EXAMPLE 3.3.1 Proof Test StressCrack Length
Relationships
For the radiallythroughthickness cracked structure
illustrated here, answer the following questions:
(a) What proof stress (s_{p }) is
required at room temperature to guarantee that the maximum crack size is less
than 0.05 inches? Also, define the ratio of proof to operating stress
conditions
(a
= s_{p}/s_{op }).
(b) For a proof test conducted at 40°F,
define the proof stress and proof stress ratio associated with finding a crack
with a length 0.05 in.
(c) If the proof test ratio is 1.5, what is the
minimum flaw size that will be detected at room temperature?
SOLUTION:
The equation that governs the
solution to all three questions is the Irwin fracture criterion, i.e.,
K
= K_{IC}
Where
_{}
with F(a/r) and the
material properties defined above.
To address the questions parts a and b, the equations are
solved for the proof stress s_{p},
i.e.
_{}
for the given K_{IC}
conditions at temperature and for a 0.05 inch long crack, i.e. a in this equation is 0.05 inch. So, for room temperature, the proof stress
is
_{}
and for 40° F the proof stress is
_{}
In both cases, the proof stress is well below the yield
strength; however, it might be noted that localized yielding at stress
concentrations could occur at these levels.
The proof stress ratios (a) are 1.23 and 1.08 for the room temperature and 40°F
proof test conditions, respectively. To
address the third part of the question, it
is necessary to solve the equations for crack length (a), i.e.
_{}
Because this equation involves crack length in the function F in a complicated fashion, the equation
is solved iteratively for the given material and stress conditions, i.e. K_{IC} = 40 ksi Öin
and s_{p}
= 1.50 x (35) = 52.5 ksi. Thus,
_{}
A series of several trials are shown in the following table,
where a match of the right and left side of the equation is achieved when a @ 0.0245 inches. Thus, 0.049 inch long cracks can be found
for a proof test ratio of 1.50.
Trial And
Error Solution
a (lefthand side)
(inch)

a/r

F(a/r)

a (righthand side)
(inch)

