ASTM C1851-2018 Standard Practice for Determining the Extent of Cracking in a Sealant using the Difference between the Compressive and Tensile Modulus《用压缩模量和拉伸模量之差测定密封剂开裂程度的标准实施规程》.pdf

《ASTM C1851-2018 Standard Practice for Determining the Extent of Cracking in a Sealant using the Difference between the Compressive and Tensile Modulus《用压缩模量和拉伸模量之差测定密封剂开裂程度的标准实施规程》.pdf》由会员分享，可在线阅读，更多相关《ASTM C1851-2018 Standard Practice for Determining the Extent of Cracking in a Sealant using the Difference between the Compressive and Tensile Modulus《用压缩模量和拉伸模量之差测定密封剂开裂程度的标准实施规程》.pdf（6页珍藏版）》请在麦多课文档分享上搜索。

1、Designation: C1851 18Standard Practice forDetermining the Extent of Cracking in a Sealant using theDifference between the Compressive and Tensile Modulus1This standard is issued under the fixed designation C1851; the number immediately following the designation indicates the year oforiginal adoption

2、 or, in the case of revision, the year of last revision. A number in parentheses indicates the year of last reapproval. Asuperscript epsilon () indicates an editorial change since the last revision or reapproval.1. Scope1.1 This practice covers a procedure for quantitativelydetermining the extent of

3、 cracking in a sealant sample byevaluating the difference between the measured compressiveand tensile modulus of a sealant relative to an unexposed oruncracked version of the same sealant. The cracks will reducethe area of the sealant in the tensile modulus, but in thecompressive modulus measurement

4、 they will not change thearea over which the modulus is determined.1.2 The values in SI units are to be regarded as standard.The values in parentheses are for information only.1.3 This standard does not purport to address all of thesafety concerns, if any, associated with its use. It is theresponsib

5、ility of the user of this standard to establish appro-priate safety, health, and environmental practices and deter-mine the applicability of regulatory limitations prior to use.1.4 This international standard was developed in accor-dance with internationally recognized principles on standard-ization

6、 established in the Decision on Principles for theDevelopment of International Standards, Guides and Recom-mendations issued by the World Trade Organization TechnicalBarriers to Trade (TBT) Committee.2. Referenced Documents2.1 ASTM Standards:2C717 Terminology of Building Seals and SealantsC719 Test

7、Method for Adhesion and Cohesion of Elasto-meric Joint Sealants Under Cyclic Movement (HockmanCycle)C1735 Test Method for Measuring the Time DependentModulus of Sealants Using Stress RelaxationE631 Terminology of Building Constructions3. Terminology3.1 Definitions:3.1.1 For definitions of terms used

8、 in this practice, refer toTerminologies E631 and C717.4. Summary of Practice4.1 This practice consists of measuring the modulus of thesealant using Test Method C1735 in both tension and compres-sion. Once these values have been determined the formulapresented in this practice will be used to determ

9、ine the degreeof cracking in the sealant relative to an unexposed or uncrackedsample of the same sealant.4.2 The motivation for this practice is to quantitativelydetermine the extent of cracking in the sealant. This measure-ment is currently determined with a qualitative measure ofcracking determine

10、d by visual inspection. The degree ofcracking has been used as a measure of performance forsealant.4.3 This practice will enable determination of the percent ofcracking of a sealant relative to an unexpected or uncrackedvesion of the same sealant.5. Significance and Use5.1 The intent of this practic

11、e is to quantitatively determinethe amount of cracking of a sealant relative to an unexposed oruncracked sample. Some samples of sealant have been ob-served to exhibit some degree of cracking some period afterinstallation. The degree of cracking is assessed visually in aqualitative manner that takes

12、 into account the area of the cracksat the exposed surface of the sealant, but does not take intoaccount the depth or profile of the cracks. The degree ofcracking in the sealant has been used as an indication ofperformance change.6. Procedure6.1 The modulus of the sealant is determined in compres-si

13、on and separately in tension using Test Method C1735. Thesevalues are determined from a new sample (the unexposedreference without any cracking), and the sample of interest.6.2 From the four modulus measurements obtained fromTest Method C1735, determine the value of the modulus at100 s for each of t

