ISO/TR 10771-2:2008

TECHNICAL ISO/TR
REPORT 10771-2
First edition
2008-12-01
Hydraulic fluid power — Fatigue pressure
testing of metal pressure-containing
envelopes —
Part 2:
Rating methods
Transmissions hydrauliques — Essais de fatigue des enveloppes
métalliques sous pression —
Partie 2: Méthodes de classement

Reference number
©
ISO 2008
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©  ISO 2008
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ii © ISO 2008 – All rights reserved

Contents Page
Foreword. iv
Introduction . v
1 Scope . 1
2 Normative references . 1
3 Terms and definitions. 2
4 Selection of material factors. 2
5 Determination of cyclic test pressure. 2
6 Conduct of fatigue test. 3
7 Rating by similarity. 4
8 Rating declaration. 4
9 Identification statement (reference to this part of ISO 10771) . 4
Annex A (informative) Material factor database. 5
A.1 Values of coefficient of variation, k , for commonly used metals . 5
o
A.2 Procedures used to establish values of coefficient of variation, k , for the metals listed in
o
Table A.1 . 5
Annex B (normative) Calculation of variability factor K . 13
V
B.1 General. 13
B.2 Method . 13
Annex C (informative) Proposal for an acceleration factor. 15
C.1 General. 15
C.2 Extrapolating data to 10 cycles . 15
C.3 Examples . 19
[9]
Annex D (informative) Basis of fatigue pressure rating . 21
D.1 Basis of pressure rating. 21
D.2 Statistical analysis theory. 22
D.3 Fatigue distribution data. 28
D.4 Data calculation example. 33
D.5 Raw data points for sample problem. 39
Bibliography . 40

Foreword
ISO (the International Organization for Standardization) is a worldwide federation of national standards bodies
(ISO member bodies). The work of preparing International Standards is normally carried out through ISO
technical committees. Each member body interested in a subject for which a technical committee has been
established has the right to be represented on that committee. International organizations, governmental and
non-governmental, in liaison with ISO, also take part in the work. ISO collaborates closely with the
International Electrotechnical Commission (IEC) on all matters of electrotechnical standardization.
International Standards are drafted in accordance with the rules given in the ISO/IEC Directives, Part 2.
The main task of technical committees is to prepare International Standards. Draft International Standards
adopted by the technical committees are circulated to the member bodies for voting. Publication as an
International Standard requires approval by at least 75 % of the member bodies casting a vote.
In exceptional circumstances, when a technical committee has collected data of a different kind from that
which is normally published as an International Standard (“state of the art”, for example), it may decide by a
simple majority vote of its participating members to publish a Technical Report. A Technical Report is entirely
informative in nature and does not have to be reviewed until the data it provides are considered to be no
longer valid or useful.
Attention is drawn to the possibility that some of the elements of this document may be the subject of patent
rights. ISO shall not be held responsible for identifying any or all such patent rights.
ISO/TR 10771-2 was prepared by Technical Committee ISO/TC 131, Fluid power systems, Subcommittee
SC 8, Product testing.
ISO/TR 10771 consists of the following parts, under the general title Hydraulic fluid power — Fatigue pressure
testing of metal pressure-containing envelopes:
⎯ Part 1: Test method
⎯ Part 2: Rating methods
iv © ISO 2008 – All rights reserved

Introduction
In hydraulic fluid power systems, power is transmitted and controlled under pressure within a closed circuit. It
is important for the manufacturer and user of hydraulic components to have information on their global
reliability because of the importance of the fatigue failure mode and the relationship with their functional safety
and service life. This part of ISO 10771 provides a method for fatigue-testing in order to verify the rating of a
pressure-containing envelope.
During operation, components in a system can be subjected to loads that arise from:
⎯ internal pressure;
⎯ external forces;
⎯ inertia and gravitational effects;
⎯ impact or shock;
⎯ temperature changes or gradients.
The nature of these loads can vary from a single static application to continuously varying amplitudes,
repetitive loadings and even shocks. It is important to know how well a component can withstand these loads,
but this part of ISO 10771 addresses only the loads due to internal pressure.
There are several International Standards already in existence for pressure rating of individual components
(e.g. for determining maximum allowable rated pressure) and this part of ISO 10771 is not intended to replace
them. Instead, a method of fatigue verification is provided.
This part of ISO 10771 describes a universal verification test to give credibility to the many in-house and other
methods of determining the pressure rating of the components. Credibility is based upon the fundamental
nature of metal fatigue with its statistical treatment and a mathematical theory of statistical verification.
Nevertheless, it is necessary to have design knowledge of the component and its representative specimens to
maximize accuracy of the verification method. The use of this test method can reduce the risk of fatigue failure
for a hydraulic component regardless of sample size.
In order to rate components in accordance with this part of ISO 10771, it is necessary to propose a rating for
the component, select test specimens and select a test pressure. A fatigue test is then conducted in
accordance with ISO 10771-1. If the test is successful, the proposed rating is verified for the family of
components represented by the sample.
This part of ISO 10771 is based on ANSI/(NFPA) T 2.6.1, a standard which was developed and has been
used in the United States for over 25 years and has been adopted for use in Japan as JSME S006-1985. If
sufficient experience is gained in other parts of the world, and additional data on materials are obtained, this
part of ISO 10771 might be re-drafted as an International Standard in the future.
It should be noted that the test factors in Annex A are based on material data obtained from sources
originating in the USA. One of the objectives in issuing this part of ISO 10771 is to obtain material data from
other countries. The test factors are based only on the material properties and not on any tolerances of the
elements in the pressure-containing envelope.
Annex C describes a possible method for accelerating testing. The example shows how material property data
can be used to determine an acceleration factor and shows that they have to be carefully chosen. Another
objective of this part of ISO 10771 is to seek additional data as described in Annex C. Contributors are asked
to submit any available data to the secretary of ISO TC 131/SC 8.
TECHNICAL REPORT ISO/TR 10771-2:2008(E)

