ISO 10928:2016

Plastics piping systems - Glass-reinforced thermosetting plastics (GRP) pipes and fittings - Methods for regression analysis and their use

ISO 10928:2016 specifies procedures suitable for the analysis of data which, when converted into logarithms of the values, have either a normal or a skewed distribution. It is intended for use with the test methods and referring standards for glass-reinforced thermosetting plastics (GRP) pipes or fittings for the analysis of properties as a function of time. However, it can be used for the analysis of other data. Depending upon the nature of the data, two methods are specified. The extrapolation using these techniques typically extends the trend from data gathered over a period of approximately 10 000 h to a prediction of the property at 50 years, which is the typical maximum extrapolation time. ISO 10928:2016 only addresses the analysis of data. The test procedures to collect the data, the number of samples required and the time period over which data are collected are covered by the referring standards and/or test methods. Clause 6 discusses how the data analysis methods are applied to product testing and design.

Systèmes de canalisation en matières plastiques — Tubes et raccords plastiques thermodurcissables renforcés de verre (PRV) — Méthodes pour une analyse de régression et leurs utilisations

General Information

• Status: Withdrawn
• Publication Date: 13-Dec-2016
• ICS: 23.040.20 - Plastics pipes
• ICS: 23.040.45 - Plastics fittings
• Technical Committee: ISO/TC 138/SC 6 - Reinforced plastics pipes and fittings for all applications
• Drafting Committee: ISO/TC 138/SC 6/WG 1 - Methods of test
• Current Stage: 9599 - Withdrawal of International Standard
• Start Date: 20-Mar-2024
• Completion Date: 13-Dec-2025

Relations

• Revised: ISO 10928:2024 - Plastics piping systems - Glass-reinforced thermosetting plastics (GRP) pipes and fittings - Methods for regression analysis and their use
• Effective Date: 06-Jun-2022
• Revises: ISO 10928:2009/Amd 1:2013 - Plastics piping systems - Glass-reinforced thermosetting plastics (GRP) pipes and fittings - Methods for regression analysis and their use - Amendment 1
• Effective Date: 05-Nov-2015
• Revises: ISO 10928:2009 - Plastics piping systems - Glass-reinforced thermosetting plastics (GRP) pipes and fittings - Methods for regression analysis and their use
• Effective Date: 04-Nov-2015

Standards Content (Sample)

INTERNATIONAL ISO
STANDARD 10928
Third edition
2016-12-15
Plastics piping systems — Glass-
reinforced thermosetting plastics
(GRP) pipes and fittings — Methods
for regression analysis and their use
Systèmes de canalisation en matières plastiques — Tubes et raccords
plastiques thermodurcissables renforcés de verre (PRV) — Méthodes
pour une analyse de régression et leurs utilisations
Reference number
©
ISO 2016
© ISO 2016, Published in Switzerland
All rights reserved. Unless otherwise specified, no part of this publication may be reproduced or utilized otherwise in any form
or by any means, electronic or mechanical, including photocopying, or posting on the internet or an intranet, without prior
written permission. Permission can be requested from either ISO at the address below or ISO’s member body in the country of
the requester.
ISO copyright office
Ch. de Blandonnet 8 • CP 401
CH-1214 Vernier, Geneva, Switzerland
Tel. [phone]
Fax [phone]
[e-mail]
www.iso.org
ii © ISO 2016 – All rights reserved

Contents Page
Foreword .iv
Introduction .v
1 Scope . 1
2 Normative references . 1
3 Terms and definitions . 1
4 Principle . 1
5 Procedures for determining the linear relationships — Methods A and B .2
5.1 Procedures common to methods A and B . 2
5.2 Method A — Covariance method . 2
5.2.1 General. 2
5.2.2 Suitability of data . 3
5.2.3 Functional relationships . 3
5.2.4 Calculation of variances . 4
5.2.5 Check for the suitability of data for extrapolation . 5
5.2.6 Validation of statistical procedures by an example calculation . 6
5.3 Method B — Regression with time as the independent variable .10
5.3.1 General.10
5.3.2 Suitability of data .10
5.3.3 Functional relationships .11
5.3.4 Check for the suitability of data for extrapolation .11
5.3.5 Validation of statistical procedures by an example calculation .11
6 Application of methods to product design and testing .12
6.1 General .12
6.2 Product design.13
6.3 Comparison to a specified value .13
6.4 Declaration of a long-term value .13
Annex A (informative) Second-order polynomial relationships .14
Annex B (informative) Non-linear relationships .19
Annex C (normative) Calculation of lower confidence and prediction limits for method A .45
Bibliography .47
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.
The procedures used to develop this document and those intended for its further maintenance are
described in the ISO/IEC Directives, Part 1. In particular the different approval criteria needed for the
different types of ISO documents should be noted. This document was drafted in accordance with the
editorial rules of the ISO/IEC Directives, Part 2 (see www.iso.org/directives).
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. Details of
any patent rights identified during the development of the document will be in the Introduction and/or
on the ISO list of patent declarations received (see www.iso.org/patents).
Any trade name used in this document is information given for the convenience of users and does not
constitute an endorsement.
For an explanation on the meaning of ISO specific terms and expressions related to conformity assessment,
as well as information about ISO’s adherence to the World Trade Organization (WTO) principles in the
Technical Barriers to Trade (TBT) see the following URL: www.iso.org/iso/foreword.html.
The committee responsible for this document is ISO/TC 138, Plastics pipes, fittings and valves for the
transport of fluids, Subcommittee SC 6, Reinforced plastics pipes and fittings for all applications.
This third edition cancels and replaces the second edition (ISO 10928:2009), which has been
technically revised and includes the following changes. It also incorporates the Amendment
ISO 10928:2009/Amd 1:2013:
— Annex A (GRP pressure pipe design procedure) has been removed from the document;
— several bibliographical references have been removed.
iv © ISO 2016 – All rights reserved