14、hese four conditions:1This practice is under the jurisdiction of ASTM Committee C24 on BuildingSeals and Sealants and is the direct responsibility of Subcommittee C24.20 onGeneral Test Methods.Current edition approved Feb. 1, 2018. Published March 2018. DOI: 10.1520/C1851-18.2For referenced ASTM sta

15、ndards, visit the ASTM website, www.astm.org, orcontact ASTM Customer Service at serviceastm.org. For Annual Book of ASTMStandards volume information, refer to the standards Document Summary page onthe ASTM website.Copyright ASTM International, 100 Barr Harbor Drive, PO Box C700, West Conshohocken,

16、PA 19428-2959. United StatesThis international standard was developed in accordance with internationally recognized principles on standardization established in the Decision on Principles for theDevelopment of International Standards, Guides and Recommendations issued by the World Trade Organization

17、 Technical Barriers to Trade (TBT) Committee.16.2.1 New Sample (the unexposed reference without anycracking), compression (E100,bo,C),6.2.2 New Sample (the unexposed reference without anycracking), Tension (E100,bo,T),6.2.3 Test Sample, Compression, (E100,C),6.2.4 Test sample, Tension (E100,T).6.3 I

18、nsert the four values determined in 6.2 into the follow-ing two empirically derived relationships:3E100,T5 dE100,bo,T$1 2 a2f 2 1 2 a2!f2% (1)E100,C5 dE100,bo,C1 2 a1f! (2)6.4 In these expressions, the experimentally determinedvalues should be used, a1= 0.118 and a2= 0.562. With theseexpressions and

19、 the values determined in 6.3, there are twoequations and two unknowns.6.5 Solve for the two remaining undermined values f and d.The unknown f represents the fraction of cracks and has a valuebetween 0 and 1. Since Eq 1 is quadratic, f will have twovalues. Only the positive value should be used. The

20、 d param-eter measures the change in modulus that is not due to crackformation. The details of the derivation of these two expres-sions (Eq 1 and Eq 2) are detailed in Appendix X1.Additionally, a worked example is presented in Appendix X2.7. Report7.1 Report the following information:7.1.1 Identific

21、ation of the sealant measured, including type,source, manufacturer code number, curing conditionsemployed,7.1.2 Identification of the substrates,7.1.3 Name and description of primers that were used, ifany,7.1.4 Number of specimens tested,7.1.5 Description of the sealant appearance.7.1.6 The value ca

22、lculated for f. This can be reported as thecalculated 0-1 value or if multiplied by 100 as % cracked.7.1.7 The value calculated for modulus change not attrib-uted to cracking, d.8. Keywords8.1 compression; cracks; modulus; sealant; tensionAPPENDIXESX1. ANALYSIS OF A CRACKED SEALANTX1.1 A Test Method

23、 C719 sealant sample is characterizedusing Test Method C1735. This standard yields a modulusversus time curve for the sealant. This performed in bothtension and compression. See Fig. X1.1.X1.2 The black circles in this plot are the pre-exposureresults from the Test Method C1735 tensions tests. Theco

24、mpression baseline results would look very similar.X1.3 The sample experiences some type of exposure. Thesample is removed from the exposure and once again TestMethod C1735 is performed in two tests, compression modulusand tensile modulus. In Fig. X1.1, a series of Test MethodC1735 tensile modulus r

25、esults after different exposures areplotted.At this point it is important to note that the curve shapeis the same but the exposure has caused the modulus to belower.X1.4 Instead of keeping the entire time dependence, thechanges to the entire curve can be represented by a single timepoint. An arbitra

26、ry choice of 100 s after the stress relaxationcomponent of the Test Method C1735 test is chosen to be farenough away from the complications associated with imposedstrain and not too long to start to see extensive relaxation of thesealant affecting the sealant. The modulus value recorded at100 s is r