Hydraulic fluid power — Fatigue pressure testing of metal
pressure-containing envelopes —
Part 2:
Rating methods
1 Scope
This part of ISO 10771 specifies a test method for fatigue rating of the pressure-containing envelopes of
components used in hydraulic fluid power systems, as tested under steady internal cyclic pressure loads in
accordance with ISO 10771-1.
This part of ISO 10771 is only applicable to components whose failure mode is the fatigue of any element in
the pressure-containing envelope, and that:
⎯ are manufactured from metals;
⎯ are operated at temperatures that exclude creep and low-temperature embrittlement;
⎯ are only subjected to pressure-induced stresses;
⎯ are not subjected to loss of strength due to corrosion or other chemical action;
⎯ can include gaskets, seals and other non-metallic components; however, these are not considered part of
the pressure-containing envelope being tested (see note 3 of 5.5 of ISO 10771-1:2002).
This part of ISO 10771 does not apply to piping as defined in ISO 4413 (i.e. connectors, hose, tubing, pipe).
NOTE See ISO 19879, ISO 6803 and ISO 6605 for methods of fatigue testing of tube connectors, hoses and hose
assemblies.
This part of ISO 10771 establishes a general rating method that can be applied to many hydraulic fluid power
components. In addition, EN 14359 has been developed for accumulators.
2 Normative references
The following referenced documents are indispensable for the application of this document. For dated
references, only the edition cited applies. For undated references, the latest edition of the referenced
document (including any amendments) applies.
ISO 5598, Fluid power systems and components — Vocabulary
ISO 10771-1:2002, Hydraulic fluid power — Fatigue pressure testing of metal pressure-containing
envelopes — Part 1: Test method
3 Terms and definitions
For the purposes of this document, the terms and definitions given in ISO 5598, ISO 10771-1 and the
following apply.
3.1
rated fatigue pressure
P
RF
maximum pressure that a component pressure-containing envelope, selected at random, has been verified to
sustain for the rated cycle life without failure, with a known probability
3.2
assurance level
probability that the fatigue strength of a randomly selected test specimen exceeds its rated fatigue pressure
3.3
verification level
probability that the fatigue strength of a randomly selected test specimen is not less than its cyclic test
pressure
3.4
coefficient of variation
k
o
standard deviation of the fatigue strength distribution of a material at a given fatigue life, divided by its mean
[1]
NOTE Adapted from ISO 3534-1:2006 .
3.5
variability factor
K
V
ratio of cyclic test pressure to rated fatigue pressure
3.6
element
part of a component; for example, tie rods on a cylinder, end caps on a valve, bolts on a pump housing
4 Selection of material factors
4.1 Select a coefficient of variation, k , for each type of material in the pressure-containing envelope. The k
o o
factor should be obtained from fatigue tests on coupons for the particular temper of material used in the
pressure-containing envelope. The fatigue test method used to obtain this data should be in accordance with
a recognized national or International Standard.
4.2 As an alternative to testing the specific material, coefficients described in Annex A can be used for the
k factor.
o
5 Determination of cyclic test pressure
5.1 Select an assurance level for the fatigue pressure rating. A nominal value is 90 %.
5.2 Select a verification level for the fatigue pressure rating. A nominal value is 90 %.
NOTE See Annex D for a tutorial that describes these terms.
5.3 Select a number of component specimens to be tested, then determine the number of element
specimens that will be tested in the components.
NOTE The verification is independent of sample size because the test pressure compensates for different quantities.
2 © ISO 2008 – All rights reserved

5.4 Determine the variability factor, K , for each element in the component using Table 1 and the procedure
V
described in the example given in Annex B. Use the largest K factor so obtained, for the calculations
V
described in the example.
5.5 Propose a rated fatigue pressure for the pressure-containing envelope of the component.
5.6 Calculate the cyclic test pressure, P , using Equation (1):
CT
PK=×P (1)
CT V RF
where
K is the variability factor;
V
P is the rated fatigue pressure of the component pressure-containing envelope.
RF
Table 1 — Variability factor, K (at a verification level of 90 %)