Introduction
This document describes the procedures intended for analysing the regression of test data, usually with
respect to time and the use of the results in design and assessment of conformity with performance
requirements. Its applicability is limited to use with data obtained from tests carried out on samples.
The referring standards require estimates to be made of the long-term properties of the pipe for such
parameters as circumferential tensile strength, long-term ring deflection, strain corrosion and creep or
relaxation stiffness.
A range of statistical techniques that could be used to analyse the test data produced by destructive
tests was investigated. Many of these simple techniques require the logarithms of the data to
a) be normally distributed,
b) produce a regression line having a negative slope, and
c) have a sufficiently high regression correlation (see Table 1).
While the last two conditions can be satisfied, analysis shows that there is a skew to the distribution
and hence this primary condition is not satisfied. Further investigation into techniques that can handle
skewed distributions resulted in the adoption of the covariance method of analysis of such data for this
document.
However, the results from non-destructive tests, such as long-term creep or relaxation stiffness, often
satisfy all three conditions and hence a simpler procedure, using time as the independent variable, can
also be used in accordance with this document.
These data analysis procedures are limited to analysis methods specified in ISO product standards or
test methods. However, other analysis procedures can be useful for the extrapolation and prediction
of long-term behaviour of some properties of glass-reinforced thermosetting plastics (GRP) piping
products. For example, a second-order polynomial analysis is sometimes useful in the extrapolation
of creep and relaxation data. This is particularly the case for analysing shorter term data, where the
shape of the creep or relaxation curve can deviate considerably from linear. A second-order polynomial
analysis is included in Annex A. In Annex B, there is an alternative non-linear analysis method. These
non-linear methods are provided only for information and the possible use in investigating the
behaviour of a particular piping product or material therefore they might not be generally applicable to
other piping products.
INTERNATIONAL STANDARD ISO 10928:2016(E)
Plastics piping systems — Glass-reinforced thermosetting
plastics (GRP) pipes and fittings — Methods for regression
analysis and their use
1 Scope
This document specifies procedures suitable for the analysis of data which, when converted into
logarithms of the values, have either a normal or a skewed distribution. It is intended for use with the
test methods and referring standards for glass-reinforced thermosetting plastics (GRP) pipes or fittings
for the analysis of properties as a function of time. However, it can be used for the analysis of other data.
Depending upon the nature of the data, two methods are specified. The extrapolation using these
techniques typically extends the trend from data gathered over a period of approximately 10 000 h to a
prediction of the property at 50 years, which is the typical maximum extrapolation time.
This document only addresses the analysis of data. The test procedures to collect the data, the number
of samples required and the time period over which data are collected are covered by the referring
standards and/or test methods. Clause 6 discusses how the data analysis methods are applied to
product testing and design.
2 Normative references
There are no normative references in this document.
3 Terms and definitions
No terms and definitions are listed in this document.
ISO and IEC maintain terminological databases for use in standardization at the following addresses:
— IEC Electropedia: available at http://www.electropedia.org/
— ISO Online browsing platform: available at http://www.iso.org/obp
4 Principle
Data are analysed for regression using methods based on least squares analysis which can accommodate
the incidence of a skew and/or a normal distribution. The two methods of analysis used are the
following:
— method A: covariance using a first-order relationship;
— method B: least squares, with time as the independent variable using a first-order relationship.
The methods include statistical tests for the correlation of the data and the suitability for extrapolation.
5 Procedures for determining the linear relationships — Methods A and B
5.1 Procedures common to methods A and B
Use method A (see 5.2) or method B (see 5.3) to fit a straight line of the form given in Formula (1):
ya=+ bx× (1)
where
y is the logarithm, lg, of the property being investigated;
a is the intercept on the y-axis;
b is the slope;
x is the logarithm, lg, of the time, in hours.
5.2 Method A — Covariance method
5.2.1 General
For method A, calculate the following variables in accordance with 5.2.2 to 5.2.5, using Formulae (2),
(3) and (4):
yY−
()
∑
i
Q = (2)
y
n
xX−
()
∑
i
Q = (3)
x
n
 
xX− ×−yY
() ()
∑
ii
 
Q = (4)
xy
n
where
Q is the sum of the squared residuals parallel to the y-axis, divided by n;
y
Q is the sum of the squared residuals parallel to the x-axis, divided by n;
x
Q is the sum of the squared residuals perpendicular to the line, divided by n;
xy
Y is the arithmetic mean of the y data, i.e. given as Formula (5):
y
∑
i
Y = (5)
n
X is the arithmetic mean of the x data, i.e. given as Formula (6):
x
∑
i
X = (6)
n
2 © ISO 2016 – All rights reserved

x , y are individual values;
i i
n is the total number of results (pairs of readings for x , y ).
i i
NOTE If the value of Q is greater than zero, the slope of the line is positive and if the value of Q is less than
xy xy
zero, then the slope is negative.
5.2.2 Suitability of data
Calculate the linear coefficient of correlation, r, using Formulae (7) and (8):
Q
xy
r = (7)
QQ×
xy
0,5
rr= (8)
()
Student's tf()
If the value of r is less than , then the data are unsuitable for analysis.
 
nt−+2 Student's f
()
 
