW-11(a)(5) are not met, a value of E not greater than that given in column (b) of Table UW-
12 shall be used. * Known as Full Radiography
So now we are sent back to UW-11(a)(5)…….
UW-11(a)(5) All Category A and D butt welds in vessel sections and heads where the design of the joint or
part is based on a joint efficiency permitted by UW -12(a), in which case:
(a) Category A and B welds connecting the vessel sections or heads shall be of Type No. (1) or
Type No. (2) of Table UW-12; * (simply means it can be radiographed)
(b) Category B or C butt welds [but not including those in nozzles or communicating chambers
except as required in (2) above *(excludes small/thin nozzles)] which intersect the Category A
butt welds in vessel sections or heads or connect seamless vessel sections or heads shall, as a
minimum, meet the requirements for spot radiography in accordance with UW-52.
UW-11(a)(5) explained: This rule is pointed toward Code manufacturers who buy parts from other “Code
Shops” and basically assemble a vessel. The concern is as follows;
Code Shop A buys a rolled and welded shell from Code Shop B, Shop B fully radiographs
the Type 1 weld and the shell part will be delivered to Shop A with a joint E of 1.0. which is
essentially equal to a seamless shell.
Code Shop A welds on two seamless formed heads. Unless Shop A performs at least Spot
RT on the Category B welds connecting the heads to the shell there will have been no
radiographic testing of Code Shop A’s welders. A graphical representation follows.

22
Example 1: The longitudinal seam weld is of Type 1. It has received Full RT at Code Shop B. Shop A has
not performed the required Spot RT on the head to shell welds.
Fully Radiographed Type 1 by Shop B
Heads welded on by Shop A. Without the
spot RT as described in UW-11(a)(5)(b) the
shell would be calculated at E= .85
Example 2: Now the Spot RT has been performed by Shop B. Therefore and E = 1.0 is allowed for the
shell.
Fully Radiographed Type 1 by Shop B
Heads welded on by Shop A. With the
spot RT as described in UW-11(a)(5)(b)
the shell would be calculated at E= 1.0
UW-11(a)(5) So this means that Shop A cannot simply weld the heads, nozzles etc. and never do any
radiographic testing of the Shop A welders. To make things consistent this rule applies
even if the entire vessel is made by one Code Shop.
So no matter what the circumstances this Spot RT must be performed to take a joint
efficiency from Col. A of to Table UW-12 for seamed shell course.

23
Example 3: One last comment. On the shop floor these two shells both have the potential for a Joint E
of 1.0 . You will see this again in UW-12(d) Seamless Shells and Heads.
Seamed Shell Course Type 1 Full RT
Seamless Shell Course
(b) A value of E not greater than that given in *column (b) of Table UW-12 shall be used in the
design calculations for spot radiographed butt welded joints [see UW-11(b)].
* Known as Spot Radiography
(c) A value of E not greater than that given in * column (c) of Table UW-12 shall be used in the design
calculations for welded joints that are neither fully radiographed nor spot radiographed [see UW-
11(c)]. * No Radiography
Now let’s examine the first three Types listed on Table UW-12 and examine the joint types, the
amount of radiography and the resulting Joint Efficiencies.

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UG-116
Joint Efficiencies based on RT Marking
Next we will discuss Nameplate RT markings and how to determine the joint E to be used in the thickness
or pressure calculations to follow. These RT markings and their descriptions are found in paragraph UG-
116. We will now discuss these accompanied by graphical representations.
"RT 1" When all pressure retaining butt welds, other than B and C associated with nozzles and
communicating chambers that neither exceed NPS 10 nor 1-1/8 inch thickness have been
radiographically examined for their full length in a manner prescribed in UW 51, full
radiography of the above exempted Category B and C butt welds if performed, may be
recorded…...
MW-1
RT 1
Shell and Heads
E= 1.0
Lethal
"RT 2" Complete vessel satisfies UW-11(a)(5) and UW- 11(a)(5)(b) has been applied. The spot
RT rules of UW-52 must be applied to the spot RT and the Full RT rules of UW-51 to the
long seams. So the 50’ increments apply and all welders in that increment must be
examined by radiography.
MW-1
RT 2
Shell and Heads
E= 1.0
Desire
for E=1.0
"RT 2" Complete vessel satisfies UW-11(a)(5) and UW- 11(a)(5)(b) has been applied.
This is the second Case of RT 2 resulting in E = 1.0, again the rules of UW-52 apply.
RT 2
Shell and Heads
E= 1.0
Seamless Shell
Seamless Heads

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"RT 3" Complete vessel satisfies spot radiography of UW-11(b). The simplest example, one
welding operator and only three radiographs in 122’ of weld. The following assumes Type
1 welds for all weld seams.
"RT 4" When only part of the vessel satisfies any of the above. * Only part of the vessel has
been radiographed due to a thickness limit being exceeded as listed in UCS 57 or the
desire to use E = 1.0 .
The next consideration is the shells and heads of vessels which are considered seamless. The
Efficiencies used to calculate these vessel parts are not found on Table UW-12 but are instead listed in
paragraph UW-12(d).
(d) Seamless vessel sections or heads shall be considered equivalent to welded parts of the same
geometry in which all Category A welds are Type No. 1. For calculations involving
circumferential stress in seamless vessel sections or for thickness of seamless heads, E=1.0
when the spot radiography requirements of UW-11(a)(5)(b) are met. E= 0.85 when the spot
radiography requirements of UW-11(a)(5)(b) are not met, or when the Category A or B welds
connecting seamless vessel sections or heads are Type No. 3, 4, 5, or 6 of Table UW-12.
* Note this rule applies to the Code Shop A and B issue.

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MW-1
RT 2
Shell and Heads
E= 1.0
(d) Seamless vessel sections or heads shall be considered equivalent to welded parts of the same
geometry in which all Category A welds are Type No. 1. For calculations involving
circumferential stress in seamless vessel sections or for thickness of seamless heads, E=1.0
when the spot radiography requirements of UW-11(a)(5)(b) are met.
Shell and Heads
E= 1.0
Seamless Shell
Seamless Heads
(d) Seamless vessel …………. E= 0.85 when the spot radiography requirements of UW-11(a)(5)(b)
are not met, or when the Category A or B welds connecting seamless vessel sections or heads
are Type No. 3, 4, 5, or 6 of Table UW-12.
* Weld Type 3 to 6 can not be radiographed by Code rules.
Shell and Heads
E= 0.85
Seamless Shell
Seamless Heads
(e) Welded pipe or tubing shall be treated in the same manner as seamless, but with allowable
tensile stress taken from the welded product values of the stress tables, and the requirements of
UW-12(d) applied.
* If the spot RT is applied use E = 1.0, if not E = 0.85
Seamed Pipe Shell
Seamless Heads

28
UW-12 Joint Efficiencies
For the purposes of choosing joint efficiencies when doing vessel section or head calculations on the API
510 Examination the following can be said.
RT 1
Full Use 1.0 if joints are of Type 1 or 0.90 if Type 2
RT 2
Case 1: Use 1.0 with Seamless Heads and Shells
Case 2: Seamed Shells/Seamless Heads
• Shells Use 1.0 if joints are Type 1or if Type 2 Use 0.90
• Use 1.0 for seamless heads
RT 3
Use 0.85 if Joints are of Type 1 or 0.80 if of Type 2
Use 0.85 for Seamless heads
RT 4
* Special case of selective radiography *
Use Table UW-12 based on Joint Type and RT described in the exam question
No RT
Go to Table UW-12 and look up the E to be used for the type of weld under consideration.
Case1: Type 1 Use 0.70
Case 2: Type 2 Use 0.65
Seamless heads use 0.85
Remember that there only two (2) joint efficiencies possible for Seamless Shell and Seamless Heads they
are;
1.0 or 0.85
1.0 When the rules of UW-11(a)(5)(b) have been applied (UW-52 Spot RT applied).
0.85 When the rules have not been applied (UW-52 Spot RT not applied).
DO NOT GO TO TABLE UW-12 FOR THE E TO USE IN SEAMLESS HEADS OR SEAMLESS SHELLS

29
Lesson 2
Head and Shell Calculations
Objectives
Learn to Calculate:
• The required thickness of a cylindrical shell based on circumferential stress given a pressure (UG-
27(c)(1).
• The vessel part Maximum Allowable Working Pressure (MAWP) for a cylindrical shell based on
circumferential stress given a metal thickness (UG-27(c)(1).
• The required thickness of a head (ellipsoidal, torispherical and hemispherical) given a pressure.
(UG-32 (d), (e),& (f).
• The vessel part MAWP for a head (ellipsoidal, torispherical and hemispherical) given a metal
thickness using paragraphs UG-32 (d), (e),& (f).
• Whether a head (ellipsoidal, torispherical or hemispherical) meets Code requirements given
pressure and metal thickness UG 32(d), (e), and (f).
In any of the above you must also be able to:
• Compensate for the corrosion allowance: add or subtract based on requirements of the exam
problem. The Appendix 1* formula for cylinders, which is based on outside diameter, can be used.
• * The Appendix 1 formulas for non-standard heads will not be required.
Overview
This lesson will start with straight forward new construction calculations from Section VIII Div.1 for internal
dimension (I.D.) and progress on to consider;
• Corrosion Allowances
• Outside dimension calculation of cylindrical shells.*
• O.D. calculations will come from Appendix 1 formula 1-1 this is only issue from Appendix 1 on the
exam.

