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OCR for page 65

A-11
25 ure A-17 shows the results of SIM tests at 2,000, 1,500, and
1,000 psi.
20 Subsequent tests revealed that better results were obtained
at applied stresses of 1,500, 1,250, and 1,000 psi. The failure
times at 2,000 psi were thought to be too quick for good
STRAIN (%)
15
extrapolations.
The results from these test show that the 50-year strength
10 of this material will be about 1,074 psi.
5
A.4.3 BFF Test for Long-Term Stress Crack
Resistance (TRI Method)
0
0 20000 40000 60000 80000 100000 Three sets of test conditions are required to perform a ser-
TIME (sec) vice lifetime prediction with the use of the BFF test. The con-
ditions selected are the same ones used in the FDOT 100-year
Figure A-13. Raw SIM data.
service-lifetime protocol. They are the following:
It should be mentioned here that the transition from one 80°C at 650 psi applied load in water,
temperature to the next is an important variable in SIM test- 80°C at 450 psi applied load in water, and
ing. The time it takes for the specimen to equilibrate at the new 70°C at 650 psi applied load in water.
temperature should be just a few minutes. Other things that
occur during the transition time are thermal expansion or con-
Service Lifetime Prediction
traction of the specimen as well as re-equilibration of the grips
and extensometer. TRI excludes the data from the transition The service lifetime is found by application of the rate
region, but keeps the time scale in place. The transitions then process method (RPM), described in ASTM D2837. The basic
show up as blank spots in any plot with time as the abscissa. equation is:
The next step in the analysis is to determine what we refer
to as the virtual starting time for each step above the first one, log t = A + B T + C log S T (1)
or t'. This accounts for the effects of the creep that occurred
where
at the lower temperature. This step is necessary because the
specimen "remembers" what had occurred at the previous t = failure time (hrs)
creep step. This also allows one to rescale the individual creep T = temperature (°K)
curves and get them all on a common time scale. S = stress (psi), and
The virtual starting time is found by plotting creep modu- A, B and C are constants.
lus vs. log time for the end of one step and the beginning of the
Under the three sets of test conditions, the values of the
next step. Then, one can adjust the virtual starting time until
constants are found by solving three equations with three
the slopes of the two steps align. A vertical shift is also added
unknowns. The three equations are the following:
at this time to aid in the alignment process. This is illustrated
in Figures A-14a to A-14d for the end of the 41°C step and the 1. 80°C/650 psi log t1 = A + B/353 + C log 650/353
beginning of the 48°C step. 2. 80°C/450 psi log t2 = A + B/353 + C log 450/353
Once this is done for each step, master curves can be pre- 3. 70°C/650 psi log t3 = A + B/343 + C log 650/343
sented as either creep modulus or strain. Master curves for this
data set are shown in Figures A-15 and A-16. Constant A can be found by subtracting equation 3 from
From these two curves, one can obtain both the 50-year equation 1:
creep modulus and 50-year creep strain. In this case, they are
10,000 psi and 8.9% respectively. These represent the behavior 353 log t1 - 343 log t 3 = 353A - 343A = 10 A
of the material when placed under a 1,000 psi load for 50 years. A = ( 353 log t1 ) - 343 log t 3 10
Constant C can be found by subtracting equation 2 from
A.4.2 Stepped Isothermal Method (SIM) for
equation 1:
Creep Rupture (ASTM D6992)
353 log t1 - 343 log t 2 = C log 650 - C log 450
Additionally, if one runs separate experiments at several
loads, a master curve of stress vs. time can be obtained. Fig- C = 353 ( log t1 - log t 2 ) 2.813 - 2.653