Table 1 gives the minimum acceptable values of the correlation coefficient, r, as a function of the number
of variables, n. The Student’s t value is based on a two-sided 0,01 level of significance.
Table 1 — Minimum values of the correlation coefficient, r, for acceptable data from n pairs of data
Number of Degrees of Student’s Minimum Number of Degrees of Student’s Minimum
variables freedom variables freedom
n n − 2 t(0,01) r n n − 2 t(0,01) r
13 11 3,106 0,683 5 26 24 2,797 0,495 8
14 12 3,055 0,661 4 27 25 2,787 0,486 9
15 13 3,012 0,641 1 32 30 2,750 0,448 7
16 14 2,977 0,622 6 37 35 2,724 0,418 2
17 15 2,947 0,605 5 42 40 2,704 0,393 2
18 16 2,921 0,589 7 47 45 2,690 0,372 1
19 17 2,898 0,575 1 52 50 2,678 0,354 2
20 18 2,878 0,561 4 62 60 2,660 0,324 8
21 19 2,861 0,548 7 72 70 2,648 0,301 7
22 20 2,845 0,536 8 82 80 2,639 0,283 0
23 21 2,831 0,525 6 92 90 2,632 0,267 3
24 22 2,819 0,515 1 102 100 2,626 0,254 0
25 23 2,807 0,505 2
5.2.3 Functional relationships
Find a and b for the functional relationship line using Formula (1).
First, set Γ as given in Formula (9):
Q
y
Γ = (9)
Q
x
then calculate a and b using Formulae (10) and (11):
05,
b =− Γ (10)
()
aY=−bX× (11)
5.2.4 Calculation of variances
If t is the applicable time to failure, then set x as given in Formula (12):
u u
xt= lg (12)
uu
Using Formulae (13), (14) and (15) respectively, calculate for i = 1 to n, the following sequence of
statistics:
— the best fit x ′ for true x ;
i i
— the best fit y ′ for true y ;
i i
— the error variance, σ for x.
δ
Γ ×+xb×−ya
()
ii
x ′= (13)
i
2× Γ
ya′′=+ bx× (14)
ii
 
yy− ′′+×Γ xx−
() ()
∑ ∑ 
ii ii
2  
σ = (15)
δ
n− 2 × Γ
()
Calculate quantities E and D using Formulae (16) and (17):
b×σ
δ
E = (16)
2×Q
xy
2××Γσb×
δ
D = (17)
nQ×
xy
Calculate the variance, C, of the slope b, using Formula (18):
CD=× 1+E (18)
()
4 © ISO 2016 – All rights reserved

5.2.5 Check for the suitability of data for extrapolation
If it is intended to extrapolate the line, calculate T using Formula (19):
b b
T = = (19)
05,,05
C
var b
()
If the absolute value, |T| (i.e. ignoring signs), of T is equal to or greater than the applicable value for
Student’s t, t , shown in Table 2 for (n − 2) degrees of freedom, then consider the data suitable for
v
extrapolation.
Calculation of confidence limits is not required by the test methods or referring standards; however,
the calculation of lower confidence limit, LCL, and lower prediction limit, LPL, is given in Annex C.
Table 2 — Percentage points of Student’s t distribution
(upper 2,5 % points; two-sided 5 % level of confidence; t for 97,5 %)
v
Degree of Student’s t Degree of Student’s t Degree of Student’s t
freedom value freedom value freedom value
(n − 2) t (n − 2) t (n − 2) t
v v v
1 12,706 2 36 2,028 1 71 1,993 9
2 4,302 7 37 2,026 2 72 1,993 5
3 3,182 4 38 2,024 4 73 1,993 0
4 2,776 4 39 2,022 7 74 1,992 5
5 2,570 6 40 2,021 1 75 1,992 1
6 2,446 9 41 2,019 5 76 1,991 7
7 2,364 6 42 2,018 1 77 1,991 3
8 2,306 0 43 2,016 7 78 1,990 8
9 2,262 2 44 2,015 4 79 1,990 5
10 2,228 1 45 2,014 1 80 1,990 1
11 2,201 0 46 2,012 9 81 1,989 7
12 2,178 8 47 2,011 2 82 1,989 3
13 2,160 4 48 2,010 6 83 1,989 0
14 2,144 8 49 2,009 6 84 1,988 6
15 2,131 5 50 2,008 6 85 1,988 3
16 2,119 9 51 2,007 6 86 1,987 9
17 2,109 8 52 2,006 6 87 1,987 6
18 2,100 9 53 2,005 7 88 1,987 3
19 2,093 0 54 2,004 9 89 1,987 0
20 2,086 0 55 2,004 0 90 1,986 7
Table 2 (continued)
Degree of Student’s t Degree of Student’s t Degree of Student’s t
freedom value freedom value freedom value
(n − 2) t (n − 2) t (n − 2) t
v v v
21 2,079 6 56 2,003 2 91 1,986 4
22 2,073 9 57 2,002 5 92 1,986 1
23 2,068 7 58 2,001 7 93 1,985 8
24 2,063 9 59 2,001 0 94 1,985 5
25 2,059 5 60 2,000 3 95 1,985 3
26 2,055 5 61 1,999 6 96 1,985 0
27 2,051 8 62 1,999 0 97 1,984 7
28 2,048 4 63 1,998 3 98 1,984 5
29 2,045 2 64 1,997 7 99 1,984 2
30 2,042 3 65 1,997 1 100 1,984 0
31 2,039 5 66 1,996 6
32 2,036 9 67 1,996 0
33 2,034 5 68 1,995 5
34 2,032 2 69 1,994 9
35 2,030 1 70 1,994 4
5.2.6 Validation of statistical procedures by an example calculation
The data given in Table 3 are used in the following example to aid in verifying that statistical procedures,
as well as computer programs and spreadsheets adopted by users, will produce results similar to those
obtained from the formulae given in this document. For the purposes of the example, the property
in question is represented by V, the values for which are of a typical magnitude and in no particular
units. Because of rounding errors, it is unlikely that the results will agree exactly, so for a calculation
procedure to be acceptable, the results obtained for r, r , b, a, and the mean value of V, and V , shall
m
agree to within ±0,1 % of the values given in this example. The values of other statistics are provided to
assist the checking of the procedure.
Sums of squares:
Q = 0,798 12;
x
Q = 0,000 88;
y
Q = −0,024 84.
xy
Coefficient of correlation:
r = 0,879 99;
r = 0,938 08.
Functional relationships:
Г = 0,001 10;
b = −0,033 17;
a = 1,627 31.
6 © ISO 2016 – All rights reserved