30
UG-27
Thickness of Shells under Internal Pressure
Here we find the formula and definitions for calculation of cylindrical shells under internal pressure. The
paragraph begins as follows;
(a) The thickness of shells under internal pressure shall be not less than that computed by the
following formulas. In addition, provision shall be made for any of the other loadings listed in UG-
22, when such loadings are expected…… The provided thickness of shells shall also meet UG-
16 * (addresses minimum thickness allowed).
(b) The symbols defined below are used in the formulas of this paragraph.
t = minimum required thickness of shell, in.
P = internal design pressure (see UG-21), psi
R = inside radius of the shell course under consideration
S = maximum allowable stress value, psi
E = joint efficiency for, or the efficiency of, appropriate joint in cylindrical or spherical shells, or
efficiency of ligaments are… not on the exam.
For welded vessels, use the efficiency specified in UW-12.
For ligaments between openings, …….are not on the exam.
(c) Cylindrical Shells. The minimum thickness or maximum allowable working pressure of cylindrical
shells shall be the greater thickness or lesser pressure as given by (1) or (2) below.
(1) Circumferential Stress Longitudinal Joints. When the thickness does not exceed one-half of
the inside radius, or P does not exceed 0.385SE, the following formulas shall apply: *(The
above is a test to see if these formula apply, they always do on this examination)
With the exception of Appendix 1 formula, all cylindrical shell calculations on the exam will use one of
these two formula! * Highlight these two!
P
0.6
-
SE
PR
=
t
t
0.6
+
R
SEt
=
P
OR
(2) Longitudinal Stress (Circumferential Joints). When the thickness does not exceed one-half
of the inside radius, or P does not exceed 1.25SE, the following formulas shall apply:
These formulas are not used on the exam.
* DO NOT USE or highlight them!
P
0.2
-
2SE
PR
=
t
t
0.2
+
R
2SEt
=
P
Not on the
Exam

31
Foot Note 14
Formulas in terms of the outside radius and for thicknesses and pressures beyond the limits fixed in this
paragraph are given in 1-1 to 1-3.
Of Appendix 1, only the first two shell formulas from paragraph 1-1 (a) (1) are on the Body of Knowledge!
Let’s have a look at those.
Appendix 1 Supplementary Design Formulas
1-1 THICKNESS OF CYLINDRICAL AND SPHERICAL SHELLS
(a) The following formulas, in terms of the outside radius, are equivalent to and may be used instead
of those given in UG-27 (c) and (d).
(1) For cylindrical shells (circumferential stress),
t
0.4
-
Ro
SEt
=
P
OR (1)
P
0.4
SE
PRo
=

t
(2) Longitudinal stress NOT ON EXAM cross out.
Example: Given a cylindrical shell with the following variables, solve for the MAWP of the cylinder using
both formulas.
P = ? * The question mark defines what is being solved for.
t = 0.500"
S = 15,000 psi
E = 1.0
R = 18.0“ and Routside = 18.5"
psi
t
R
SEt
8
.
409
18.3
7500
=
0.500)
x
(0.6
+
18.0
x0.500
1.0
x
15,000
=
6
.
0
P
27(c)(1)
-
UG 


psi
8
.
409
3
.
18
7500
0.500)
x
(0.4
-
5
.
18
0.500
x
1.0
x
000
,
15
0.4t
-
R
SEt
=
P
1)
-
(1
1
App
o




32
Which formula you use is determined by how the question is asked?
Example 1:
Internal Formula
A vessel shell has corroded to an inside radius of 23.58” its working pressure is 500 psi and its stress
allowed is 17,500 psi. What is the required thickness?
Other terms sometimes used:
Corroded internally, found to have an inside diameter/radius, etc.
In these cases we must use the inside formula of UG-27
Example 2:
External Formula
A vessel shell has corroded to an outside diameter of 23.58” its working pressure is 500 psi and stress
allowed is…..what is the required thickness?
Other terms sometimes used:
found externally corroded, attacked by corrosion under insulation (CUI) etc.
Here we would have to use the formula of Appendix 1.
• You can use either formula in some situations.
Example 3:
Internal Formula or External Formula
A vessel shell has corroded to an inside radius of 23.58 ” its working pressure is 500 psi and its stress
allowed is ….its original thickness was .500” and the original inside radius was 24.0 ” ** for inside
calculations use R = 23.58 (actual)
To use the outside formula we can add the original thickness to the original inside radius.
24” + .500 = 24.5” = Ro Original radius outside
Now we can use the formula of Appendix 1-1 if we chose to.
• Also there is the situation where you are given only the Outside Dimension (O.D.) and asked to solve
for the thickness required or maximum allowable working pressure.
Example 4: External Formula for Thickness
A vessel shell has an outside radius of 24.0 ” its working pressure is 500 psi and its stress allowed is
15,000 psi. The joint efficiency, E = 1.0. The shell has corroded internally to a thickness of 0.343”. What is
its present Maximum Allowable Working Pressure?
Here you must use the O.D. Formula since you cannot determine the present internal corroded radius, not
having the original thickness you cannot determine the original I.D.!

33
Here is an example of working a problem using both inside and outside dimensions having all the
information needed.
A cylindrical shell has been found to have a minimum thickness of .353". Its original thickness was .375“
with an original inside radius of 12.0”. What is its present MAWP ?
Pulling the information from the stated problem, we have:
P = 300 psi
t = .353"
S = 13,800 psi
E = .85
R = 12.0" + (.375-.353) = 12.022 corroded inside radius
Ro= 12.0" + 0.375 (orig. t) =12.375” original outside radius
Here is a graphical representation of the problem:
Which formula? R inside corroded = 12 + (0.375 – 0.353) = 12.022”
R outside = 12.0" + 0.375 (orig. t) = 12.375”
Radius Inside for MAWP using UG-27(c)(1).
psi
t
R
SEt
46
.
338
12.2338
4140.69
=
.353)
x
(0.6
+
12.022
.353
x
.85
x
13,800
=
6
.
0
P
27(c)(1)
-
UG 


Radius Outside for MAWP using App: 1 (1-1).
psi
46
.
338
2338
.
12
69
.
4140
.353)
x
(0.4
-
375
.
12
.353
x
.85
x
800
,
13
0.4t
-
R
SEt
=
P
1)
-
(1
1
App
o



You Choose!
t Corroded = 0.353"
t Corroded = 0.353"
t Orig. = 0.375"

34
Let's do a simple internal shell calculation now. We will use a shell which is seamless. You may find the
following approach helpful in keeping track of the data. As the problems become more difficult, it becomes
harder to track the variables if you are not organized.
• Make a simple sketch of the shell and label its dimensions.
• List what is required to know. We will call these givens.
• State the code paragraph that applies, i.e., UG-27, etc.
Use this approach for all calculations.
Givens: Sketch
t =
P =
R =
S =
E =
Code Paragraph UG-27 (c) (1) t
0.6
+
R
SEt
=
P
I.D. ?
t = ?

35
Problem # 1
Find the Maximum Allowable Working Pressure (MAWP) of a 12 inch inside diameter shell. This shell is
seamless and is stamped RT 2. It has an allowable stress value of 16,600 psi and the wall thickness is
.406”. No corrosion is expected.
I.D. 12.0"
SKETCH
t = .406"
R = 6.0"
Givens: Plug in from the values given in the question!
P =?
t = .406
R = D/2 = 12/2 = 6.0” this formula uses the Radius.
S = 16,600 psi
E = 1.0 per UW-12(d) Seamless shells and heads
From UG-27 (c) (1) Circumferential Stress
t
0.6
+
R
SEt
=
P
psi
44
.
1079
2436
.
6
6
.
6739
.406)
x
0.6
(
+
)
(6.0
x.406
1.0
x
16,600
= 

