• DTDHandbook
• Contact
• Contributors
• Sections
• 1. Introduction
• 2. Fundamentals of Damage Tolerance
• 3. Damage Size Characterizations
• 0. Damage Size Characterizations
• 1. NDI Demonstration of Crack Detection Capability
• 2. Equivalent Initial Quality
• 3. Proof Test Determinations
• 0. Proof Test Determinations
• 1. Description of the Proof Test Method
• 2. Examples
• 4. References
• 4. Residual Strength
• 5. Analysis Of Damage Growth
• 6. Examples of Damage Tolerant Analyses
• 7. Damage Tolerance Testing
• 8. Force Management and Sustainment Engineering
• 9. Structural Repairs
• 10. Guidelines for Damage Tolerance Design and Fracture Control Planning
• 11. Summary of Stress Intensity Factor Information
• Examples

# Section 3.3.2. Examples

Two examples are now presented to illustrate how the proof test might be used.  The first example describes how a proof stress condition might be chosen to find specific crack sizes.  The second example describes a typical situation whereby the proof test must be designed to guarantee a service life interval.

EXAMPLE 3.3.1         Proof Test Stress-Crack Length Relationships

(a)  What proof stress (sp ) 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 = sp/sop ).

(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 = KIC

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 sp, i.e. for the given KIC 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. KIC = 40 ksi Öin and sp = 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 (left-hand side) (inch) a/r F(a/r) a (right-hand 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 first-order 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 on-off 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.  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 stress-intensity 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 stress-intensity factor at all depths through the thickness.  As a first step, determine if the structure will leak before it breaks by calculating the stress-intensity 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 KIC = 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 KIC.  Thus, given this situation, the vessel will fail catastrophically. Change in Crack Geometry to Through-Thickness 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 one-year life interval (a two-year 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 (Dai) and crack growth rate values (da/dtêi) are chosen to be compatible.

On the basis of the given material data, one must assume that both a fatigue and a stress-corrosion 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 stress-corrosion cracking rate data can be described with the power law: which is valid for sustained loading conditions when Kmax is between the threshold of stress corrosion cracking (KIscc = 65 ksi Öin) and the fracture toughness level (KIC = 90 ksi Öin).

As a first approximation of the effect of combined stress corrosion action and fatigue crack growth, the linear summation hypothesis of Wei-Landes is suggested (see Section 5): where the time based fatigue crack growth rate is obtained from whereby the cycle-dependent component from the power law equation is multiplied by the cyclic frequency (f).  It is also to be noted that the stress-corrosion cracking rate contribution for a day in service is one-half 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 length-life relationship.  The method for this example will be to choose equal increments of Kmax between the crack size at failure and the other crack lengths established to obtain the Dai values.  The next table describes the relationships between the maximum stress-intensity factor and the crack length, the crack length increment, the average values of the maximum stress-intensity factor ( ) and stress-intensity factor range ( ).

Crack Interval Table

 Kmax (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 Kmax through the equation which when solved for a typical value of Kmax, 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 Kmax values, e.g. 62.5 ksi Öin  = 0.5 (60 + 65). The values of are computed from the relationship DK = (1-R) Kmax, 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

Stress-Corrosion 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 KIscc 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 stress-intensity 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 Stress-Corrosion 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 one-half of the time (on-off cycling) so the numbers are one-half 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 Stress-Corrosion 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 (pp) 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