0.020

0.08

2.835

0.0230

0.025

0.10

2.733

0.0247

0.030

0.12

2.641

0.0265

0.0255

0.102

2.723

0.0249

0.0245

0.098

2.743

0.0246

In the above
solutions, it is seen that in some cases the proof stress is sufficiently large
such that yielding can be expected at the edge of the hole and other stress
concentration sites. The reader is
cautioned that linear elastic fracture mechanics (LEFM) techniques such as
applied in these equations should not be utilized when extensive local yielding
occurs except to obtain firstorder estimates of the crack length. From a proof test standpoint, the LEFM
estimates of the minimum crack length will be actually larger than those
screened by loading the structure to the proof condition, assuming load control
conditions, and thus conservative.
EXAMPLE 3.3.2 Proof
Test Conditions to Guarantee Life
The pressure vessel shown here has a semicircular surface crack
of unknown size located in the longitudinal direction. This vessel is subjected to an onoff
pressure loading condition of the type illustrated below and is made of a
structural steel with the mechanical properties shown.
Pressure Vessel Structure
with Semicircular Surface Crack
For economic purposes, it has been decided that the structure
will only be inspected yearly and the inspection procedure has been chosen to
be a proof test. You have been asked to
select the proof pressure level that will guarantee that this vessel will not
fail during the interval between proof test inspections subject to the crack/loading/material
property assumptions.
Pressure/Time Loading
Cycle
Material Properties for Steel Pressure Vessel
SOLUTION:
It is first necessary to calculate the gross stress in the
section of the structure where the crack is located. From any standard strength of materials text, it is determined
that for a pressure (p) of 2,000 psi,
the maximum operating stress (s)
for the vessel with an outside diameter of 40 inch and a thickness (B) of 0.4 inch is given by
_{}
or 100 ksi, and the range of stress is
_{}
For the semicircular crack partly through the vessel wall, the
stressintensity factor is given by
_{}
neglecting the back surface correction factor. Assume for
illustrative purposes that the equation can be considered a reasonable estimate
of the true stressintensity factor at all depths through the thickness. As a first step, determine if the structure
will leak before it breaks by calculating the stressintensity factor for the
condition where the crack depth is equal to the thickness. Thus, with s
= 100 ksi and a = 0.4 in.,
_{}
which is less than K_{IC}
= 90 ksiÖin
and thus the vessel might leak before fracturing. Consider, however, the potential cracking situation that occurs
if the semicircular crack penetrates the wall and immediately transitions to a
through thickness crack as shown. An
analysis indicates that K @ 112
ksiÖin,
which is greater than K_{IC}. Thus, given this situation, the vessel will
fail catastrophically.
Change in Crack Geometry to ThroughThickness Crack
After the Semicircular Crack Grows to the Inside Wall
To establish the crack size associated with the proof test, one
must conduct a life analysis which works from the final crack size (a = 0.4 inch) backwards until the
oneyear life interval (a twoyear life interval with the factor of two life
margin) is guaranteed. The life
analysis that is conducted illustrates an incremental crack length method that
uses the iterative equation
_{}
where the increments of crack length (Da_{i})
and crack growth rate values (da/dtê_{i}) are chosen to be compatible.
On the basis of the given
material data, one must assume that both a fatigue and a stresscorrosion
cracking mechanism are active (see Section 5 for discussion on these mechanisms). The fatigue crack growth rate behavior can
be described using the power law
_{}
On the basis of the material data, this equation is restricted
to the range 10 < DK
< 90 ksi Öin,
and to the stress ratio (R) of 0.25,
which is compatible with the given loading cycle.
The stresscorrosion cracking rate data can be described with
the power law:
_{}
which is valid for sustained loading conditions when K_{max} is between the threshold
of stress corrosion cracking (K_{Iscc}
= 65 ksi Öin)
and the fracture toughness level (K_{IC}
= 90 ksi Öin).
As a first approximation of the effect of combined stress
corrosion action and fatigue crack growth, the linear summation hypothesis of
WeiLandes is suggested (see Section 5):
_{}
where the time based fatigue crack growth rate is obtained from
_{}
whereby the cycledependent
component from the power law equation is multiplied by the cyclic frequency (f).
It is also to be noted that the stresscorrosion cracking rate
contribution for a day in service is onehalf that established by the da/dt equation since the vessel is only
loaded to the maximum pressure only half the time.
There are a number of ways that the
Life equation can be used to
establish the crack lengthlife relationship.
The method for this example will be to choose equal increments of K_{max} between the crack size
at failure and the other crack lengths established to obtain the Da_{i} values. The next table describes the relationships between the maximum
stressintensity factor and the crack length, the crack length increment, the
average values of the maximum stressintensity factor (_{}) and stressintensity factor range (_{}).
Crack Interval Table
K_{max} (ksi
Öin)

55

60

65

70

75

80

a
(inch)

0.189

0.225

0.264

0.307

0.352

0.400

Da
(inch)

0.036

0.039

0.043

0.045

0.048

_{}

57.5

62.5

67.5

72.5

77.5

_{}

43.1

46.9

50.6

54.4

58.1












*Average values for the interval
The calculations of crack length a in this table are directly related to K_{max} through the equation
_{}
which when
solved for a typical value of K_{max},
say 55 ksi Öin,
the crack length becomes
_{}
The difference in crack lengths (Da) comes from
subtracting the two corresponding crack lengths. The values of _{}_{max} are computed by averaging the two
corresponding K_{max} values,
e.g. 62.5 ksi Öin = 0.5 (60 + 65). The values of _{} are computed from
the relationship DK = (1R) K_{max},
where R is the stress ratio (0.75).
The next table presents the fatigue crack growth rate
contribution and the following table presents the stress corrosion cracking
contribution.
Fatigue Crack Growth Rate
Contribution
_{} (ksi
Öin)