V
b
Assurance No. of Material coefficient of variation, k
o
level specimens
0 0,02 0,04 0,06 0,08 0,10 0,12 0,14 0,16 0,18 0,20 0,22 0,24 0,26 0,28 0,30
a
n
1 1,001,09 1,20 1,321,461,631,832,082,382,773,29 — — — — —
2 1,001,08 1,16 1,261,381,521,681,882,132,452,87 — — — — —
99,9 %
3 1,00 1,07 1,15 1,23 1,34 1,46 1,61 1,78 2,01 2,29 2,66 3,18 — — — —
4 1,00 1,06 1,13 1,22 1,31 1,42 1,56 1,72 1,93 2,19 2,54 3,02 — — — —
5 1,00 1,06 1,13 1,20 1,29 1,40 1,53 1,68 1,87 2,12 2,44 2,89 — — — —
1 1,00 1,08 1,16 1,25 1,35 1,47 1,60 1,75 1,92 2,12 2,35 2,63 2,96 — — —
2 1,00 1,06 1,12 1,20 1,28 1,37 1,47 1,58 1,72 1,87 2,05 2,26 2,52 2,85 — —

99 %
3 1,00 1,05 1,11 1,17 1,24 1,32 1,40 1,50 1,62 1,75 1,90 2,09 2,31 2,59 2,94 —
4 1,00 1,05 1,10 1,15 1,21 1,28 1,36 1,45 1,55 1,67 1,81 1,98 2,18 2,43 2,74 3,16
5 1,00 1,04 1,09 1,14 1,20 1,26 1,33 1,41 1,51 1,62 1,75 1,90 2,08 2,31 2,60 2,98
1 1,00 1,05 1,11 1,17 1,23 1,29 1,36 1,44 1,52 1,60 1,69 1,79 1,89 2,00 2,12 2,25
2 1,00 1,04 1,07 1,11 1,16 1,20 1,25 1,30 1,35 1,41 1,47 1,54 1,61 1,69 1,77 1,86
90 % 3 1,00 1,03 1,06 1,09 1,12 1,16 1,19 1,23 1,28 1,32 1,37 1,42 1,48 1,54 1,60 1,67
4 1,00 1,02 1,05 1,07 1,10 1,13 1,16 1,19 1,23 1,26 1,30 1,34 1,39 1,44 1,49 1,55
5 1,00 1,02 1,04 1,06 1,08 1,11 1,13 1,16 1,19 1,22 1,25 1,29 1,33 1,37 1,41 1,46
a
Test twice the number of specimens if a 99 % verification level is chosen.
b
Use an interpolation of k values between those tabulated here, or calculate K from Equation (D.14) in Annex D.
o V
6 Conduct of fatigue test
5 7
6.1 Determine the number of cycles, between 1 × 10 and 1 × 10 , for which the component will be rated.
6.2 Subject the test specimens to a fatigue pressure test in accordance with ISO 10771-1 for the number of
cycles determined in 6.1, using the P calculated from Equation (1).
CT
6.3 The fatigue pressure test is successful if all of the element specimens selected in 5.3 do not fail as
described in ISO 10771-1:2002, Clause 8.
7 Rating by similarity
It is permitted to extend a verified P to other components of similar shape if it can be shown that differences
RF
between those components and the components tested do not result in any reduction of their fatigue strength
capabilities. Examples of this are components that have smaller ports or different axial lengths but are
otherwise identical in geometry to the component tested.
8 Rating declaration
The P proposed in 5.5 will be verified if the requirements of 6.3 are met. A code should be applied to the
RF
component to declare its rating as:
P = P (in megapascals)/assurance level/verification level/K in the test/number of test cycles
RF RF V
EXAMPLE The rated fatigue pressure (12,5 MPa) of a component’s pressure-containing envelope that was tested at
an assurance level of 99 %, a verification level of 90 %, a K of 1,36 for 2 × 10 cycles, would be declared as:
V
P = 12,5 MPa/ 99 %/ 90 %/ 1,36/ 2 × 10 cycles
RF
9 Identification statement (reference to this part of ISO 10771)
Use the following statement in test reports, catalogues and sales literature when complying with this part of
ISO 10771:
“Method for fatigue pressure rating conforms to ISO TR 10771-2:2008, Hydraulic fluid power — Fatigue
pressure testing of metal pressure-containing envelopes — Part 2: Rating methods”.
4 © ISO 2008 – All rights reserved