P

36
As can be seen the calculations are simple, it is more a matter of deciding on the correct formula to use,
inside or outside, and transferring the givens accurately to the formula. Once again use the approach;
Givens: SKETCH
P =
t =
ect.
About rounding answers. In the ASME Code and for the exam you must round DOWN for pressure
allowed so in our solution below we would round down to 1079 psi. Even if our solution had been
1079.999 we cannot round to 1080, we still round down to 1079 psi.
This is the conservative approach taken by the Codes in general and of course is different for the normal
rules of rounding.
Problem # 2
Find the minimum required thickness of a cylindrical shell designed for a working pressure of 100 psi. The
shell's inside radius is 2'-0". The longitudinal joint is type 1 (table UW-12) and no radiography was
performed. The shell is made of carbon steel rolled plate with an allowable stress of 15,000 psi.
SKETCH
t = ?
Type 1 Category
A
No RT
Givens:
t = ?
P = 100 psi
R = 24"
S = 15000
E = .70 ( From Table UW-12 column C)
From UG-27 (c) (1) Circumferential Stress
When rounding thickness required we must round up. The most conservative thing to do. So our example
below would round to .230”. Even it had been .2291 we would still round up to .230”.
P
0.6
-
SE
PR
=
t
"
.2298
=
10440
2400
=
100)
x
(0.6
-
)
.70
x
(15,000
24
x
100
=
t

37
We have now calculated the pressure allowed on a seamless shell in Problem #1 and have calculated the
thickness required of a seamed shell in Problem #2.
Now for one more example.
Problem # 3
Determine the minimum required thickness of a cylindrical shell designed for an internal pressure of 50
psi, no corrosion is expected. The shell’s inside diameter is 10’-0”.
The shell’s Category A and B, Type 1 welds have been fully radiographed. The material’s stress allowable
is 17,500 psi. The vessel will be stamped RT 1.
Long Joint (Circumferential Stress)
SKETCH:
I.D. 10'- 0"
t = ?
Type 1 Category
A & B
Full RT
Givens:
t = ?
P = 50
R = 10’ x 12” = 120”/2 = 60"
S = 17500
E = 1.0 (RT 1)
From UG-27 (c) (1) .172"
to
rounds
"
.1717
=
17470
3000
=
50)
x
(0.6
-
)
1.0
x
(17,500
60
x
50
=
t

38
You are now familiar with the basic cylindrical shell formula from UG-27. However that formula in its
published form is only useful for the calculation of vessel shells that are designed without a corrosion
allowance. Usually during design a corrosion allowance will be given to vessel part.
Example:
A vessel is being designed for a specific volume of water. The designer determines the optimum inside
diameter and length of the vessel to obtain that volume.
The engineer set the inside diameter at 48” so it must be constructed with that inside diameter, resulting in
an inside radius of 24” to be used in the calculation.
In the design calculation the engineer adds the corrosion allowance to the radius. The basic formula of
UG-27 would be modified to be;
P
0.6
-
SE
.)
.
(R
P
=
a
c
t

t
0.6
+
c.a)
(R
SEt
=
P

or
t
0.6
+
)
125
.
0
(24
SEt
=
P

Inside diameter = 48.0”
Inside radius = 24.0”
c.a. = 1/8” = 0.125”
Inside radius used in calculations = 24.0 + 0.125 =
24.125” resulting in the following;
The vessel shell would be constructed of the required calculated thickness and then rolled to an inside
radius of 24”, it retirement radius would be 24.125”
This is no different from what occurs during the evaluation of an in service vessel that has corroded.
However here we use actual measurements. Suppose the vessel shell above was built with a thickness of
0.500” and rolled to the 24.0” inside radius. Corrosion has occurred and the new minimum wall thickness
is 0.450”. To calculate we would use a radius of 24.0 + (0.500 – 0.450) or 24.050”. This would leave a
remaining corrosion allowance of .125 -.050 = .075”

39
UG-32
Internal Pressure On Formed Heads
There are three types of calculations for formed heads listed in the Body of Knowledge: Ellipsoidal,
Torispherical and Hemispherical. A sketch and the formula for thickness of each kind are below.
P
0.2
-
2SE
PD
=
t
0.1P
-
SE
0.885PL
=
t
0.2P
-
2SE
PL
=
t
(a) The required thickness at the thinnest point after forming of ellipsoidal, torispherical,
hemispherical, conical, and toriconical (not on exam) heads under pressure on the concave
side (plus heads) shall be computed…….
(b) The thickness of an unstayed ellipsoidal or torispherical head shall in no case be less than….this is
a test to see if you should use this formula or the ones given in Appendix 1. Not on Exam!
(c) The symbols defined below are used in the formulas of this paragraph:
t = minimum required thickness of head after forming, in. (mm)
P = internal design pressure (see UG-21), psi (kPa)
D = inside diameter of the head skirt; or inside length of the major axis of an ellipsoidal head; in.
(mm)
S = maximum allowable stress value in tension.
E = lowest efficiency of any joint in the head; for hemispherical heads this includes head-to-
shell joint; for welded vessels, use the efficiency specified in UW-12
L = inside spherical or crown radius, in. (mm)
There are 5 formed heads listed in UG-32. You will be responsible for the calculations of these 3 only;
(a) ElliEllipsoidal (b) Spherically Dished
(Torispherical)
(c) Hemispherical

40
Hemispherical, Ellipsoidal and, Torispherical.Heads
The next series of slides are example calculations of all three types for thickness required. These
calculations will use the exact same conditions for service, stress allowed, Joint E, dimensions, and
pressure.
• With all things being equal, which do you suspect will be the thinnest allowed?
• Which do you think will be the thickest required?
• Which is in the middle?
Examples:
Givens: The same pressure, stress and, dimension values will be used for all heads. Let’s determine
which type of head will be the thickest required and which will be the thinnest allowed.
Given:
P = 100 psi
S = 17500 PSI
E = .85 for spot RT of Hemi-head joint to shell
E = 1.0 for seamless heads ( Ellipsoidal and Torispherical )
L = 48" for the inside spherical radius for the hemi-head
L = 96" for the inside crown radius of the torispherical head
D = 96" inside diameter of the ellipsoidal
t = ? Required wall thickness, inches
Problem # 1
Given the above data find the required thickness of a seamless ellipsoidal head.
P
0.2
-
2SE
PD
=
t
From UG-32 (d)
"
.2744
=
34980
9600
100)
x
(0.2
-
1.0)
x
17,500
x
(2
96
x
100
=
t

41
Problem # 2
Using the same data, calculate the required thickness of a hemispherical head.
0.2P
-
2SE
PL
=
t
From UG-32 (f)
"
1614
.
0
29730
4800
)
100
2
.
0
(
)
85
.
0
500
,
17
2
(
48
100




x
x
x
x
t
Problem # 3
Determine the required t of this torispherical head. (These are also called ASME flanged and dished
heads, by the way).
0.1P
-
SE
0.885PL
=
t
From UG-32(e)
"
.4857
=
17490
8496
=
100)
x
(0.1
-
1.0)
x
(17,500
96
x
100
x
0.885
=
t
So we have from thickest to thinnest, all things equal:
Torispherical = .4857” (Rounds to .486”)
Ellipsoidal = .2744 (Rounds up to .275”)
Hemispherical = .1614 (Rounds to .162”)
There have been several exams where the question was asked, “Which is required to be thickest” or
“Which can be the thinnest” Remember this.

42
One Last Important Comment:
Hemispherical heads while they can be formed seamless are not considered seamless heads by Section
VIII. As mentioned previously they essentially form a Category A seam between the head and the other
part. The spot RT of UW-12(d) does not apply to the Joint E used to calculate a Hemispherical head. They
are never seamless; their Joint E comes from Table UW-12 based on the Type of weld and the extent of
Radiography applied. * Remember This.*

43
Lesson 3
Maximum Allowable Working Pressure
UG-98
(a) The maximum allowable working pressure for a vessel is the maximum pressure permissible at the top
of the vessel in its normal operating position at the designated coincident temperature specified for
that pressure. It is the least of the values found for maximum allowable working pressure for any of
the essential parts of the vessel by the principles given in (b) below, and adjusted for any difference in
static head that may exist between the part considered and the top of the vessel (See 3-2).
(b) The maximum allowable working pressure for a vessel part is the maximum internal or external
pressure, including the static head thereon, as determined by the rules and formulas in this Division,
together with the effect of any combination of loadings listed in UG-22 which are likely to occur, for the
designated coincident temperature, excluding any metal thickness specified as corrosion allowance.
(c) Maximum allowable working pressure may be determined for more than one designated operating
temperature, using for each temperature the applicable allowable stress value. In the Code there are
two types of Maximum Allowable Working Pressures (MAWP). One is for the vessel itself, the one
most think of and refer to all the time. The other is the one for each part of a vessel referred to in UG-
98 as the part MAWP. Think of it in this way: a vessel has a shell, heads, chambers… etc., and
pressure allowed or thickness required calculations must be performed for each one to determine the
MAWP of the vessel. When doing these calculations, you cannot take credit for any extra thickness
designed into the vessel as a corrosion allowance. The weakest of the vessels parts, considering
other loadings such as the static head of the contents, weight of insulation, wind, etc., will determine
the MAWP of the entire vessel. It is the weakest link in the chain…The pressure referred to here can
be internal or external.