_{} (in/cycle)

_{}

43.1

2.26 x 10 ^{5}

1.13 x 10 ^{–4}

46.9

2.91 x 10 ^{5}

1.46 x 10 ^{–4}

50.6

3.64 x 10 ^{5}

1.82 x 10 ^{–4}

54.4

4.51 x 10 ^{5}

2.25 x 10 ^{4}

58.1

5.48 x 10 ^{5}

2.74 x 10 ^{4}

StressCorrosion Cracking Rate Contribution.
_{}

_{}

_{}

57.5

0*

0

62.5

0*

0

67.5

3.73 x 10 ^{4}

1.86 x 10 ^{4}

72.5

5.65 x 10 ^{4}

2.82 x 10 ^{4}

77.5

8.3 x 10 ^{4}

4.16 x 10 ^{4}

* _{} is below K_{Isc}_{c}
and therefore no growth occurs
In the Fatigue Crack Growth Rate Table, the _{} values are taken
from the Crack Interval Table and cover each of the consecutive intervals of
crack length. From the da/dN equation the crack growth fatigue
rate for a stressintensity range of 43.1 is
_{}
The calculations of _{} follow directly from
multiplying the fatigue crack growth rates by the frequency of load application
(5 cycles/day).
In the StressCorrosion Cracking Rate Table, the _{} values are taken
from Crack Interval Table and cover each of the consecutive intervals of crack
length. From the da/dt equation, the sustained load stress corrosion cracking growth
rate is
_{}
The calculations of the corrosion contribution to the total da/dt equation are also given in the
table. These come directly from the
fact that the structure is only loaded into the range where stress corrosion
cracking occurs for onehalf of the time (onoff cycling) so the _{} numbers are onehalf
those given in the middle column.
The total contribution to cracking behavior is calculated from
the total da/dt equation, and the
individual crack increments in the Life
equation are used to establish the time that it takes to grow the crack through
the successive intervals. The
appropriate calculations are reported in the next table.
Estimating the Time To Growth Through Successive
Intervals.
Da
(inch)

_{} (in/day)

Dt
(days)

A
(inch)

t =
SDt
(days)

0.036

1.13 x 10 ^{4}

318.6

0.189

861.1

0.039

1.46 x 10 ^{4}

267.8

0.225

542.5

0.042

3.68 x 10 ^{4}

114.9

0.264

274.7

0.045

5.07 x 10 ^{4}

89.5

0.307

159.8

0.048

6.9 x 10 ^{4}

70.3

0.352

70.3




0.400

0

The crack length increment (Da)
and the crack length (a) values given
in this table come from the Crack Interval Table. The total crack growth rate _{} values come from the
total da/dt equation, where the
individual contributions come from the Fatigue Crack Growth Rate and
StressCorrosion Cracking Rate Tables, e.g.
_{}
for Da
= 0.045 inch and a between 0.307 and
0.352 inch. The increment of time
required to propagate the crack through this interval is obtained from
_{}
The total time that it takes to grow through successive
intervals is obtained by summing the results from this equation for each
interval using the Life equation.
The data from the table that relates crack length (a) to the total time (t) to failure shows that the proof test
must find a crack length between 0.189 and 0.225 inch to guarantee the
integrity of the vessel with a factor of two life margin. The crack length versus total time to
failure data have been graphically displayed in the next figure, where it can
be seen that for one year of growth the crack length is 0.245 inch (and for a
factor of two life margin the crack length is 0.20 inch). The required proof stress for the 0.20 inch
long crack length is obtained from the Irwin criterion:
_{}
which is about 80 percent of the yield strength and therefore,
the proof pressure (p_{p})
must be at least
_{}
to ensure that all semicircular cracks longer than 0.2 inch are
removed from the center section of the vessel prior to operation.
Graphical Procedure for Interpreting Crack Length