Table 3 — Basic data for example calculation and statistical analysis validation
n V Y Time X
lg V h lg h
1 30,8 1,488 6 5 184 3,714 7
2 30,8 1,488 6 2 230 3,348 3
3 31,5 1,498 3 2 220 3,346 4
4 31,5 1,498 3 12 340 4,091 3
5 31,5 1,498 3 10 900 4,037 4
6 31,5 1,498 3 12 340 4,091 3
7 31,5 1,498 3 10 920 4,038 2
8 32,2 1,507 9 8 900 3,949 4
9 32,2 1,507 9 4 173 3,620 4
10 32,2 1,507 9 8 900 3,949 4
11 32,2 1,507 9 878 2,943 5
12 32,9 1,517 2 4 110 3,613 8
13 32,9 1,517 2 1 301 3,114 3
14 32,9 1,517 2 3 816 3,581 6
15 32,9 1,517 2 669 2,825 4
16 33,6 1,526 3 1 430 3,155 3
17 33,6 1,526 3 2 103 3,322 8
18 33,6 1,526 3 589 2,770 1
19 33,6 1,526 3 1 710 3,233 0
20 33,6 1,526 3 1 299 3,113 6
21 35,0 1,544 1 272 2,434 6
22 35,0 1,544 1 446 2,649 3
23 35,0 1,544 1 466 2,668 4
24 35,0 1,544 1 684 2,835 1
25 36,4 1,561 1 104 2,017 0
26 36,4 1,561 1 142 2,152 3
27 36,4 1,561 1 204 2,309 6
28 36,4 1,561 1 209 2,320 1
29 38,5 1,585 5 9 0,954 2
30 38,5 1,585 5 13 1,113 9
31 38,5 1,585 5 17 1,230 4
32 38,5 1,585 5 17 1,230 4
Means: Y = 1,530 1 X = 2, 930 5
Calculated variances (see 5.2.4):
−2
E = 3,520 2 × 10 ;
−6
D = 4,842 2 × 10 ;
−6
C = 5,012 7 × 10 (the variance of b);
2 −2
σ = 5,271 1 × 10 (the error variance of x).
δ
Check for the suitability for extrapolation (see 5.2.5):
n = 32;
t = 2,042 3;
v
−6 0,5
T = −0,033 17 / (5,012 7 × 10 ) = −14,816 7;
|T| = 14,816 7 > 2,042 3.
The estimated mean values for V at various times are given in Table 4 and shown in Figure 1.
Table 4 — Estimated mean values, V , for V
m
Time
V
m
h
0,1 45,76
1 42,39
10 39,28
100 36,39
1 000 33,71
10 000 31,23
100 000 28,94
438 000 27,55
8 © ISO 2016 – All rights reserved

Key
1 438 000 h (50 years)
2 regression line from Table 4
3 data point
X-axis lg scale of time, in hours
Y-axis lg scale of property
Figure 1 — Regression line from the results in Table 4
5.3 Method B — Regression with time as the independent variable
5.3.1 General
For method B, calculate the sum of the squared residuals parallel to the Y-axis, S , using Formula (20):
y
Sy=−Y (20)
()
∑
y i
Calculate the sum of the squared residuals parallel to the X-axis, S , using Formula (21):
x
Sx=− X (21)
()
∑
x i
Calculate the sum of the squared residuals perpendicular to the line, S , using Formula (22):
xy
 
Sx=− Xy×−Y (22)
() ()
∑
xy ii
 
where
Y is the arithmetic mean of the y data, i.e.
y
∑
i
Y = ;
n
X is the arithmetic mean of the x data, i.e.
x
∑
i
X = ;
n
x , y are individual values;
i i
n is the total number of results (pairs of readings for x , y ).
i i
NOTE If the value of S is greater than zero, the slope of the line is positive and if the value of S is less than
xy xy
zero, then the slope is negative.
5.3.2 Suitability of data
Calculate the squared, r , and the linear coefficient of correlation, r, using Formulae (23) and (24):
S
xy
r = (23)
SS×
xy
0,5
rr= (24)
()
If the value of r or r is less than the applicable minimum value given in Table 1 as a function of n,
consider the data unsuitable for analysis.
10 © ISO 2016 – All rights reserved