44
The MAWP of a vessel is the pressure allowed in a vessel at its top in its normal operating position
and at its maximum operating temperature. The MAWP can be determined for more than one
designated operating temperature, using for each temperature the applicable allowable stress value.

45
Lesson 4
Hydrostatic Head Pressure
Overview
What is Hydrostatic Head Pressure? Let’s examine the words to better understand the meaning of
hydrostatic.
• Hydro meaning liquid
• Static meaning unchanging.
• Pressure is a force exerted over an area.
Which of leads us to the following;
It is a pressure that is generated by the weight of the liquid due to gravity. The taller the height of a liquid
column the greater the force, which is expressed as pounds per square inch (psi) for our purposes. The
Hydro (liquid) of interest on the exam is water, since it is the primary liquid we use for Hydrostatic testing.
Other liquids can be and are used.
The hydrostatic head of water is part of our everyday lives. For example the water tower that supplies your
home uses the principle of “Hydrostatic Head” or gravity to push the water into your home and out of your
faucets. Let’s have a look at a graphic of a water tower that will detail this principle.
Hydrostatic Head of a Water Tower
140’ x 0.433 = 60.6 psig and 100’ x 0.433 = 43.3

46
Hydrostatic Head of Water Basic Principle
The hydrostatic head of water is equal to 0.433 psi per vertical foot above the point where the pressure
will measured. For example the hydrostatic head of water at a point in a vessel with 10 feet of water above
it is calculated by multiplying 10 x 0.433 psi.
10 x 0.433 = 4.33 psi
The 4.33 psi is being exerted totally by the weight of the water. No other external pressure having been
applied. If an external source of pressure is applied it would be added to the hydrostatic head pressure of
the water at any given point in the vessel, more on this later.
Now for a pressure vessel, no external pressure, filled with water only. 0 psi at top, the bottom is 100 x
0.433 = 43.3 psi
0 psi
43.3
psi
100 Feet
100
psi
143.3
psi
100 Feet
External pressure of 100 psi is now applied resulting in a gage pressure at the bottom of 143.3 psi. The
43.3 psi is static, never changing. From these simple water tower and pressure vessel examples the
following can be understood and applied to a pressure vessel. For a pressure vessel the MAWP is always
measured at the top in its normal operating position. Here are the issues on the exam that must be
understood to work a H.H. problem.
Case 1: How do you determine hydrostatic head based on a given elevation?
Case 2: When do you add the hydrostatic head pressure in vessel calculations?
Case 3: When do you subtract the hydrostatic head pressure in vessel calculations?
Case 1: To determine hydrostatic head based on an elevation from a stated problem it must be
understood that elevations are normally taken from the ground level to a vessel’s very top. You
must subtract the Given elevation from the Total elevation to determine vertical feet of
hydrostatic head above the given elevation.

47
Example: A vessel has an elevation of 18 feet and is mounted on a 3 foot base. What is the hydrostatic
head pressure of water at the 11 foot elevation which is located at the bottom of the top shell
course? Remember it is the number of vertical feet above the given elevation in question
which causes the hydrostatic head at that point. To find the hydrostatic head you must subtract
the elevation of the Given point from the Total elevation given for the vessel.
18' feet total
-11' desired point
7' total hydrostatic head
Hydrostatic head pressure at 11' elevation is:7 x 0.433psi = 3.03 psi
=
Case 2: Hydrostatic head at a point in a vessel must be added to the pressure used (normally vessel
MAWP) when calculating the required thickness of the vessel component at that elevation.
Example: Determine the required thickness of the shell course in Case 1. The vessel's MAWP (Always
measured at the top in the normal operating position) is 100 psi. The following variables apply:
Givens:
t = ? Circumferential stress from UG-27(c)(1)
P = 100 psi + Hydrostatic Head
S = 15,000 psi
E = 1.0
R = 20"
Since the bottom of this shell course is at the 11 foot elevation the pressure it will see is 100 psi + the
hydrostatic head.
100 + 3.03 = 103.03 psi
Also our basic formula becomes;
1379
.
18
.
14938
6
.
2060
)
3
.
103
6
.
0
(
)
0
.
1
000
,
15
(
20
03
.
103



x
x
x
x
t
.)
.
(
6
.
0
.)
.
(
H
H
P
SE
R
H
H
P
t



-

48
Case 3 You must subtract hydrostatic head pressure when determining the MAWP of a vessel. If
given a vessel of multiple parts and the MAWP for each of the parts, the MAWP of the entire
vessel is determined by subtracting the hydrostatic head pressure at the bottom of each part
to find the part which limits the MAWP of the vessel.
Example: A vessel has an elevation of 40 feet including a 4 foot base. The engineer has calculated the
following part’s MAWP to the bottom of each part based on each part's minimum thickness
and corroded diameter. Determine the MAWP of the vessel as measured at the top.
Calculated Part MAWP at the bottom of:
Top Shell Course 28' Elev. 406.5 psi
Middle Shell Course 16.5' Elev. 410.3 psi
Bottom Shell Course 4' Elev. 422.8 psi
Bottom of top shell course:
40.0' elev.
-28.0' elev.
12.0' of hydrostatic head 12' x 0.433 psi = 5.196 psi of Static
Bottom of the middle shell course:
40.0' elev.
-16.5' elev.
23.5' of hydrostatic head
23.5' x 0.433 psi = 10.175 psi of Hydrostatic HeadBottom of bottom shell course:
40.0' elev.
-4.0' elev.
36.0' of hydrostatic head
36' x 0.433 psi = 15.588 psi of Hydrostatic Head
The final step in determining the MAWP of the vessel at its top is to subtract the hydrostatic head of water
from each of the calculated Part MAWP. The lowest pressure will be the maximum gauge pressure
permitted at the top of the vessel.
Bottom of top shell course 406.5 - 5.196 = 401.3 psi
Bottom of mid shell course 410.3 - 10.175 = 400.125 psi
Bottom of btm. shell course 422.8 - 15.588 = 407.212 psi

49
Therefore the bottom of the middle shell course’s MAWP limits the pressure at the top and, determines
the MAWP of the vessel.
The MAWP of the vessel is 400.125 psi
One thing to remember is this pressure is static. In our example the if the applied external pressure at the
top were raised above 400.125 psi, then down at the 16.5’ elevation the gage would exceed that shell
course’s MAWP of 410.3.

50
One last example using a vessel which is horizontal, just to reinforce the concept that it is the Vertical
Height that must be considered. The 6.928 psi total H.H. must be considered at the bottom when
calculating the sump head.
MW-1
10'
3'
3'
Total 16' x 0.433 psi = 6.928 psi H.H.
16 feet
Depth of a Hemispherical and Ellipsoidal heads
Hydrostatic Head of Water
One final thing the determination of H.H. for two formed heads, Hemispherical and Ellipsoidal.
Hemispherical Head
For this example we will use a hemispherical head that has an inside diameter of 48 inches which means
it has a radius of 24 inches. The radius is the depth of the hemispherical head
Ellipsoidal Head
An ellipsoidal head's I. D. will be the same as the shell. The inside diameter of an ellipsoidal head is also
its major axis. This fact is the basis of finding the depth of a 2 to 1 ellipsoidal head. Notice that we are
strictly talking about 2 to 1 ellipsoidal heads. The 2 to 1 refers to the ratio of the Major Axis to the Minor
Axis of an ellipse which is used to form the head
Of course only half of the Minor Axis is used for the head.

51
Now add the 2 inch flange to the dish.
Therefore, our 2 to 1 Ellipsoidal head has a depth of 14 inches. Hint: To find the depth of a 2 to 1
ellipsoidal head divide the major axis by 4. In our example 48/4 = 12 then add the 2” flange.
Ellipsoidal
Converting to feet: 18" divided by 12 =
1.5' x 0.433 psi = 0.6495 psi
Hemispherical
Converting to feet. 32" divided by 12 =
2.666' x 0.433 psi = 1.1543 psi
Adding H.H. and Corrosion Wall Loss in Calculations
Increasing internal or decreasing external dimensions due
to corrosion was introduced in Lesson 2, “Shell and Head
Calculations”. In actual practice Hydrostatic Head would
also need to be considered. The following demonstrates
the principals involved.
Example:
A vertical vessel shell course has an MAWP of 200 psi,
and an allowable stress of 14,800 psi. The inside radius is
84”. The nameplate is stamped RT1. The shell has
corroded down to 1.28 inches. Its original t was 1.375".
There exists 21.9964 psi H.H. at the bottom of the shell
course.
What is its current calculated minimum thickness of this
shell course in accordance with rules of Section VIII
Division 1 considering both corrosion and hydrostatic
head?