Annex A
(informative)
Material factor database
A.1 Values of coefficient of variation, k , for commonly used metals
o
Table A.1 tabulates data calculated from the sources listed in the bibliography.
Table A.1 — Values of coefficient of variation, k , for commonly used metals
o
k
Metal
o
Alloy, low 0,14
Carbon, plain 0,08
Steel
Nickel 0,10
Stainless 0,09
Tool 0,10
Iron 0,14
Aluminium (except unalloyed) 0,13
Unalloyed aluminium 0,23
Cobalt 0,13
Nonferrous Copper 0,09
Magnesium 0,17
1)
0,27
Monel
Titanium 0,12
A.2 Procedures used to establish values of coefficient of variation, k , for the metals
o
listed in Table A.1
A.2.1 Values of k were calculated from fatigue test data on test coupons that were published in the
o
references cited in the bibliography. The types of data taken from these references were one of the following:
a) Means, µ, and standard deviations, σ, of normal distributions;
b) parameters of Weibull distributions;
c) raw data points on S-N curves. From these data, individual coefficients of variation, k , were calculated at
o
10 cycles for:
1) normal distributions; k equals the standard deviation divided by the mean;
o
1) This is an example of a suitable product available commercially. This information is given for the convenience of users
of this document and does not constitute an endorsement by ISO of this product.
2) Weibull distributions; k were calculated from a formula given in Reference [12]. The formula includes
o
a gamma function, the value of which was selected as a constant at 0,89 because its variations were
generally less than ± 2 % in the range of interest (a few data points went to a difference of ± 4 %);
3) S-N curves; the references had either included limit bands (assumed to be two sigma from the mean)
or actual standard deviation points. These were then used to calculate k in the same manner as a
o
normal distribution.
A.2.2 The resulting k values (shown as individual values in Table A.2 to Table A.13) include a mix of
o
notched and unnotched specimens, several different tempers, plus different methods of testing (e.g. axial,
rotating beam). However, only those tested at room temperature were used. No attempt was made to
segregate these data. It is reasoned that the components to be tested will have a variety of tempers and
notches, so an application of these published data to components can only be justified if the data are treated
statistically at a conservative value.
A.2.3 Therefore, the values given in Table A.1 were derived by assuming that all k data for a particular
o
metal group are part of a normal distribution, and a value that is greater than 90 % of this distribution was
selected. This ensures that the selection is substantially conservative. However, this part of ISO 10771 allows
the use of a more accurate k value, which is representative of the specific alloy and temper of the elements
o
being tested, if sufficient testing is performed to obtain those data, as described in 4.1. This approach will
likely yield a value that would be more advantageous for a particular application, but less than the
conservative values presented in Table A.1.
A.2.4 Table A.2 to Table A.13 describe all of the k calculations made from the data obtained from
o
Reference [10], Reference [11], and Reference [13] to Reference [17]. Most of the data are based on the
strength distribution at 10 cycles, but some data are at the endurance limit and these are identified in each
table, if applicable.
Table A.2 — Summary of k calculations for iron
o
a
k values
Type Reference Number of distributions
o
1)
[11] 1 0,0220
Armco
a a a a
0,0402; 0,042 ; 0,044 ; 0,0652; 0,126 ; 0,146 ;
Pearlitic (grey) [15];[17] 8
a
0,137
a a a a
Ferritic (malleable) [15];[17] 7 0,055 ; 0,0596; 0,0649; 0,065 ; 0,075 ; 0,109
a a a a a a
0,029 ; 0,040 ; 0,049 ; 0,065 ; 0,086 ; 0,094 ;
Nodular [15] 10
a a a a
0,095 ; 0,098 ; 0,173 ; 0,185
Fe; 5,5 % Mo; 2,5 %
[13] 2 0,0286; 0,0477
Cr; 0,5 % C
(k ) 90 % = 0,1335; (µ = 0,0771; σ = 0,0440)
Summary of all data
o
[13];[15];[11];[17] 28
above Conclusion: k value selected = 0,14
o
a
Data from reference [15] are at endurance limit.

6 © ISO 2008 – All rights reserved

Table A.3 — Summary of k calculations for aluminium
o
a
k values
Alloy Reference Number of distributions
o
1) 0,0720
[13] 1
Duraluminum
356 [15] 2 0,038; 0,042
355 [14] 1 0,0766
1100 [14] 2 0,1742; 0,2377
0,017; 0,0400; 0,0527; 0,0534; 0,0541; 0,0556;
2014 [14];[15] 13 0,0702; 0,0732; 0,1152; 0,1164; 0,1215; 0,1386;
0,1400
0,026; 0,0498; 0,0542; 0,0561; 0,0613; 0,0708;
2024 [14];[15] 14 0,0717; 0,0765; 0,0825; 0,0974; 0,1039; 0,1190;
0,1404; 0,1840
2025 [14] 3 0,0347; 0,0549; 0,0947
2026 [14] 2 0,0507; 0,0834
2219 [14] 2 0,0701; 0,0705
5052 [14] 2 0,0845; 0,0914
5056 [14] 1 0,0947
5086 [14] 1 0,0640
5154 [14] 1 0,0662
5456 [14];[15] 2 0,012; 0,0708
6061 [14];[15] 4 0,018; 0,027; 0,0478; 0,087
7039 [14] 1 0,1405
0,040; 0,0505; 0,059; 0,0615; 0,0689; 0,0906;
7075 [14];[15] 8
0,1686; 0,2157
a a
0,0413 ; 0,0593
7076 [10] 2
7079 [14] 3 0,0560; 0,0942; 0,1486
7178 [14] 2 0,0484; 0,0881
R303 [14] 1 0,0434
0,0934; 0,1302
5 Mg Al [14] 2
7,5 Zn 2,5 Mg Al [14] 1 0,0570
Summary of all (k ) 90 % = 0,1390; µ = 0,0811; σ = 0,0452
o
[13];[14];[15];[10] 71
data above
(k ) 90 % = 0,1288; µ = 0,0775; σ = 0,0400
o
Conclusion: select k = 0,13 for alloyed
Remove the o
aluminium;
1100 data
select k = 0,23 for unalloyed
o
aluminium
a
Data from reference [10] are at endurance limit.