5.3.3 Functional relationships
Calculate a and b for the functional relationship line [see Formula (1)], using Formulae (25) and (26):
S
xy
b= (25)
S
x
aY=−bX× (26)
5.3.4 Check for the suitability of data for extrapolation
If it is intended to extrapolate the line, calculate M using Formula (27):
tS××SS−
()
S vx yxy
x
M =− (27)
2 2
S nS−2 ×
()
xy y
where
t is the applicable value for Student’s t determined from Table 2.
v
If M is equal to or less than zero, consider the data unsuitable for extrapolation.
5.3.5 Validation of statistical procedures by an example calculation
The data given in Table 5 are used in the following example to aid in verifying that statistical
procedures, as well as computer programs and spreadsheets adopted by users, will produce results
similar to those obtained from the formulae given in this document. Use the data given in Table 5 for the
calculation procedures described in 5.3.2 to 5.3.4 to ensure that the statistical procedures to be used in
conjunction with this method will give results for r, r , a, b and V to within ±0,1 % of the values given
m
in this example.
Table 5 — Basic data for example calculation and statistical validation
n Time X V Y
T in h lg T lg V
1 0,10 −1,000 0 7 114 3,852 1
2 0,27 −0,568 6 6 935 3,841 0
3 0,50 −0,301 0 6 824 3,834 1
4 1,00 0 6 698 3,825 9
5 3,28 0,515 9 6 533 3,815 1
6 7,28 0,862 1 6 453 3,809 8
7 20,0 1,301 0 6 307 3,799 9
8 45,9 1,661 8 6 199 3,792 3
9 72,0 1,857 3 6 133 3,787 7
10 166 2,220 1 5 692 3,755 2
11 219 2,340 4 5 508 3,741 0
12 384 2,584 3 5 393 3,731 8
13 504 2,702 4 5 364 3,729 5
14 3 000 3,477 1 5 200 3,716 0
15 10 520 4,022 0 4 975 3,696 8
Means: X = 1,445 0 Y = 3,781 9
Sums of squares:
S = 31,681 1;
x
S = 0,034 7;
y
S = −1,024 2.
xy
Coefficient of correlation:
r = 0,955 6;
r = 0,977 5.
Functional relationships (see 5.3.3):
a = 3,828 6;
b = −0,032 3.
Check for the suitability for extrapolation (see 5.3.4):
t = 2,160 4;
v
M = 942,21.
The estimated mean values, V , for V at various times are given in Table 6.
m
Table 6 — Estimated mean values, V , for V
m
Time
V
m
h
0,1 7 259
1 6 739
10 6 256
100 5 808
1 000 5 391
10 000 5 005
100 000 4 646
438 000 4 428
6 Application of methods to product design and testing
6.1 General
The referring standards specify limiting requirements for the long-term properties and performance
of a product. Some of these are based on destructive tests, for example, hoop tensile strength, while
others are based on actual or derived physical properties, such as creep stiffness.
These properties require an extrapolated long-term (e.g. 50 years) value for the establishment of a
product design or comparison with the requirement. This extrapolated value is determined by inserting,
as necessary, the values for a and b, determined in accordance with 5.2 and 5.3 as appropriate, into
Formula (28).
lg ya=+ bt× (28)
L
12 © ISO 2016 – All rights reserved

where
t is the logarithm, lg, of the long-term period, in hours [for 50 years (438 000 h), t = 5,641 47].
L L
Solving Formula (28) for y gives the extrapolated value.
The use of the data and the specification of requirements in the product standards are in three distinct
categories.
6.2 Product design
In the first category, the data are used for design or calculation of a product line. This is the case for
[1]
long-term circumferential strength testing (ISO 7509) .The long-term destructive test data are
analysed using method A.
6.3 Comparison to a specified value
The second category is where the long-term extrapolated value is compared to a minimum requirement
[3]
given in the product standard. This is the case for long-term ring bending (ISO 10471) and strain
[4]
corrosion (ISO 10952) . The long-term destructive test data are analysed using method A and can be
used to establish a value to compare to the product standard requirement. As the analysis of the data
does provide a long-term stress or strain value, this value may also be utilized in analysis of product
suitability for a range of installation and application conditions.
6.4 Declaration of a long-term value
The third category is when the long-term extrapolated value is used to calculate a long-term property
[2]
and this value is then declared by the manufacturer. This is the case for long-term creep (ISO 10468)
stiffness. The long-term non-destructive test data are analysed using method B.
Annex A
(informative)
Second-order polynomial relationships
A.1 General
This method fits a curved line of the form given in Formula (A.1):
yc=+dx×+ex× (A.1)
where
y is the logarithm, lg, of the property being investigated;
c is the intercept on the y-axis;
d, e are the coefficients to the two orders of x;
x is the logarithm, lg, of the time, in hours.
A.2 Variables
Calculate the following variables:
x , the sum of all individual x data;
• i
x , the sum of all squared x data;
• i
x , the sum of all x data to the third power;
• i
x , the sum of all x data to the fourth power;
• i
y , the sum of all individual y data;
• i
y , the squared sum of all individual y data;
()
∑ i
y , the sum of all squared y data;
• i
xy× , the sum of all products x ,y ;
() i i
∑ ii
xy× , the sum of all products x y ;
i i
( ii)
∑
Sx=−X , the sum of the squared residuals parallel to the X-axis for the linear part;
()
x ∑ i
14 © ISO 2016 – All rights reserved

Sx=−X , the sum of the squared residuals parallel to the X-axis for the quadratic part;
( )
xx ∑ i
Sy=−Y , the sum of the squared residuals parallel to the Y-axis;
()
y ∑ i
 
Sx=−Xy×−Y , the sum of the squared residuals perpendicular to the line for the
() ()
xy ∑ ii
 
linear part;
 