52
Basic Formula: UG-27 (c) (1)
Modified to consider Hydrostatic Head and increased radius due to internal corrosion.
Givens:
H.H.)
(P
0.6
-
SE
)
(R
H.H.)
(P


 corrosion
t =?
P = 200
S = 14,800 psi
E = 1.0 RT 1
R = 84” = 84” + (1.375-1.28) = 84.095”
H.H..= 21.9964 rounded to 22 psi




)
22
(200
(0.6)
-
.0)
(14,800)(1
)
095
.
(84
22)
(200
=
t 
)
22
(2
(0.6)
-
.0)
(14,800)(1
)
095
(84.
(222)

133.2
-
14,800
18669.09
"
273
.
1
14,666.8
18669.09

Its present thickness is 1.28” and its minimum calculated thickness is 1.273, very close to repair or
retire.
P
0.6
-
SE
PR
=
t

53
Lesson 5
Hydrostatic, Pneumatic Tests
UG-99 Standard Hydrostatic Test
(a) A hydrostatic test shall be conducted on all vessels after:
(1) All fabrication has been completed, except for operations which could not be performed prior
to the test such as weld end preparation….
(2) All examinations have been performed, except those required after the test…...
(b) Except as otherwise permitted in (a) above and 27-3, vessels designed for internal pressure shall be
subjected to a hydrostatic test pressure which at every point in the vessel is at least equal to 1.3 times
the maximum allowable working pressure to be marked on the vessel multiplied by the lowest ratio
(for the materials of which the vessel is constructed) of the stress value S for the test temperature on
the vessel to the stress value S for the design temperature…Stress at Test TempStress at Design
Temp
(c) A hydrostatic test based on a calculated pressure may be used by agreement between the user and
the manufacturer… A New and Cold Test relates to a statement in API 510.
(d) The requirements of (b) above represent the minimum standard hydrostatic test pressure required by
this Division…(g) Following the application of the hydrostatic test pressure, an inspection shall be
made of all joints and connections. This inspection shall be made at a pressure not less than the test
pressure divided by 1.3. Except for leakage that might occur at temporary test closures for those
openings intended for welded connections, leakage is not allowed at the time of the required visual
inspection…. The visual inspection of joints and connections for leaks at the test pressure divided by
1.3 may be waived provided:
(1) a suitable gas leak test is applied;(2) substitution of the gas leak test is by agreement reached
between Manufacturer and Inspector;(3) all welded seams which will be hidden by assembly
be given a visual examination for workmanship prior to assembly;(4) the vessel will not contain
a “lethal” substance.(h) Any non-hazardous liquid at any temperature may be used for the
hydrostatic test if below its boiling point. Combustible liquids having a flash point less than 110°F,
such as petroleum distillates, may be used…
It is recommended that the metal temperature during hydrostatic test be maintained at least 30°F
above the minimum design metal temperature, but need not exceed 120°F, to minimize the risk of
brittle fracture.
API 510 has a different rule for this, it recommends that the temperature be 10°F above for 2”
(50mm) thickness and under and 30°F above for over 2 inches (50mm).
Footnote Caution: A small liquid relief valve set to 1-1/3 times the test pressure is recommended for
the pressure test system in case a vessel, while under test, is likely to be warmed up materially with
personnel absent.
(i) Vents shall be provided at all high points of the vessel in the position in which it is to be tested to
purge possible air pockets…
(j) Before applying pressure, the test equipment shall be examined to see that it is tight and that all
low pressure filling lines and other appurtenances…(k) Vessels, except for those in lethal service,
may be painted or otherwise coated either internally or externally, and may be lined internally, prior
to the pressure test. However, the user is cautioned that such painting / coating / lining may mask
leaks that would otherwise have been detected during the pressure test.

54
API 510 Hydrostatic Test Procedures
1. If the test is required it shall be conducted after welded repairs.
2. The test pressure must at least be1.3 times the MAWP or whatever the original design Code specified
i.e. 1.5.
3. The test pressure shall be adjusted for lowest ratio of stresses.
4. Any non-hazardous fluid may be used if below its boiling point.
5. It is recommended that the metal temperature during hydro test be maintained at least 10 °F above
MDMT for vessels 2” (51mm) and less and 30 °F above for vessels over 2” (51mm) to minimize the
risk of brittle fracture.
6. Following the application of hydro pressure a visual inspection shall be performed at no less than the
test pressure divided by 1.3 or whatever was originally used.
Problem: Calculate the required hydro test pressure for a vessel using the following conditions:
Material Carbon Steel
Design Temp. 700 °F
Test Temp 85 °F
MAWP 350 psi
Step 1 Determine the ratio of stresses for the test and design temperatures.
(a) From Table 1A Section II Part D.
Stress allowed at 700 °F = 15,500 psi
Stress allowed at 85 °F = 16,300 psi
b) Per UG-99 the ratio equals
Temp.
Design
at
Stress
Temp.
Test
at
Stress
Step 2 UG-99(b) Test pressure equals 1.3 x MAWP x ratio
1.3 x 350 psi x 1.05 = 477.75 psi at the top of the vessel

55
UW-50 NDE of Welds for Pneumatically Tested Vessels
Look at the reference next to UG-100 (See UW-50)
This is what is referred to as a parenthetical reference in the ASME Codes. You must read these to see
what modifiers the Code has placed on subject paragraph.
On welded pressure vessels to be pneumatically tested in accordance with UG-100, the full length of the
following welds shall be examined for the purpose of detecting cracks:
(a) all welds around openings;(b) all attachment welds, including welds attaching non-pressure parts to
pressure parts, having a throat thickness greater than 1/4 in….
UG-100 Standard Pneumatic
(a) Subject to the provisions of UG-99(a)(1) and (a)(2), a pneumatic test prescribed in this paragraph may
be used in lieu of the standard hydrostatic test prescribed in UG-99 for vessels:
(1) That are so designed and/or supported that they cannot safely be filled with water;
(2) Not readily dried, that are to be used in services where traces of the testing liquid cannot be
tolerated and the parts of which have, where possible, been previously tested by hydrostatic
pressure to the pressure required in UG-99.
(b) Except for enameled vessels, for which the pneumatic test pressure shall be at least equal to, but
need not exceed, the maximum allowable working pressure to be marked on the vessel, the
pneumatic test pressure shall be at least equal to 1.1 times the maximum allowable working pressure
to be stamped on the vessel multiplied by the lowest ratio of the stress value S for the test
temperature of the vessel to the stress value S for the design temperature. In no case shall the
pneumatic test pressure exceed 1.1 times…
(c) The metal temperature during pneumatic test shall be maintained at least 30°F (17°C) above the
minimum design metal temperature to minimize brittle fracture risk. [See UG-20 and General Note (6)
to Fig. UCS-66.2]
* API 510 states as minimum Pneumatic tests shall meet all the safety requirements of ASME Section VIII.
(d) The pressure in the vessel shall be gradually increased to not more than one-half of the test pressure.
The test pressure shall be increased in steps of approximately one-tenth of the test pressure until the
required test pressure has been reached. Pressure shall be reduced to a value equal to the test
pressure divided by 1.1 and held for a sufficient time to permit inspection of the vessel…
UG-100 Standard Pneumatic Test
1. If the test is required it shall be conducted after welded repairs.
2. The welded repairs shall be subjected to the tests required by UW-50.
3. The test pressure must at least be 1.1 times the MAWP or whatever the original design Code
specified i.e.1.25.
4. Test pressure is adjusted for lowest ratio of stresses. (Same method as hydrostatic testing)
5. Metal must be maintained at least 30 °F over MDMT.
6. The test pressure shall be raised at a gradual rate to not more than 1/2 the test pressure and then
raised by 1/10th of the test pressure until the test is reached.
7. Visual inspection must be made at test pressure divided by 1.1 or whatever was originally used.. The
visual may be waived if the requirements in UG-100 are met.

56
Problem: Calculate the required pneumatic test pressure for a vessel using the following conditions.
Material Carbon Steel
Design Temp. 700 o F
Test Temp 85°F
MAWP 350 psi
Step 1: Determine the ratio of stresses for the test and design temperatures.
(a) From Table 1A Section II Part D.
Stress allowed at 700 o F= 15,500 psi
Stress allowed at 85 o F= 16,300 psi(b) Per UG-100 the ratio equals
Step 2 Per UG-100(b) Test pressure equals
1.1 x MAWP x
Temp.
Design
at
Stress
Temp.
Test
at
Stress
1.1 x 350 psi x 1.05 = 404.25 psi
Pneumatic Test Procedure
1. Slowly raise the pressure to approximately one-half 404.25 psi which equals 202.125. Next raise the
pressure in steps of one-tenth of the test pressure.
2. 202.125 + 40.425 = 242.55 psi
3. 242.55 + 40.425= 282.975 psi
4. 282.975 + 40.425 = 323.40 psi
5. 323.40 + 40.425 = 363.825 psi
6. 363.825 + 40.425 = 404.25 psi
There are a total of 6 steps when raising up to pneumatic test pressure. Finally lower to the inspection
pressure of 404.25/1.1 = 367.5 psi
UG-102 Test Gauges
Overview
The Code has some definite requirements for the selection and uses of gages for the tests described in
UG-99 and UG-100. Directions for location, number of, range of and the calibration of the indicating
gage(s) is located in UG-102. The high points of UG-102 are below.
1. An indicating gage shall be connected directly to the vessel. If it is not readily visible to the operator of
the test equipment an additional gage shall be used which is visible....
2. When doing large vessel pressure tests it is recommended to have a recording gage in addition to the
indicating gage.
3. Dial type indicating gages shall have a range of about double the maximum test pressure, but in no
case shall the range of the gage be less than 1 1/2 times nor more than 4 times the maximum test
pressure.
4. Digital gages having a wider range may be used as long as they provide the same or greater accuracy
of the dial type.
5. All gages shall be calibrated against a standard deadweight tester or a calibrated master gage.
6. Gages must be calibrated any time their accuracy is in doubt.