Table A.4 — Summary of k calculations for low alloy steels
o
(containing silicon at less than 1 % and 1 or more of the following: nickel - less than 4 %;
chrome - less than 2 %; molybdenum - less than 0,5 %)
Number of
a
Alloy Reference k values
o
distributions
2340 [13] 6 0,0190; 0,0296; 0,0311; 0,0374; 0,0622 0,0696
3140 [13] 4 0,0145; 0,0283; 0,0435; 0,0919
4140 [13] 2 0,0650; 0,1102
4330 [10] 6 0,0313; 0,0372; 0,0498; 0,0644; 0,1063; 0,1129
0,0525; 0,0819; 0,1023; 0,1037; 0,1219; 0,1285;
4340 [13] 14 0,1301; 0,1335; 0,1428; 0,1438; 0,1476; 0,1484;
0,1600; 0,2035
4340 [11] 4 0,0253; 0,0301; 0,0321; 0,0627
0,0497; 0,0509; 0,0607; 0,0608; 0,0640; 0,0795;
4340 [10] 10
0,0822; 0,0833; 0,0880; 0,0966
0,0759; 0,0837; 0,0878; 0,0895; 0,1025; 0,1055;
4350 [10] 8
0,1209; 0,1213
AMS 5727 [13] 3 0,0341; 0,0385; 0,1002
(k ) 90 % = 0,1348; µ = 0,0812; σ = 0,0418
o
Summary of all data
[13];[11];[10] 57
above
Conclusion: k value selected = 0,14
o
a
Data from reference [10] are at endurance limit.

Table A.5 — Summary of k calculations for cobalt
o
Metal and alloy Reference Number of distributions k values

o
1)
Stellite 31 [14] 3 0,0771; 0,1202; 0,1472
S-816 (AMS5765) [13];[14] 4 0,0448; 0,0456; 0,0730; 0,0777
S-816 (AMS5534) [13] 1 0,0646
(k ) 90 % = 0,1269; µ = 0,0813; σ = 0,0355
Summary of all data
o
[13];[14] 8
above
Conclusion: k value selected = 0,13
o
8 © ISO 2008 – All rights reserved

Table A.6 — Summary of k calculations for copper
o
Number of
a
Metal and alloy Reference k values

o
distributions
100 % Cu [14] 2 0,0522; 0,0804
70/30 Brass [14] 2 0,0153; 0,0735
Cu-7,3 AI BRNZ [14] 2 0,0496; 0,1037
a
AI Ni BRNZ [10] 1
0,0938
a
Beryllium [10] 1
0,0740
0,0153; 0,0173; 0,0181; 0,0185; 0,0200; 0,0211;
Copper casting alloys [16] 8
0,0234; 0,0514
(k ) 90 % = 0,0853; µ = 0,0455; σ = 0,0311
o
Summary of all data
[14];[16];[10] 16
above
Conclusion: k value selected = 0,09

o
a
Data from reference [10] are at endurance limit.

Table A.7 — Summary of k calculations for magnesium
o
Metal and alloy Reference Number of distributions k values

o
AZ31A [14] 1 0,0331
AZ31B [14] 1 0,0714
AZ61A [14] 3 0,0854; 0,1705; 0,1735
AZ80A-F [14] 1 0,0457
AZ81 [14] 2 0,0458; 0,0761
ZK60A [14] 1 0,1053
2,5 Al Mg [14] 3 0,1152; 0,1444; 0,1702
(k ) 90 % = 0,1694; µ = 0,1031; σ = 0,0517
Summary of all o
[14] 12
data above
Conclusion: k value selected = 0,17
o
Table A.8 — Summary of k calculations for plain carbon steel
o
k values
Group and alloy Reference Number of distributions
o
1045 [13] 3 0,0273; 0,0581; 0,0682
1050 [11] 1 0,0171
(k ) 90 % = 0,0739; µ = 0,0427; σ = 0,0244
o
Summary of all
[13];[11] 4
data above
Conclusion: k value selected = 0,08

o
Table A.9 — Summary of k calculations for stainless steel
o
k values
Metal and alloy Reference Number of distributions
o
321 [13] 3 0,0439; 0,0606; 0,0755
A-286 [13] 2 0,0958; 0,1303
347 [13] 4 0,0302; 0,0491; 0,0802; 0,1162
0,0163; 0,0180; 0,0230; 0,0313; 0,0313; 0,0315;
1)
[13] 14 0,0325; 0,0367; 0,0381; 0,0407; 0,0427; 0,0544;
Multimet N-155
0,0547; 0,0574
PH 15-7 [13] 2 0,0676; 0,0936
17-7 PH [13] 4 0,0135; 0,0145; 0,0168; 0,0505;
403 [13] 2 0,0160; 0,0381
(k ) 90 % = 0,0868; µ = 0,0484; σ = 0,0300
o
Summary of all data
[13] 31
above
Conclusion: k value selected = 0,09

o
Table A.10 — Summary of k calculations for tool steel
o
a
Metal and alloy Reference Number of distributions k values

o
a a a
1)
[10] 3 0,0350 ;0,0483 ; 0,0513
Tricent
a
1)
[10] 1 0,0909
Ferrovac
a
H-23 [10] 1 0,0648
a a
0,0672; 0,0704 ; 0,0707; 0,0714 ; 0,1032;
M-10 [13];[10] 6
0,1087
0,0400; 0,0546; 0,0572; 0,1178; 0,0258;
D (6AC) [13] 4
0,0266
0,0307; 0,0325; 0,0399; 0,0444; 0,0544;
0,0555; 0,0633; 0,0635; 0,0690; 0,0726;
H-11 [13] 19
0,0760; 0,0766; 0,0798; 0,0843; 0,0923;
0,1195; 0,1219
1)
[13] 2 0,0251; 0,0724
Thermold J
1)
[13] 1 0,0231
Timken 16-25-6
(k ) 90 % = 0,0999; µ = 0,0649; σ = 0,0273
o
Summary of all data
[13];[10] 37
above
Conclusion: k value selected = 0,10

o
a
Data from reference [10] are at endurance limit.