Sx=−Xy×−Y , the sum of the squared residuals perpendicular to the line for the
()
xxy ( ii)
∑
 
 
quadratic part
where
y
Y i
∑
is the arithmetic mean of the y data, i.e. Y = ;
n
x
∑ i
X
is the arithmetic mean of the x data, i.e. X= .
n
A.3 Solution system
Determine c, d and e using the following matrix:
yc=×nd+× xe+× x (A.2)
∑∑ii ∑ i
xy× =×cx +×dx +×ex (A.3)
()
∑∑ii ii∑∑ i
2 34
xy× =×cx +×dx +×ex (A.4)
( )
∑∑ii ii∑∑ i
A.4 Suitability of data
Calculate the squared, r , and the linear coefficient of correlation, r, using Formulae (A.5) and (A.6):
 
cy×+dx×× ye+× xy× − yn/
()
()
ii ii( ii)
∑ ∑ ∑ ∑
 
 
r = (A.5)
 
yy− /n
()
∑ ii∑
 
 
20,5
rr= (A.6)
( )
If the value of r or r is less than the applicable minimum value given in Table 1 as a function of n,
consider the data unsuitable for analysis.
A.5 Check for the suitability of data for extrapolation
If it is intended to extrapolate the line, calculate M using Formula (A.7):
22 2
2 2 tS××SS−+SS×−S
vx( yxyxxy xxy )
S S
x xx
M =+ − (A.7)
2 2 2
S S nS−2 ×
()
xy xxy y
If M is equal to or less than zero, consider the data unsuitable for extrapolation.
A.6 Validation of statistical procedures by an example calculation
Use the data given in Table 5 for the calculation procedures described in A.1 to A.5 to ensure that the
statistical procedures to be used in conjunction with this method will give results for r, r , a, b and V
m
to within ±0,1 % of the values given in this example (n = 15).
= 21,671;
x
i
•
= 62,989;
x
• i
= 180,623;
x
• i
= 584,233;
x
• i
= 56,728;
y
i
•
= 3 218,09;
y
()
∑ i
= 214,574;
y
• i
= 80,932;
xy×
()
∑ ii
= 235,175;
xy×
( ii)
∑
= 31,681;
Sx=−X
()
x ∑ i
22 = 386,638;
Sx=−X
)
xx ∑( i
= 0,034 7;
Sy=−Y
()
y ∑ i
= −1,024 2;
Sx=− Xy×−Y 
() ()
xy ii
∑
 
 
Sx=−Xy×−Y
() = −3,041 8.
xxy ∑ ( ii)
 
 
Solution system:
c = 3,828 8;
d = −0,026 2;
e = −0,002 2.
Coefficient of correlation:
16 © ISO 2016 – All rights reserved

r = 0,964 7;
r = 0,982 2.
Check for the suitability for extrapolation:
t = 2,160 4;
v
M = 15 859,6.
The estimated mean values, V , for V at various times are given in Table A.1 and shown in Figure A.1.
m
Table A.1 — Estimated mean values, V , for V
m
Time
V
m
h
0,1 7 125
1 6 742
10 6 315
100 5 856
1 000 5 375
10 000 4 884
100 000 4 393
438 000 4 091
Key
1 438 000 h (50 years)
2 regression line from Table A.1
3 data point
X-axis lg scale of time, in hours
Y-axis lg scale of property
Figure A.1 — Regression line from the results in Table A.1
18 © ISO 2016 – All rights reserved

Annex B
(informative)
Non-linear relationships
B.1 General
Given a model for a set of data obtained from a long-term stiffness test on GRP test pieces, the objective
of this annex is to produce explicit formulae for
a) estimating all four parameters in the model, i.e. a, b, c and d, and
b) calculating confidence and prediction intervals about the curve.
These formulae are presented in this annex, along with associated graphical displays for a typical data set.
NOTE While the data and procedures refer to a long-term stiffness test, the method can also be applied to
data which fit the mathematical model and require extrapolation to 50 years.
B.2 Model
As described in B.4, the model shown can be re-expressed as two linked straight-line regression models
called Line 1 and Line 2. Answers obtained from the procedures described for Line 1 are then applied in
the Line 2 procedures to obtain the four parameters in the model, which is then used to obtain the long-
term value for the property under investigation.
B.2.1 Procedure for Line 1
The formulae given in the following subclauses are used to develop the models for Lines 1 and 2.
NOTE Many of the formulae make reference to a subscript i, which is the count value shown in the tables.
For the data used in the tables, i runs from 0 to 16, where data indexed 1 to 15 are the experimental measured
values, and those indexed 0 or 16 are the calculated values. When using the procedures described in this annex,
the amount of data that can be analysed is not limited to 15 data sets but can be any number.
B.2.1.1 Determination of derived values for Y , x and y
i i i
Calculate Y , x , y , x and y using Formulae (B.1), (B.2), (B.3), (B.4) and (B.5):
i i i
YS= lg (B.1)
()
ii10
x =+lg (minutes1) (B.2)
i 10
 
ya=+ln bY− / Ya− (B.3)
() ()
ii i
 
xx= / n (B.4)
()
∑
i
yy= / n (B.5)
()
∑
i
Formulae (B.1), (B.2), (B.3), (B.4) and (B.5) relate to values obtained for the property, S , after various
i
periods of time under test, x .
i
B.2.1.2 Determination of parameters a and b
Calculate the following using Formulae (B.6), (B.7) and (B.8):
aY= 0,995 min. (B.6)
()
0 i
ab+= 1,005 max. Y (B.7)
()
00 i
ba=+ ba− (B.8)
()
00 00
Formulae (B.6), (B.7) and (B.8) produce the initial estimates for two of the parameters in the model,
a and b .
0 0

B.2.1.3 Determination of the least squares estimates for A and B and unbiased estimate for σ
Calculate the following using Formulae (B.9), (B.10), (B.11) and (B.12):
ˆ ˆ
Ay=− Bx (B.9)
ˆ
Bx=− xy − yx/ − x (B.10)
()() ()
∑ ∑
ii i
 ˆ
σ =−yy //nn−22=−RSS (B.11)
() () ()
∑
1 ii
where
ˆ ˆ
RSS =−yA×−yB××xy (B.12)
∑ ∑ ∑
ii ii
NOTE RSS is the residual sum of squares.
B.2.1.4 Determination of estimates for parameters c and d
Calculate the following using Formulae (B.13) and (B.14):
−1
ˆˆ
ˆ
cA=+B lg 60 (B.13)
()
ˆ ˆ−1
dB=− (B.14)
B.2.2 Procedure for Line 2
B.2.2.1 Determination of X , Y , X and Y
i i
Calculate X using Formula (B.15):
i
ˆ
 ˆ
Xc=+11/ expT−lg ime − / d (B.15)
()
{}
i 10
 