57
Lesson 6
Post Weld Heat Treatment
UW-40 Definition of Nominal Thickness for Butt Welds
The Post Weld Heat Treatment mandatory requirements and time at temperature are based on the base
metal’s thickness. The Code defines thickness at a welded joint in a very specific way, which is as follows:
(f) The term nominal thickness as used in Tables UCS-56, UCS-56.1, UHA-32 and UHT-56, is the
thickness of the welded joint as defined below. For pressure vessels or parts of pressure vessels
being postweld heat treated in a furnace charge, it is the greatest weld thickness in any vessel or
vessel part which has not previously been postweld heat treated..
(1) When the welded joint connects parts of the same thickness, using a full penetration butt weld,
the nominal thickness is the total depth of the weld exclusive of any permitted weld
reinforcement.
Depth of
weld
(5) When a welded joint connects parts of unequal thicknesses, the nominal thickness shall be the
following:
(a) the thinner of two adjacent butt-welded parts including head to shell connections;……
UCS-56 Requirements for Post Weld Heat Treatment
PWHT is performed to specific rules based on the thickness of the weld to be heat treated. We will now
examine those rules.
(a) Before applying the detailed requirements and exemptions in these paragraphs, satisfactory weld
procedure qualifications of the procedures to be used shall be performed in accordance with all the
essential variables of Section IX including conditions of postweld heat treatment or lack of
postweld heat treatment.
Question: What must always be present prior to welding?
Answer: A Section IX qualified welding procedure.

58
UCS 56
(d) The operation of postweld heat treatment shall be carried out by one of the procedures given
in UW-40 in accordance with the following requirements:
(1) The temperature of the furnace shall not exceed 800°F (425°C) at the time the vessel or part
is placed in it.
(2) Above 800°F (425°C ), the rate3 of heating shall be not more than 400°F/hr (222°C) divided
by the maximum metal thickness of the shell or head plate in inches, but in no case more
than 400°F/hr (222°C). During the heating period there shall not be a greater variation in
temperature throughout the portion of the vessel being heated than 250°F (120°C) within any
15 ft (4.6 m) interval of length.
400°F / 2 inches so no more than 200°F/hr
UCS-56 Requirements for Post-Weld Heat Treatment
We will now examine the requirements for PWHT using the tabular form of UCS-56 for P-Number 1 base
metal.
This is just one Table, there are many more based on the material’s P-Number. The others follow the
same format and once we have learned to use this one the others will be much easier to understand. You
are responsible for all of the tables on this examination however most questions come from the P-Number
1 Table!
The tables cannot be interpreted with out reading the notes that are beneath them. Let’s have a look.

60
We can gather the following from the Table for P-Number 1 materials that;
• The normal holding temperature is 1100
o
F for P-No. 1.
• The minimum time at holding temperature is based on the thickness of the part.
• Note 1 references alternative PWHT holding temperatures
• Note 2 determines the thickness at which PWHT is mandatory.
There are three cases of thickness listed in the table P-1.
• Up to 2 inches (51 mm) the PWHT is held for 1 hour per inch (25 mm) of thickness with 15
minutes minimum in all cases. The 15 minute minimum applies to cases where;
1. The vessel is in lethal service and requires PWHT in all thicknesses.
2. The vessel is being heat treated voluntarily to prevent a service induced problem such as
cracking i.e. Amine service.

61
Note (2) Postweld heat treatment is mandatory under the following conditions:
(a) For welded joints over 1-1/2 in. (38 mm) nominal thickness; ( So 1- 9/16” (40 mm) and greater
would require PWHT)
(b) For welded joints over 1-1/4 in. (32 mm) nominal thickness through 1-1/2 in. (38 mm) nominal
thickness unless preheat is applied at a minimum temperature of 200°F (93°C) during
welding;
Based on the various thicknesses up to 2 inches we have the following graphical representation of these
rules.
The Code sets the minimum thickness of a vessel at 1/16” (1.6 mm) in paragraph UG-16, one exception is
for an Unfired Steam Boiler which has a 1/4” (6 mm) minimum.
1/16" to ¼” 15 min.
½” 30 min.
¾” 45 min.
1” - 1 hour.
> 1-¼” 1:15 min. No Preheat
1-½” 1:30 min. No Preheat
1-¾” 1:45 min.
2” 2 hours.
Exceeds 1-¼” up to
1-½” if no preheat
applied.
1/16" to ¼” Lethal
or Service reason
> 1-½” Thickness
Reason For PWHT 1/16 to 2 inches
The second thickness range:
• Over 2 in. (51 mm) to 5 in. (127 mm) the PWHT is held for a flat 2 hours for the first 2 inches (51
mm) of thickness with an additional 15 minutes per inch over 2 inches. Let’s look at a graphic of
this thickness range.
First 2 in. (51 mm) 2 hours
OVER 2 in. to 5 in.
(51 mm to 127 mm)
Additional 1 in. (25 mm) 15
min. - Total 2:15 min.
3 in (75 mm)

62
The third thickness range:
• Over 5 in. (127 mm) the PWHT is held for a flat 2 hours for the first 2 inches (51 mm) of
thickness with an additional 15 minutes per inch over 2 inches. For P-Number 1 there is no
change from the previous example. This third range does changes for some of the other P-
Numbers. Look at the P-Number 4 Table for example;
Alternative Post-Weld Heat Treatment Requirements for Carbon and Low Alloy Steels
By Note 1 of Table UCS 56 for P-Number 1 materials, it is possible to PWHT at a temperature lower than
that given in Table UCS-56. It involves heat treating for a longer periods of time, based on the amount of
reduction in temperature below the stated minimum in Table UCS-56.
The following Table UCS-56.1 outlines these rules.

63
Before going any further it must be cautioned that you cannot use this alternate unless you have been
referred to it by a note in one of the material tables. In our example we are using Note 1 referenced by P-
Number 1.
From the table we find that there are three columns.
• The left column lists the decrease in temperature below that given on the appropriate table in
UCS-56 based on material.
• The center lists the Minimum Holding Time at a decreased temperature and;
• The third lists references to notes below the table.
Reading Note 1 found below the chart and referenced up in the Minimum Holding Time column we see;
(1) Minimum holding time for 1 in. (25 mm) thickness or less. Add 15 minutes per inch(25 mm) fro
thickness greater tan 1 in. (25 mm).
Reading Note 2 listed in the Notes column;
(2) These lower postweld heat treatment temperatures permitted only for P-No. 1 Gr. 1 and 2 materials.
As regards Note 2, there is a P-No. 1 Group 3 material. So be cautious on the exam this could be a
trick question.

64
Note 1 is best addressed using a graphic as follows;
We will first examine a 50°F (28°C) drop from 1100 to1050°F. Below is the holding time from our previous
3” coupon based on 1100°F. How long would we be required to hold it at 1050°F?
First 1 in. (25 mm) 2 hours
Lower PWHT at 1050 F
Additional 1 in. (25 mm) add
15 min. Total 2:30
3 in (75 mm)
Additional 1 in. (25 mm) add
15 min.
Which leads to this total time, up from 2:15 min. to 2:30 min.
Now how about 100°F reduction to 1000°F?

65
Lesson 7
UG-28 External Pressure
UG – 28 Thicknesses of Shells and Tubes under External Pressure
(a) Rules for the design of shells and tubes under external pressure given in this Division are limited to
cylindrical shells, with or without stiffening rings, tubes, and spherical shells…..
(b) The symbols defined below are used in the procedures of this paragraph:
A = Factor determined from Fig. G in Subpart 3 of Section II, Part D and used to enter the
applicable material chart in Subpart 3 of Section II, Part D……
B = Factor determined from the applicable material chart in Subpart 3 of Section II, Part D for
maximum design metal temperature, psi .
DO = Outside diameter of cylindrical shell course or tube, in.
E = Not on exam.
L = Total length, in. (mm), of a tube between tubesheets, or design length of a vessel section
between lines of support (see Fig. UG-28.1). A line of support is:
(1) A circumferential line on a head…. Not on exam!
(2) A stiffening ring……. Not on exam!
(3) A jacket closure ……. Not on exam!
(4) A cone-to-cylinder Not on exam!
P = External design pressure, psi
Pa = Calculated value of maximum allowable external working pressure for the assumed value of t,
psi
RO = Outside radius of spherical shell, in.
t = Minimum required thickness of cylindrical shell or tube, or spherical shell, in.
ts = Nominal thickness of cylindrical shell or tube, in.