10 © ISO 2008 – All rights reserved

Table A.11 — Summary of k calculations for titanium
o
k values
Alloy Reference Number of distributions
o
Ti-140 (AMS 493) [13] 1 0,0956
0,0385; 0,0480; 0,0514; 0,0514; 0,0556; 0,0557;
Ti-6Al-4V [13];[14] 13 0,0616; 0,0741; 0,0792; 0,0841; 0,0882; 0,1086;
0,1226
0,0272; 0,0507; 0,0637; 0,0794; 0,0837; 0,1003;
Ti-A55 [14] 8
0,1033; 0,1515
Ti-75A [14] 6 0,0298; 0,0380; 0,0413; 0,0508; 0,0621; 0,0626
Ti-150A [14] 2 0,0707; 0,0882
Ti-0,2 (0 )
[14] 3 0,0422; 0,0461; 0,0906
Ti-0,2C [14] 1 0,0691
Ti-4Al-3Mo-1V [14] 5 0,0961; 0,0979; 0,1338; 0,1450; 0,2041
Ti-5Al-2,5Sn-0,07(N )
[14] 2 0,0581; 0,0697
Ti-5Al-2,5Sn-0,2(0 )
[14] 1 0.07760
Ti-6Al [14] 4 0,0442; 0,0472; 0,0480; 0,0821
(k ) 90 % = 0,1203 ; µ = 0,0754; σ = 0,0350
o
Summary of all data
[13];[14] 46
above
Conclusion: k value selected = 0,12

o
Table A.12 — Summary of k calculations for nickel steel (nickel content at least 40 %)
o
Alloy Reference Number of distributions k values

o
GMR-235 [13] 1 0,0301
1)
Udimet 500 [13] 1 0,0642
1)
Hastelloy C [14] 1 0,0753
Hasteloy R235 [14] 2 0,0592; 0,1166
1)
Incoloy 901
[14] 1 0,0570
(AMS 5560A)
1)
Inconel 718 [14] 3 0,0705; 0,0933; 0,0952
1)
Waspaloy [14] 1 0,0714
Rene-41
[14] 3 0,0336; 0,0741; 0,0876
1)
(AMS 5713)
1)
6 Mo Waspaloy [14] 2 0,0327; 0,0741
(k ) 90 % = 0,1003; µ = 0,0690; σ = 0,0244
Summary of all data o
[13];[14] 15
above
Conclusion: k value selected = 0,10

o
Table A.13 — Summary of k calculations for Monel
o
Alloy Reference Number of distributions k value

o
0,2643
Monel [14] 1
Conclusion: k value selected = 0,27

o
12 © ISO 2008 – All rights reserved

Annex B
(normative)
Calculation of variability factor K
V
B.1 General
Annex B provides an example of how to calculate variability factor K .
V
B.2 Method
Consider a simple pressure-containing envelope consisting of a cylindrical tube with a square head on each
end, held together by tie rods and nuts in each of the square’s four corners (see Figure B.1).

Key
1 square head
2 cylinder tube
3 tie rod
4 tie rod nuts
Figure B.1 — Pressure-containing envelope
The test operator has chosen an assurance level of 90 % and a verification level of 99 %. Information needed
to obtain K values from Table 1 is shown in Table B.1.
V
Table B.1 — Other information needed to calculate K
V
Number of
Number of actual
k K
Element Material
specimens chosen
o v
test specimens
from Table 1
Cylinder tube Aluminium 0,13 2 1 1,40
End head Magnesium 0,17 4 2 1,38
Tie rod Steel 0,08 8 4 1,10
Tie rod nut Steel 0,08 16 See example 1,05
The value of K for the last row in Table B.1 is calculated as described in the example that follows.
V
Equation (D.6) in Annex D is used, and is shown here as Equation (B.1).
1+kZ
o 4
K = (B.1)
V
1−kZ
o2
EXAMPLE The value of K for the tie rod nut was calculated as:
V
For an assurance level of 90 %, tail area A = 0,10; then, Z = 1,282.
2 2
For a verification level of 99 %, tail area A = 0,01, n = 16.
1/n 1/16
(A + A ) = A = (0,01) = 0,7499, and Z = − 0,674
1 4 1 4
1+−(0,08)( 0,674)
K = = 1,054 which is rounded to 1,05
V
1− (0,08)(1,282)
NOTE 1 Z = 0 at (A + A ) = 0,5 and is negative above 0,5.
4 1 4
In the case described above, the highest value of K for any of the elements is 1,40, therefore, this is the
V
value of K that is used to calculate the P .
V CT
NOTE 2 The highest K value does not always correspond to the highest k value.
V o
NOTE 3 When a verification level of 99 % is chosen, the number of test specimens is always an even number.
14 © ISO 2008 – All rights reserved