NOTE Values for c and d are given as Formulae (B.13) and (B.14).
Y = lg stiffness (B.16)
()
i 10
20 © ISO 2016 – All rights reserved

XX= /n (B.17)
()
∑
i
YY= /n (B.18)
()
∑
i
ˆ
ˆ 
B.2.2.2 Determination of the least squares estimates for a and b and unbiased estimate for σ
The estimates are derived using Formulae (B.19), (B.20), (B.21) and (B.22):
ˆ
bX=− XY −YX/ − X =
()() ()
∑ ∑
ii i
nX×−YX × Yn/ ×−XX × X (B.19)
()
∑∑∑ ∑ ()∑ ∑
ˆ
ˆ
aY=−bX× (B.20)
ˆ

σ =−YY //nn−22=−RSS (B.21)
() ()
∑()
1 ii
where
ˆ
ˆ
RSS =−Ya×−Yb××XY (B.22)
∑ ∑ ∑
ii ii
ˆ
ˆ ˆ
Check that the following constraints are met: ab+> Ya> .
i
B.2.2.3 Determination of confidence and prediction intervals
ˆ
ˆ
Calculate the variances of a and b using Formulae (B.23) and (B.24):
2 2
   
22 2 22
ˆ 
var aX= σσnX − XX= nX − X (B.23)
() () ()
()∑∑∑∑()  ∑ 
2 ii ii 2 i
   
2 2
   
ˆ 22 2

var bn= σσnX − Xn= nX − X (B.24)
() ()
( )
()  ∑∑∑  ()  
2 ii 2 i
   
ˆ
ˆ
Calculate the estimated standard error, ε, of a and b given as Formulae (B.25) and (B.26):
ˆˆ
ε aa= var (B.25)
() ()
ˆˆ
ε bb= var (B.26)
( ) ( )
Formulae for 100 % confidence and prediction intervals about the fitted Line 2 as functions of X are
given as Formulae (B.27) and (B.28), respectively:
 
XX−
()
1 
ˆ 
Confidence interval μμ=± t σ + (B.27)
 
XX P 2
n
 
XX−
()
∑
i
 
 
XX−
()
 
ˆ 
Prediction interval Yt=±μσ 1++ (B.28)
 
XX P 2
n
 
XX−
()
∑
i
 
where
ˆ ˆ
ˆ ˆ
Ya==μ +bX (B.29)
XX
B.2.2.4 Student’s t test for a and b
ˆ
ˆ
To test whether a or b is equal to 0, perform the following calculations:
Pr tt< = P;
()
P
t has Student’s t-distribution on (n − 2) degrees of freedom.
From Statistical tables t boundary for P = 90 % = 1,771.
From Statistical tables t boundary for P = 95 % = 2,160.
If Formulae (B.30) and (B.31) give t values greater than the applicable one of the above bounds, it is
ˆ
ˆ
certain that a and b are not equal to 0.
ˆˆ ˆ
t for aa=ε (a) (B.30)
ˆˆ ˆ
t for bb=ε (b) (B.31)
B.2.2.5 Calculation of long-term (50-year) stiffness
All the formulae in B.2.1 and B.2.2 are standard formulae for straight-line regressions. From the values
given in Tables B.4 and B.5, the estimated long-term stiffness and its confidence and prediction limits
can be determined using Formulae (B.32), (B.33), (B.34), (B.35) and (B.36):
Using Formula (B.29), the extrapolated long-term stiffness is given as Formula (B.32):
ˆ ˆ
ˆ ˆ
Ya==μ +bX (B.32)
XX
Using Formula (B.27), the confidence limits for μ are given as Formula (B.33):
50 years
ˆ
μμ=± μ (B.33)
50 yearsbX ounds
Using Formula (B.28), the prediction limits are given as Formula (B.34):
ˆ
YY= ±Y (B.34)
50 yearsbXY ounds
Transforming these values back to stiffness gives:
ˆ
Y
50 years
Extrapolated long-term stiffness = 10
22 © ISO 2016 – All rights reserved

90 % confidence limits for extrapolated long-term stiffness are given as Formula (B.35):
ˆ
μμ±
X bounds
1μ S = 0 (B.35)
()
50 years
90 % prediction limits for extrapolated long-term stiffness are given as Formula (B.36):
YY±
XY bounds
S =10 (B.36)
50 years
B.3 Validation of statistical procedures by an example calculation
Use the data given in Table B.1 for the calculation procedures described in B.2.1 to B.2.2.5 to ensure
that the statistical procedures to be used in conjunction with this method will give results to within
±0,1 % of the values given in this example (n = 15).
B.3.1 Procedure for Line 1
B.3.1.1 Determination of derived values for Y , x and y
i i i
YS= lg (B.37)
()
ii10
NOTE 1 See derived values in Table B.1.
x =+lg (minutes1) (B.38)
i 10
NOTE 2 See derived values in Table B.1.
 
ya=+ln bY− / Ya− (B.39)
() ()
ii i
 
NOTE 3 See derived values in Table B.1.
xx= / n ==48,465 540/15 3,231 036 (B.40)
()
∑
i
yy= / n =−2,512828/15 =−0,167522 (B.41)
()
∑
i
B.3.1.2 Determination of parameters a and b
For values of Y , see Table B.1.
i
aY= 0,995 min. =×0,995 3,696 793 = 3,678 309 (B.42)
()
0 i
ab+= 1,005 max. Y =×1,005 3,852 114 = 3,871 375 (B.43)
()
00 i
ba=+ ba− =−=3,871 375 3,678 309 0,193 066 (B.44)
()
00 00