66
Beginning with UG-28(c) there are step by step instructions for working these problems. We will go
through these steps one at a time.
(c) Cylindrical Shells and Tubes. The required minimum thickness of a cylindrical shell or tube under
external pressure, either seamless or with longitudinal butt joints, shall be…….
1. Cylinders having Do /t values > or = 10:
Step 1 Assume a value for t and determine the ratios L/Do and Do /t.
* You do not assume a value for thickness (t) on the exam, it will be given in the stated problem for
the external pressure shell or tube calculation. As will the (Do) Diameter Outside and the (L)
Length in other words all that is needed to solve the problem will be provided.
Looking at an example, we can start learning this process.
1. Cylinders having Do /t values > or = 10:
Step 1 Assume a value for t and determine the ratios L/Do and Do /t.
Example: The cylinder has corroded to a wall thickness of 0.530”, its length is 120” and the outside
diameter is 10”. It operates at 500°F.
So then;
Temperature = 500°F
t = 0.530”
L = 120”
Do = 10”
Calculate Do/t = 10/.530 = 18.8 call it 19 (no need to be exact)
Now we do L/Do = 120/10 = 12

67
Step 2 Enter Fig. G in Subpart 3 of Section II, Part at the value of L/Do determined in Step 1
For values of L/Do greater than 50, enter the chart at a value of L/Do = 50. For values of L/Do less than
0.05, enter the chart at a value of L/Do = 0.05.
In our example problem we must go up the left side of the Fig. G until we reach the value of L/Do of 12.
• Using the chart we have the following;
•

68
Step 3 Move horizontally to the line for the value Do /t determined in Step 1.. Which in our case was 19,
but we will round this to 20 since these problems are not meant to be extremely precise. So now
we have.

69
Step 4
From this point of intersection move vertically downward to determine the value of factor A
This gives us the following;

70
Step 5 Using the value of A calculated in Step 3, enter the applicable material chart in Subpart 3 of
Section II, Part D for the material under consideration. Move vertically to an intersection with the
material/temperature line for the design temperature see UG-20). Interpolation may be made
between lines for intermediate temperatures. In cases where the value of A falls to the right of
the end of the material /temperature line, assume an intersection with the horizontal projection of
the upper end of the material/temperature line.
To use the next figure we enter at the bottom at the value Factor A = .0028 and then up to our
temperature of 500°F.
A = .0028 and then up to our temperature of 500°F.
Factor B is 12,000. Plug it into the formula and we have our External Pressure allowable, Pa Which will
be;
As regards the final answers to these problems, because of the difficulty of being precise with the Fig. G
there will always be some difference from one person to the next in the determination of Factor A. This is
allowed for on the exam by listing choices of answers that are in a range of +/- 5%. In our previous
problem the answer was 842 psi, on the exam the correct choice would have been offered as 799 to 884
psi, i.e.
Answer Range: 799 – 884 psi
psi
842
57
48000
3x19
4x12000
=
Pa 


71
To Summarize UG-28
External calculations depart significantly from internal calculations simply because under external
pressure the vessel is being crushed. Internal pressure wants to tear the vessel apart.
Because of the crushing or buckling load, the Length the Outside Diameter and the Thickness of the
vessel are important. External pressure problems are based on the thickness of the shell to the outside
diameter ratios. There are two types of external pressure calculations, the type we will use is when the
O.D to (Do) thickness ratio (t) is greater than 10 and the other type, not on the test, is when it is less than
10.
In order to solve these types of problems two charts will be required. The first chart Fig. G is used to find a
value called Factor A and then Factor A is used to find a Factor B in the second material specific chart.
The value of Factor B found is the number needed to solve the problem using the formula given in
paragraph UG-28 (c)(1) step 6. As stated in the API 510 Body of Knowledge, these charts will be provided
in the exam body, IF an external calculation is given on the examination.

72
Find the allowed external pressure on an existing vessel of a known thickness with a Do/t ratio > 10.
Problem: A vessel is operating under an external pressure, the operating temperature is 500° F. The
outside diameter of the vessel is 40 inches. Its length is 70 inches. The vessel’s wall is 1.25
inches thick and is of SA-515-70 plate. Its specified min. yield is 38,000 psi. What is the
maximum external pressure allowed?
Givens:
Temp = 500° F
t = 1.25
L = 70 inches
D0 = 40 inches
From UG-28 (c) Cylindrical Shells and Tubes
The required minimum thickness of a shell or a tube under external pressure, either seamless or with
longitudinal butt joints, shall be determined by the following procedure.
(1) Cylinders having a …
Testing to see if this paragraph applies:
32
=
1.25
40
=
t
Do
Step 1 Our value of Do is 40 inches and L is 70 inches. We will use these to determine the ratio of:
1.75
=
40
70
=
o
D
L

73
Step 2 Enter the Factor A chart at the value of 1.75 previously determined.
Step 3 Then move across horizontally to the curve Do/t = 32. Then down from this point to find the value
of Factor A which is .0045

74
Step 4. Using our value of Factor A calculated in Step 3, enter the Factor B (CS-2) chart on the bottom.
Move vertically to the material temperature line given in the stated problem (in our case 500°F).
Step 5 Then across to find the value of Factor B. We find that Factor B is approximately 13000.
Step 6 Using this value of Factor B; calculate the value of the maximum allowable external pressure Pa
using the following formula:
)
t
D
3(
4B
=
Pa
o
psi
541.66
=
96
52,000
=
3(32)
4x13,000
=
Pa
+/- 5%
Answer Range: 514 – 568 psi

75
Lesson 8 Charpy Impact Testing
This is why we Impact test!
Brittle Fracture and Charpy impact Testing
Overview
The concern expressed by the Codes should now be very clear based on the previous pictures. Brittle
fracture can and does occur. So exactly what is brittle fracture? Perhaps the best way to explain this is to
contrast two materials commonly used:
• Glass and Lead
• Glass is of course very brittle at room temperature.
• Lead is very ductile at room temperature.
• Glass shatters when struck.
• Lead deforms, it flows plastically without rupture (ductility).
• Glass is hard with a high strength and has little ductility.
• Lead is soft with a low strength and deforms under load.
In pressure vessels we need something in between.
So using the term loosely, we do not want our pressure vessel in “glass like state” when it is exposed to
lower temperatures. It must be able to absorb energy in the presence of a Code acceptable size flaw such
as a small welding discontinuity or unknown crack like flaw.
Designers must evaluate a given material of construction for its acceptability, at what ASME Section VIII
refers to as the vessel’s Minimum Design Metal Temperature (MDMT).
The question becomes, is this metal in this thickness and heat treated condition, prone to brittle fracture at
the desired MDMT?
Section VIII has several paragraphs that address the acceptability for materials; these are referred to as
exemptions from Charpy Testing. For purposes of the examination we are restricted to paragraph UCS.

76
So the task becomes evaluating a given material for exemptions from testing. This is a four step process,
ending with a ‘yes you must’ or ‘no you don’t’ solution.
The four steps are;
1. The exemption given in paragraph UG-20(f).
2. The exemptions listed in UCS-66 (Table UCS-66).
3. The reduction in temperature provided by Table UCS-66.1 to Table UCS-66
4. The reduction in temperature to Table UCS-66 given in paragraph UCS-68(c).
If at the end of the 4 steps, impact testing is required, then they must be conducted in accordance with the
rules described in the paragraph UG-84.
There exist two possible categories of questions on the examination.
1. Are they required?
2. If the tests are required
• How must they be conducted and,
• What passes and what is considered to have failed the tests?
We start with are they required?
The search will begin in UG-20(f) and progress through UCS 66, and 68. If no exemption is found impact
tests are required. The best approach is to list these by steps.