Annex C
(informative)
Proposal for an acceleration factor
C.1 General
7 6
It is proposed that a product can be rated for 10 cycles, but tested for only 10 cycles if it is tested at a higher
pressure.
6 7
If material data for a fatigue strength distribution at both the 10 and 10 lives are available, then an
acceleration factor can be determined. This acceleration factor would be a simple ratio of the fatigue strength
6 7
at 10 cycles, to that at 10 cycles. The acceleration factor would be applied to the test pressure in order to
raise the stress level in the test samples.
Data at the characteristic life are often available for Weibull distributions, and ratios of the characteristic
strengths between those two levels can be used to calculate the acceleration factor. Likewise, data at the
median life are often available for Normal distributions, and ratios of the median strengths between those two
levels can be used for the acceleration factor.
For example, an AISI 4140 steel with a Weibull distribution at 10 cycles has a characteristic fatigue strength
of 530 MPa; and the distribution at 10 cycles has a characteristic fatigue strength of 486 MPa. The ratio of
these fatigue strengths is 1.09. Therefore, the test pressure would be raised by this factor if a 10 cycle rating
was desired with a 10 cycle test. This factor would be in addition to any other factors imposed on the test
pressure.
Considerable judgment would be required in proposing a pressure rating when using acceleration factors,
because the probabilities of failure during the test are greater.
C.2 Extrapolating data to 10 cycles
The number of fatigue strength distributions available at 10 cycles is not very abundant. Therefore, a method
to extrapolate data to 10 cycles is proposed if data are available at three lower levels, but spaced well apart,
4 5 6
such as 10 , 10 and 10 cycles.
Consider an S-N curve as shown in Figure C.1:
Key
X life, expressed in number of cycles
Y applied stress from testing; and material strength from failures
−D
1 S = CN + E, where S is strength, N is life in cycles; C, D and E are coefficients
Figure C.1 — S-N curve
At each level of life, there is a fatigue strength distribution. A constant probability curve joins each point of the
fatigue strength from each of the distributions, and is suggested to have an equation of the form shown in
Equation (C.1).
−D
S = CN + E (C.1)
Since Equation (C.1) has three unknown coefficients, it is theoretically possible to determine them from three
sets of data points. Therefore let these data points be the pairs:
4 5 6
S ,N @10 S ,N @10 S ,N @10
1 1 2 2 3 3
Then, knowing the coefficients, the corresponding value of S can be projected to N = 10 cycles. This value
can then be used to determine the acceleration factor discussed in the previous section.
To begin, re-arrange Equation (C.1) as:
−D
(S − E) = CN and ln(S − E) = lnC − D ln N (C.2)
Inserting data points into Equation (C.2) gives:
ln(S−E) = lnC − D ln N
1 1
ln(S−E) = lnC − D ln N
2 2
ln(S−E) = lnC − D ln N
3 3
16 © ISO 2008 – All rights reserved

Subtracting:
ln(S− E) − ln(S−E) = D ln N − D ln N = D(ln N − ln N) (C.3)
1 2 2 1 2 1
Similarly:
ln (S− E) − ln (S−E) = D ln N − D ln N = D(ln N − ln N) (C.4)
1 3 3 1 3 1
Dividing Equation (C.3) by Equation (C.4):
lnSE−− lnS−E D lnN− lnN
( ) ( ) ( ) lnNN− ln
12 2 1
= == C (C.5)
ln()SE−− lnS−E D lnN− lnN lnN− lnN
()()
1 3 31 31
It is observed that the right-hand side of Equation (C.5) is a constant.
Continuing from Equation (C.5):
lnS−−E lnSE− =C lnS−E−C lnSE−
( ) ( ) ( ) ( )
12 11 13
C
⎛⎞⎛ ⎞
⎛⎞SE−−SE SE−
11 1
ln ==C ln ln (C.6)
⎜⎟⎜ ⎟
⎜⎟
SE−−SE SE−
23 3
⎝⎠ ⎝⎠⎝ ⎠
Equating the logs of both sides, and expanding using the binomial theorem to 3 terms gives Equation (C.7).
CC−−121
C () () 2
C
1 SC−+S E C(1C−)S E+"
⎛⎞ 11 1
SE−−SE
11 2
== (C.7)
⎜⎟
CC−−121
SE−−SE C () ()
11 2
23⎝⎠
SC−+S E C(1C−)S E+"
33 1
It is observed that the coefficients of the binomial can be replaced by constants:
C−1 1 C−2 C−1 1 C−2
( ) () ( ) ()
1 1 1 1
CC= S CC=−C 1S CC= S CC=−(1C )S
()
21 311 41 511
1 1 3 1
2 2
Resulting in Equation (C.8):
C
SC−+ECE
SE−
= (C.8)
C
SE− 1
2 SC−+ECE
Cross multiplying and expanding Equation (C.8):
CC22
S−−E S C E+C E= S−E S−CE+CE
() ()
) )
14( 5 2 ( 23
CC223C 2C 23
11 1 1
S S−SC E+SC E−S E+−=C E CE S S−S C E+S CE−S E+CE−CE
114 15 4 5 2 22 23 2 3
33 1 1
Combining terms results in a cubic equation:
CC C C
11 1 1
C−+CE⎡⎤C−C+SC−SC E+SC−SC+SS− E+SSS− S= 0
() ()()
35 4 2 15 23 (22 14 ) (1 2 )
⎣⎦ 13 3 1
Dividing by the first coefficient gives Equation (C.9):
CC C C
11 1 1
SC−+SC S−S S S−S S
( 22 1 4 ) ( 1 2 )
CC−+SC−SC 13 3 1
()
3242 15 23
EE++ E+ = 0 (C.9)
CC−−CC CC−
() () ()
35 35 3 5
Substituting for the coefficients:
CC CC
11 11
SC−+SC S−S SS −S S
) )
CC−+SC−SC ( 22 1 4 (12
() 13 31
42 15 23
p= q= r=
CC− CC− CC−
() () ()
35 35 35
This results in the classic textbook cubic Equation (C.10):
3 2
E + pE + qE + r = 0 (C.10)
[18]
which can be solved with software (also available on the Internet ). There are three roots to
Equation (C.10), and one of them should be selected for continued use. This choice is made by examining the
values of the three roots and selecting the one that is a real number, and less than the value of S (it might
even be negative). It can also be necessary to try more than one root and examine the results.
Returning to Equation (C.1), and substituting the chosen value of E:
−D −D
S = CN + E and (S − E) = CN
Inserting data points and dividing gives:
−D
−D
⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞
SE− CN N SE− N
11 1 11
== and ln =−Dln
⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟
−D
SE− N SE− N
CN
⎝⎠22⎝⎠ ⎝⎠22⎝⎠
which can be solved for D:
⎛⎞SE−
ln
⎜⎟
SE−
⎝⎠2
−= D (C.11)
⎛⎞
N
ln
⎜⎟
N
⎝⎠2
Finally, substituting known values into Equation (C.1) again gives:
−D
SE−=CN and
SE−
= C (C.12)
−D
N
Thus, all three coefficients for Equation (C.1) are now known, and one of the fatigue strength values
(characteristic or mean) at 10 cycles can be determined using Equation (C.13):
−D
SC=+N E (C.13)
18 © ISO 2008 – All rights reserved