B.3.1.3 Determination of the least squares estimates for A and B and unbiased estimate for σ
Calculate the following:
ˆ ˆ
Ay=− Bx =−0,167 50−×,831 93,231 036 =−2,855 5 (B.45)
2  
ˆ 2
Bx=− xy − yx/ − xn=× xy −×xy ÷×nx − x (B.46)
()() () () ()
∑ ∑ ∑∑∑ ∑  ∑ 
ii i
 
Using values from Table B.2,
ˆ
 
B =×15 17,,84−×8472,/51 15×−187,,75 2349 34 = 0,8319
()
()
 
Using values for A and B in this subclause and from Table B.2,
 ˆ
σ =−yy //nn−22=−RSS ==0,/6653130,0512 (B.47)
() () ()
∑
1 ii
where
ˆ ˆ
RSS =−yA×−yB××xy (B.48)
∑ ∑ ∑
ii ii
B.3.1.4 Determination of estimates for parameters c and d
ˆ ˆ ˆ
ˆ
Using the values for ABand calculated above, the estimated values for c and d are:
ˆˆ−1
ˆ
cA=+B lg 60 = 1,65353 (B.49)
()
−1
ˆ ˆ
dB=− =− 1,202 (B.50)
B.3.2 Procedure for Line 2
B.3.2.1 Determination of X , Y , X and Y
i i
Calculate the following:
ˆ
 ˆ
Xc=+11/ expl− gTime − / d (B.51)
()
{}
i 10
 
NOTE 1 See derived values in Table B.3.
NOTE 2 Values for c and d are given in Formulae (B.49) and (B.50).
Y = lg stiffness (B.52)
()
i 10
NOTE 3 See derived values in Table B.3.
XX= //n ==7,966259 15 0,531084 (B.53)
()
∑
i
YY= //n ==56,728211 15 3,781881 (B.54)
()
∑
i
24 © ISO 2016 – All rights reserved

ˆ 2
ˆ 
B.3.2.2 Determination of the least squares estimates for a and b and unbiased estimate for σ
The estimates are derived using Formulae (B.55), (B.56), (B.57) and (B.58):
ˆ
bX=− XY −YX/ − X =
()() ()
∑ ∑
ii i
nX×−YX × Yn/ ×−XX × X (B.55)
()
∑∑∑ ∑ ()∑ ∑
ˆ
b =×15 30,302557 −×7,966259 56,728211 15 ×−5,146065 7,966259 ×7,,966259
() ()
==2,626734 13,729693 0,191318
ˆ
ˆ
aY=−bX× (B.56)
ˆ
a =−3,781881 0,191318×=0,53108343,680275
2 ˆ

σ =−YY //nn−22=−RSS (B.57)
() ()
()
∑
1 ii
ˆ
ˆ
RSS =−Ya×−Yb××XY =
∑ ∑ ∑
ii ii
ˆ
ˆ
Ya−× Yb−× XY =−214,573977 3,680275×−56,728211 0,191318×30,3022557 =
∑ ∑ ∑
= 0,001136

σ ==0,001136/13 0,000087 (B.58)
ˆ
ˆ ˆ
Check that the following constraints are met: ab+> Ya>
i
ˆ
ˆ ˆ
max.YY==3,852 114, min.3,696 793, 3ab+= ,680 275+=0,191 318 33,871 593 and a = 3,680 275
ii
From inspection of these values, the constraints are satisfied.
B.3.2.3 Determination of confidence and prediction intervals
ˆ
ˆ
Using Formulae (B.59) and (B.60), determine the variances of a and b:
2 2
 
 
22 2 22
ˆ 
var aX= σσnX − XX= nX − X (B.59)
()
() ()
()∑∑ ∑∑ ()  ∑ 
2 ii ii2 i
   
ˆ
var0a = ,000 033
()
2 2
   
ˆ 22 2

var bn= σσnX − Xn= nX − X (B.60)
( ) () ()
()  ∑∑∑  ()  
2 ii 2 i
   
ˆ
var0b = ,000 097
( )
ˆ
ˆ
Using Formulae (B.61) and (B.62), compute the estimated standard errors, ε, of a and b:
ˆˆ
ε aa= var (B.61)
() ()
ˆ
ε a ==0,000033 0,005756
()
ˆˆ
ε bb= var (B.62)
( ) ( )
ˆ
ε b ==0,000097 0,009 828
( )
Formulae for 100 % confidence and prediction intervals about the fitted Line 2 as functions of X are
given as Formulae (B.63) and (B.64), respectively:
 
XX−
()
 
ˆ 
Confidence interval μμ=± t σ + (B.63)
 
XX P 2
n
 
XX−
()
∑
i
 
 
XX−
()
 
ˆ 
Prediction interval Yt=±μσ 1++ (B.64)
 
XX P 2
n
 
XX−
()
∑
i
 
where
ˆ
ˆ ˆ
extrapolated long-term stiffness is μ =+abX (B.65)
X
t = 1,771 for P = 90 %
P
In Table B.4, the confidence intervals are shown as μ and μ and the prediction intervals as Y and Y .
L U L U
In Table B.5, the confidence intervals are shown as Sμ and Sμ and the prediction intervals as S and S .
L U L U
B.3.2.4 Student’s t test for a and b
ˆ
ˆ
To test whether a or b is equal to 0, the following calculations are performed:
Pr tt< = P;
()
P
t has Student’s t-distributio
...