Cover Image

Not for Sale



View/Hide Left Panel
Click for next page ( 66


The National Academies | 500 Fifth St. N.W. | Washington, D.C. 20001
Copyright © National Academy of Sciences. All rights reserved.
Terms of Use and Privacy Statement



Below are the first 10 and last 10 pages of uncorrected machine-read text (when available) of this chapter, followed by the top 30 algorithmically extracted key phrases from the chapter as a whole.
Intended to provide our own search engines and external engines with highly rich, chapter-representative searchable text on the opening pages of each chapter. Because it is UNCORRECTED material, please consider the following text as a useful but insufficient proxy for the authoritative book pages.

Do not use for reproduction, copying, pasting, or reading; exclusively for search engines.

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 80C at 650 psi applied load in water, temperature to the next is an important variable in SIM test- 80C at 450 psi applied load in water, and ing. The time it takes for the specimen to equilibrate at the new 70C 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 41C step and the 1. 80C/650 psi log t1 = A + B/353 + C log 650/353 beginning of the 48C step. 2. 80C/450 psi log t2 = A + B/353 + C log 450/353 Once this is done for each step, master curves can be pre- 3. 70C/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

OCR for page 65
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.

OCR for page 65
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 80C 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- 80C 450 psi = 430 172 hrs = t 2 log t 2 = 2.633 pletheservice lifetime at 23C and 500 psi would be found from: COV = 35% 70C 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.

OCR for page 65
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 23C and 500 psi of stress to be 176 years and also deter- 50 year strength at 23C 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.