77
Step 1
UG-20(f) lists an exemption from impact testing for materials that meet “All” of the following
requirements.
1. Material is limited to P-No.1 Gr. No.1 or 2 and the thicknesses don't exceed the following:
(a) 1/2 in. for materials listed in Curve A of Fig. UCS-66;
(b) 1 in for materials from Curve B, C or D of Fig. UCS-66;
2. The completed vessel shall be hydrostatically tested
3. Design temperature is no warmer than 650°F or colder than -20°F.
4. The thermal or mechanical shock loadings are not controlling design.
5. Cyclical loading is not a controlling design requirement.
Reminder
All of the conditions of UG-20(f) must be met to take this exemption from impact testing.
Step 2 UCS-66 (a) Turn your attention to Fig. UCS-66 Impact Test Exemption Curves and Table UCS-66.
The Graph or Table is used to determine the minimum temperature a material thickness can be operated
at without mandatory impact testing.
The graph has four curves: A, B, C and D. In Fig. UCS-66 along with the graph is a listing of carbon and
low alloy steels. This listing of materials is used to determine the curve on the Graph or in the Table for a
given material.
After finding the curve for the material, there are two choices.
You may use the graph of figure UCS 66 or the Table UCS 66 to determine the minimum temperature for
a given thickness. It is recommended to use the Table. The Table is a lot easier to use with accuracy.
If the material thickness is operated at or above the temperature listed in Table UCS-66, impact tests are
not required. If the material thickness is to operate below the given minimum temperature, impact testing
is required. The temperature found in the table is the MDMT of that material thickness without impact
testing being required.

79
Step 2 Figure UCS-66 Material Curves
Let’s take the example of material that has been assigned to Curve B which is 2 inches (51 mm) thick.
Using the more friendly table we find the column for Curve B materials, move down until we find the
thickness row for 2 inches and across to find the MDMT that this material can be used without impact
testing is 63
o
F (17
o
C).
That doesn’t seem like an acceptable minimum design temperature for most vessels. This makes a Curve
B material a poor choice at 2” thickness.
Step 3 Figure UCS 66.1 Coincident Ratio
The Coincident Ratio is based on a vessel’s extra thickness due to its design calculations which were
based on its Maximum Temperature. Meaning that;
As metal’s temperature increases its strength decreases, hotter means weaker, therefore the allowable
stress is decreased during calculations resulting in vessel that requires thicker walls when hot than when it
is operating at its coldest temperature, the MDMT.
This ratio takes credit for the extra wall thickness that is present, but not needed to resist pressure at the
MDMT. The following graphic will help.
Usually when there is a drop in temperature there is also a drop in the pressure. The two operating
conditions are calculated and the Ratio is determined. This Ratio is given on the exam and you need only
use the table to apply this rule.

80
Coincident Ratio Figure UCS-66.1
Using the Coincident Ratio given in a problem we enter the graph on the left side at that value then across
to intersect the curve now down to find a temperature given. We take that temperature back to Table
UCS-66 and reduce the temperature given there for the material of interest by the amount we found using
Table UCS-66.1.

81
Example: The Coincident Ratio is given as .60. Now using our previous Table UCS-66 2” Curve B
material that has a MDMT of 63
o
F we adjust and find a new MDMT. Like this!
63 – 40 = 23
o
F our adjusted MDMT
Step 4
UCS-68(a) Design rules for carbon and low alloy steels stipulate requirements about construction of the
vessel or part. The main points are: mandatory joint types, required post weld heat
treatments below -55 °F unless the vessel is installed in a fixed (stationary) location, and the
coincident Ratio of stress is less than 0.35.
UCS-68(b) Welded joints must be postweld heat treated when required buy other rules of this Division
or when the MDMT is colder than -55°F and for vessel installed in a fixed (stationary)
location the coincident Ratio is 0.35 or greater.
UCS-68(c) Notice a reduction of 30°F below that of Figure UCS-66 for P-1 materials if post welded
heat treatment is performed when it is not otherwise required in the Code. This means that
30°F can be subtracted from the temperature found in Table UCS-66. If the adjusted
temperature is below that desire, Impact Tests are not required. It is exempt. If a statement
about heat treatment is made in a particular problem the task becomes finding out if heat
treatment was required or not. If it is not mentioned, it must be concluded that it was not
performed and therefore the exemption cannot be taken.
Example:
Givens:
Material SA-516-70 normalized (plate)
Thickness 2"
Min. Yield 38 KSI
MDMT -25 °F
Coincident Ratio = .85
Step 1: Check for the exemptions of UG-20(f)

82
Our material applies to Curve D of Figure UCS-66 and exceeds the 1“ limit for exemption. It also
exceeds and lower temperature limits - 20°F.
Our Material 516 Normalized is on Curve D below
FIG. UCS-66 IMPACT TEST EXEMPTION CURVES [SEE NOTES (1) AND (2)] [SEE UCS-66(A)]

83
Step 2: Checking Table UCS-66 and entering at our thickness of 2 inches on the left and moving across
to Curve D column, we find the MDMT of this thickness to be -4°F. This exemption does not
apply our goal is -25°F.
Step 3: Checking Fig. UCS-66.1 and entering at our stated Coincident Ratio of .85 and then down to
read the temperature reduction permitted we find 15°F.

84
Step 3: This 15°F is subtracted directly from the table UCS-66.
So we now have -4 from Table UCS-66
And -15 from Table UCS-66.1
-19 °F
Not there yet, we need -25 °F to be exempt from testing.
Step 4
UCS-68 (c) If postweld heat treating is performed when it is not otherwise a requirement of this
Division, a 30°F (17 °C) reduction in impact testing exemption temperature may be given to
the minimum permissible temperature from Fig. UCS-66 for P-No.1 materials.
P-1 materials (only) if post welded heat treatment is performed when it is not otherwise required.
This would occur if the note 2(b) of table UCS-56 for P No. 1 materials is complied with or if the vessel is
in general service and has no mandatory heat treatment requirements in the Code.
Checking UCS-68 (c), we find that we cannot take a reduction because PWHT is a requirement of UCS-
56 for this material's thickness of 2 inches.
Answer:
Impact tests are required for the desired MDMT of -25 °F.
So how must they be done?

85
UG-84 Charpy Impact Tests
UG-84(a) General
Charpy V-notch impact tests in accordance with the provisions of this paragraph shall be made on
weldments and all materials for shells, heads, nozzles, and other vessel parts subject to stress due to
pressure for which impact tests are required by the rules in Subsection C.
UG-84(b) Test Procedures
UG-84(b)(l) Impact test procedures and apparatus shall conform to the applicable paragraphs of SA-
370 or IS0 148 (Parts 1, 2, and 3).
UG-84(c) Test Specimens
UG-84(c) (1) Each set of impact test specimens shall consist of three specimens.
UG-84(c) (1) Each set of impact test specimens shall consist of three specimens.
UG-84(c) (2) The impact test specimens shall be of the Charpy V-notch type and shall conform in all
respects to Fig. UG-84.The standard (10 mm ×10 mm) specimens, when obtainable, shall
be used for nominal thicknesses of 0.438 in. (11.13 mm) or greater, except as otherwise
permitted.
.

86
UG-84(c)(6) When the * average value of the three specimens equals or exceeds the minimum value
permitted for a single specimen and the value for more than one specimen is below the
required average value, or when the value for one specimen is below the minimum value
permitted for a single specimen, a retest of three additional specimens shall be made. The
value for each of these retest specimens shall equal or exceed the required average value.
Example: Average required is 15 ft-lb (joules 20.4)
* 15 + 16 + 14 = 45/3 = 15 Passed.
* 18 + 14 + 13 = 45/3 = 15 Failed, more than one below 15
* 18 + 18 + 9 = 45/3 = 15 Failed, one below 2/3 of 15 = 10
UG-84(d)(1) Reports or certificates of impact tests by the material manufacturer will be acceptable
evidence that the material meets the requirements of this paragraph, provided the
specimens comply with UCS-85, UHT-5, or UHT-81, as applicable.
* This means you may not have to do impact tests on your base material because the manufacturer
has performed the tests already and they meet the Code requirements.
UG-84(f)(2) All test plates shall be subjected to heat treatment, including cooling rates and aggregate
time at temperature or temperatures as established by the Manufacturer for use in actual
manufacture.
Let’s try to interpret what this statement means in practical terms.
*Charpy tests must be conducted with the test coupon having received the minimum required PWHT. So if
a vessel underwent 2 hours of PWHT, then it has been proven that the 2 hours of PWHT did not make the
vessel subject to brittle fracture.
Now suppose you need to repair the vessel after having been in service. PWHT would be required again
for two hours where the repair was made. That local area will now have seen 4 hours of PWHT. Has the
additional 2 hours made it brittle? There is only one way to find out, more Charpy testing. Where do you
get the metal from for the Charpy coupons? Out of the vessel, which is not easy and very expensive? **
API 510 has an alternative method.

87
UG-84(g) Location, Orientation, Temperature, and Values of Weld Impact Tests
All weld impact tests shall comply with the following:
UG-84(g)(1) Each set of weld metal impact specimens shall be taken across the weld with the notch in
the weld metal. Each specimen shall be oriented so the notch is normal to the surface of
the material and one face of the specimen shall be within 1/16” (1.6 mm) of the surface of
the material.
Front View

88
UG-84(g)(2) Each set of heat affected zone impact specimens shall be taken across the
weld and of sufficient length to locate, after etching, the notch in the heat affected zone.
The notch shall be cut approximately normal to the material surface in such a manner as to
include as much heat affected zone material as possible in the resulting fracture.

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