27、epresented by: E100. So now we have four values forthe 100 s modulus determined by the four Test Method C1735tests:X1.4.1 The initial (baseline modulus) in tension: E100, T, bo.X1.4.2 The value in tension after some exposure b: E100, T.X1.4.3 The initial (baseline modulus) in compression: E100,C, bo

28、.X1.4.4 The value in compression after some exposure b:E100, C.3The derivation of these relationships are shown in Appendix X1 and AppendixX2 of this practice.FIG. X1.1 Modulus versus Time CurveC1851 182X1.5 Now we can use the following expressions:E100,T5 dE100,bo,T$1 2 a2f 2 1 2 a2!f2% (X1.1)E100,

29、C5 dE100,bo,C1 2 a1f! (X1.2)X1.6 There are several factors needed to define in theseexpressions: a1, a2, f, and d. The factors a1and a2are factorsthat depend only on the geometry of the samples. These arediscussed later in this paragraph. The symbol d in the aboveexpression is the fractional of the

30、original modulus that wouldbe retained if there were no flaws or debonding and onlymolecular level changes (range from 1 to 0). So the fraction ofmodulus loss produced by molecular level changes is (1-d).The parameter f is the fraction of the cross-section that is notflawed or debonded (range 1 to 0

31、). The fraction cross-sectionthat is flowed or debonded cross-section is (1-f). The fitconstraints are a1and a2. The values for a1and a2weredetermined from a series of three different sealant chemistrysamples with a variety of induced known cracks and nodegradation. It as found that the sealant chem

32、istry of the natureof the cracks (symmetric or asymmetric, front or back, top orbottom) did not affect the determined values of a1and a2asexpected as these are purely geometric dependent values. a1and a2were determined by experimental fit to the data in Fig.X1.2 to be a1= 0.118, and a2= 0.52 (note t

33、hat this assumes fis not in percent but in the range from 1 to 0.X1.7 Table X1.1 shows the types of cracks induced in thethree different sample chemistries and the resulting modulusratio versus effective debonded area is plotted in Fig. X1.2.Inthe figure, the model predictions are from Eq X1.1 and E

34、qX1.2. The interfaces refer to the position of the induced crack,in the center of the sealant or near the interface with thesubstrate. The side refers to the location of the crack relative tothe front or back of the sealant or from the side. The symmetricor asymmetric refers to whether the crack is

35、just on one side(asymmetric) or balanced on both sides (symmetric).X1.8 From Eq X1.1 and Eq X1.2 the values of a1and a2areknown. Additionally, we have measured the four modulusvalues: E100,T,bo; E100,T; E100,C,bo; E100,C. At this point it isuseful to define the ratios RC, RTand R as follows:Let:.RC=

36、 E100,C/E100,bo,C,RT= E100,T/E100,bo,T, andR = RT/RC=(E100,TE100,bo,C)/(E100,bo,TE100,C)X1.9 With these defined relationships, the next task is tofind the f and d. By substituting in the values of RC, RT, R intoEq X1.1 and Eq X1.2, the following expression is produced:R 51 2 a2f 2 1 2 a2!f21 2 a1f(X

37、1.3)so:1 2 a2f 2 1 2 a2!f25 R 2 Ra1fand:1 2 a2!f21a22 Ra1!f1R 2 1! 5 0 (X1.4)The fraction of cross section without debonds or flaws is:f 5Ra12 a2!6=a22 Ra1!22 4 1 2 a2!R 2 1!21 2 a2!(X1.5)The values of E100if only molecular level changes werepresent are:FIG. X1.2 Effective Debonded Area (%)C1851 183

38、dE100,bo,T5 E100,T$1 2 a2f 2 1 2 a2!f2% (X1.6)dE100,bo,C5 E100,C1 2 a1f!and the degradation factor is:d 5 RT$1 2 a2f 2 1 2 a2!f2% 5 RC1 2 a1f! (X1.7)X1.10 Eq X1.6 is quadratic so there are two values for fpossible. An examination of the graphs suggests that one willbe positive and the other negative

39、. Since f must be in the rangefrom 0 to 1, the positive value is required. Generally, thismeans the plus sign on the radical. For the values of a1and a2given above, the solutions for different values of R are given inTable X1.2. Note that as the value of R goes from 1 to 0, thecorresponding f result