C.3 Examples
C.3.1 General
Weibull fatigue strength distributions for five materials (both smooth and notched specimens) were found for
4 5 6 7
cycle lives at 10 , 10 , 10 and 10 cycles. The equations in Clause C.2 were used to calculate the fatigue
7 7
strengths at 10 cycles, and then compared to the published value at 10 cycles. From this, the acceleration
factors were calculated. Results are summarized in Table C.1 and Table C.2.
Table C.1 — Smooth specimen examples
Material 4140 steel H11 steel 4340 steel Alloy steel 2024
Aluminium
Published at 10 485,5 MPa 673,1 MPa 620,7 MPa 1079,3 MPa 186,2 MPa
Calculated at 10 487,2 MPa 684,0 MPa 609,3 MPa 1082,9 MPa 186,8 MPa
Error +0,36 % +1,61 % −1,83 % +0,33 % −0,04 %
Acceleration factor
1,09 1,12 1,13 1,08 1,22
published
Acceleration factor
1,09 1,10 1,15 1,08 1,21
calculated
Error
−0,36 % −1,59 % +1,86 % −0,33 % −0,28 %
Table C.2 — Notched specimen examples
Material 4140 steel H11 steel 4340 steel Alloy steel 2024
Aluminium
Published at 10 219,3 MPa 675,9 MPa 87,6 MPa 497,2 MPa 92,3 MPa
Calculated at 10 219,7 MPa 677,4 MPa 80,6 MPa 503,1 MPa 92,2 MPa
Error +0,17 % +0,22 % −1,83 % +1,17 % −0,19 %
Acceleration factor
1,18 1,16 1,81 1,05 1,41
published
Acceleration factor
1,18 1,16 1,97 1,04 1,41
calculated
Error −0,17 % −0,22 % +8,70 % −1,15 % +0,19 %

The materials used in this comparison were all with three-parameter Weibull distributions and the data
sources were Reference [13] and Reference [14].
C.3.2 Observations
The following observations were made.
a) The accuracy of projecting the characteristic and mean life to 10 cycles was very good for all the
example materials, except the 4340 notched steel. However, data for this material are questionable
because the notched values for minimum life were very low, resulting in only one real root to its cubic
equation and that was larger than the minimum life at 10 cycles.
Characteristic life Characteristic life

b) The acceleration factor from projected life had good accuracy compared to the acceleration factor
calculated directly from published data.
c) There is a significant difference in the acceleration factor between smooth and notched specimens, for
some of the materials, for example:
1) the pressure increase would double for notched specimens over smooth ones in the 4140 steel. (For
6 2)
example, if the test pressure for a 10 cycle test were 10 MPa , the smooth specimen data would
result in a 0,9 MPa increase in test pressure, to 10,9 MPa, for rating at 10 MPa. If notched data
were used, the increase would be 1,8 MPa, resulting in a test pressure of 11,8 MPa);
2) the pressure increase is about 33 % more for notched specimens than smooth ones in the H11 steel;
3) the pressure increase is actually less for notched specimens than smooth samples in the alloy steel;
4) the acceleration factor is unreasonably high for notched specimens in aluminium and the 4340 steel.
C.3.3 Proposal
The examples justify the accuracy of the analysis technique, but selection of material data for use in a
standard should be explored further. The sources cited for the data had some questionable values (as noted
for the 4340 ste
...