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A-12
.xls, g6.4 .xls, g6.4
27500 27500
27000 27000
26500 26500
26000 26000
CREEP MODULUS
CREEP MODULUS
25500 25500
25000 25000
24500 24500
24000 24000
23500 23500
23000 23000
22500 22500
5.55 5.6 5.65 5.7 5.75 5.8 5.85 5.9 5.95 6 6.05 5.55 5.6 5.65 5.7 5.75 5.8 5.85 5.9 5.95 6 6.05 6.1
LOG TIME (sec) LOG TIME (sec)
(a) t' = 0 sec (b) t' = 10,000 sec
.xls, g6.4 .xls, g6.4
27500 27500
27000 27000
26500 26500
26000 26000
CREEP MODULUS
CREEP MODULUS
25500 25500
25000 25000
24500 24500
24000 24000
23500 23500
23000 23000
22500 22500
5.6 5.7 5.8 5.9 6 6.1 6.2 6.3 5.9 6 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8
LOG TIME (sec) LOG TIME (sec)
(c) t' = 20,000 sec (d) t' = 28,100 sec
Figure A-14. Curve alignments at different virtual starting times.
180000
10,000 Hour 50 Year
160000
140000
CREEP MODULUS (psi)
120000
100000
80000
60000
40000
20000 REFERENCE TEMPERATURE - 23C
0
-3 -2 -1 0 1 2 3 4 5 6 7
LOG TIME (hr)
Figure A-15. Creep modulus master curve under 1,000 psi of stress.

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A-13
40
REFERENCE TEMPERATURE - 23C
35
30
STRAIN (%)
25
20
15
10
5
10,000 Hour 50 Year
0
-4 -3 -2 -1 0 1 2 3 4 5 6 7 8
LOG TIME (hr)
Figure A-16. Creep strain master curve under 1,000 psi of stress.
And, constant B can be found by substituting the equations Determine C
derived above into Equation 1.
C = 353 (log t1 - log t2)/2.813 - 2.653
log t1 = A + B 353 + C log 650 353 C = -1306.10
353 log t1 = 353 A + B + C log 650
Determine B
B = 353 log t1 - 353A - C log 650
B = 353 log t1 - 353 A - C log 650
Example Problem: Given the following failure times, calcu-
B = 720.47 + 7247.09 + 3674.06
late the three constants.
B = 11,628.95
80°C 650 psi = 110 ± 33 hrs = t1 log t1 = 2.041
COV = 30% Now,withtheuseof these constants one can calculate the fail-
uretime,t,foranyother set of temperature and stress. For exam-
80°C 450 psi = 430 ± 172 hrs = t 2 log t 2 = 2.633
pletheservice lifetime at 23°C and 500 psi would be found from:
COV = 35%
70°C 650 psi = 500 ± 175 hrs = t 3 log t 3 = 2.699 log t = A + B 296 + ( C log 650 ) 296
COV = 40%
28.95 296 - (1306.10 log 500 ) 296
log t = -20.50 + 11, 62
Determine A log t = -20.50 + 39.29 - 11.98
A = (353 log t1 - 343 log t3)/10 log t = 688
A = -20.50 t = 7, 568, 640 h = 886 Years
s
2512
3.4
RUPTURE STRESS (psi)
LOG YIELD STRESS (psi)
3.3 1995
10,000 hour = 1307 psi
3.2 1585
50 year = 1074 psi
100 year = 1014 psi
3.1 1259
sim rupture Regression Line
regression
(time is dependent variable)
3.0 1000
23C Reference Temperature
2.9 794
0 1 2 3 4 5 6 7
LOG TIME (hr)
Figure A-17. Long-term yield stress by SIM.

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A-14
3.4
Average Predicted Lifetime Under 500 psi = 864
3.2 Average 50 Year Strength = 955 psi
y = -0.2266x + 4.2577
3
Log Stress (psi)
2.8
2.6
y = -0.2702x + 3.3647
2.4 R2 = 1
2.2
0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00
Log Failure Time (hrs)
Figure A-18. BFF lifetime prediction.
This value represents the average lifetime based on the test 1669.74 = -6076.88 + 11, 628.95 - 1306.10 log S
results.
8 = log S, S = 950 psi
2.98
Long-Term Strength
The rate process method predicted a service lifetime at
Once the constants are known, one can also calculate the
23°C and 500 psi of stress to be 176 years and also deter-
50 year strength at 23°C from equation 1. 50 years = 438,000 h
mined the 50-year strength would be 661 psi. This strength
represents the stress under which stress cracks will not form
log 438,000 = A + B 296 + ( C log S ) 296
for 50 years. These results can be presented graphically as in
5.641 = -20.50 + 11, 628.95 296 + ( -1306.10 log S ) 296 Figure A-18.