40、s go from 0 to 1 as must be the case.X1.11 From this procedure, the values of f and d can bedetermined.X1.11.1 The validity of this procedure was examined bydoing the following experiment. A set of aged white samplesthat had developed cracks were measured using the procedureoutlined in this appendix

41、. Values for f and d were determined.The same samples were then coated with black ink. The blackink dried. The samples were pulled to failure. This resulted ina sample where the cracked areas of the failure surface werecoated black and the sealant was white. Image analysis wasused to determine the f

42、raction of cracked sample f. The resultsare shown in Table X1.3.X1.11.2 Within the relative standard error of themeasurements, the image analysis and the procedure outlinedin this standard were statistically equivalent.TABLE X1.1 Crack CharacteristicsSample InterfaceCenterSideFront/BackSymmetricAsym

43、metric1C1 Interface Side Asymmetric1C2 Interface Side Symmetric2C1 Center Side Asymmetric3C1 Interface Back Asymmetric4C1 Center Front/Back Symmetric5C1 Interface Side Asymmetric2C2 Center Side Asymmetric5C2 Interface Both Sides Asymmetric3C2 Interface Back plus both sides Asymmetric4C2 Center F/B p

44、lus both sides SymmetricC1851 184X2. WORKED EXAMPLE CALCULATIONX2.1 In this appendix a worked example calculation ispresented.X2.1.1 To begin, a sample is characterized using TestMethod C1735 in tension and compression.At some later pointin time, a second Test Method C1735 characterization wasperfor

45、med in both tension and compression.X2.1.2 The four curves are plotted and it is determined thatthe two tension curves are the same shape and the twocompression curves are the same shape. This allows the use ofthe 100 s modulus to be used to represent the entire curve.X2.1.3 The values for the 100 s

46、 modulus are:E100, T, bo = 0.7509 MPaE100, T = 0.212 MPaE100, C,bo = 0.6191 MPaE100, C = 0.256 MPaX2.1.4 Using the following expressions from Appendix X1,the ratio values can be calculated.RC= E100,C/E100, bo, C,RC= E100,T/E100, bo, T,andR = RT/RC=(E100,TE100,bo,C)/(E100,bo,TE100,C),X2.1.5 From our

47、example:RC5 E100,CE100,bo,C, or 0.414 5 0.2560.6191RT5 E100,TE100,bo,T, or 0.281 5 0.2120.7509R 5 RTRC5 E100,TE100,bo,C!E100,bo,TE100,C!, 5 0.2810.4145 0.6802TABLE X1.2 Solutions for Different Values of RRfone f two1 0 1.0140.95 0.1012 1.1280.9 0.1861 1.2270.85 0.2605 1.3150.8 0.3273 1.3950.75 0.388

48、4 1.470.7 0.4449 1.5390.65 0.4977 1.6060.6 0.5473 1.6690.55 0.5942 1.7290.5 0.6388 1.7870.45 0.6813 1.8430.4 0.722 1.8970.35 0.7611 1.950.3 0.7987 2.0010.25 0.835 2.0510.2 0.8701 2.0990.15 0.904 2.1470.1 0.9369 2.1930.05 0.9689 2.2390 1 2.283TABLE X1.3 Fraction of Cracked Sample fSample f from modul

49、us ratio(this procedure)f from image analysis1 (477)% (525)%2 (567)% (605)%C1851 185X2.1.6 Now that we have the R value we can substitute intoEq X1.5 from Appendix X1:f 5Ra12 a2!6=a22 Ra1!22 41 2 a2!R 2 1!21 2 a2!f 5s0.6802 0.1182 0.562ds0.5622 0.6802 0.118d224s1 2 0.562ds0.68022 1d2s1 2 0.562dThis gives two values for f (0.467 and 1.56). If we take thepositive value of f it is 0.467. This matches the data tablefrom Appendix X1.X2.1.7 Now for d we can use Eq X1.7 from Appendix X1.d 5 RT$1 2 a2f 2 1 2 a2!f2