美国材料试验标准 中文翻译.doc

翻译1.scope1.1 This test method covers the determinations of denisty, percent absorption, and percent voids in hardened concrete.1.2 The text of this test method references notes and footnotes which provide explanatory information. These notes and footnotes (excluding those in tables and figures) shall not be considered as requirements of this standard.1.3 The values stated in SI units are to be regarded as the standard.1.范围1.1这个测试方法包括密度的确定,吸水率,和硬化混凝土中的孔隙率。1.2这个测试方法引用了一些提供解释信息的文本注释和脚注。这些注释和脚注(包括表和数据)，不应当视为本标准的要求。1.3以国际单位值作为标准。2. Significance and Use2.1 This test method is useful in developing the data required for conversions between mass and volume for concrete. It can be used to determine conformance with specifications for concrete and to show differences from place to place within a mass of concrete.3. Apparatus3.1 Balance, sensitive to 0.025 % of the mass of the specimen.3.2 Container, suitable for immersing the specimen and suitable wire for suspending the specimen in water.2.意义和应用2.1这个测试方法，在开发混凝土的质量和体积之间转换所需的数据方面是有帮助的。它可以用来确定符合混凝土规范，它也可以显示一个混凝土内部各处质量的差异。3. 仪器3.1样本质量精确到0.025%3.2容器（适合浸泡样本和合适的水位标线）4. Test Specimen4.1 Whenever possible, the sample shall consist of several individual portions of concrete, each to be tested separately. The individual portions may be pieces of cylinders, cores, or beams of any desired shape or size, except that the volume of each portion shall be not less than 350 cm3 (or for normal weight concrete, approximately 800 g) and each portion shall be free from observable cracks, fissures, or shattered edges.4. 样本测试4.1样品尽可能由混凝土的几个独立的部分组成,每个部分单独进行测试。各部分可能是任何所需形状或大小的一块柱面,核心,或者束, 每个部分的体积不得少于350cm3(或正常重量的混凝土,大约是800 g)，并且每个部分从可观察到的裂缝、裂缝或破碎的边缘都应当是自由。5. Procedure5.1 Oven-Dry MassDetermine the mass of the portions, and dry in an oven at a temperature of 100 to 110 C for not less than 24 h. After removing each specimen from the oven, allow it to cool in dry air (preferably in a desiccator) to atemperature of 20 to 25 C and determine the mass. If the specimen was comparatively dry when its mass was first determined, and the second mass closely agrees with the first, consider it dry. If the specimen was wet when its mass was first determined, place it in the oven for a second drying treatment of 24h and again determine the mass. If the third value checks the second, consider the specimen dry. In case of any doubt, redry the specimen for 24h periods until check values of mass are obtained. If the difference between values obtained from two successive values of mass exceeds 0.5 % of the lesser value, return the specimens to the oven for an additional 24-h drying period, and repeat the procedure until the difference between any two successive values is less than 0.5 % of the lowest value obtained. Designate this last value A.5. 步骤5.1烘干质量确定其质量，再将其放置在100110C的烘箱温度下干燥，且不少于24小时。从烘箱移出每个标本后,让它们在干燥的空气冷却(最好放在干燥器里)，控制温度在20到25C之间，然后确定质量。如果样品在第一次测定质量时比较干燥的话,并且第二次测定的数据非常接近第一次的数据，那么我们认为样品是干燥的。如果样品在第一次测定质量时是潮湿的话，则将它放在烤箱里进行第二次干燥处理，时间为24小时，再确定质量。如果第三次测定值核对上第二次的数值，那么我们认为样品干燥。如有疑问，再将样本干燥24小时，直到获得核对的数值。如果连续测得的两个值中，它们相差超过较小值的0.5%，那么把样本放回烘箱，额外进行一次干燥处理，时间为24h，并且重复这个过程，直到连续两个测定值之间的差别小于最小值的0.5%时为止，然后选取最后一个测值，定为A。5.2 Saturated Mass After ImmersionImmerse the specimen, after final drying, cooling, and determination of mass, in water at approximately 21C for not less than 48h and until two successive values of mass of the surface-dried sample at intervals of 24h show an increase in mass of less than 0.5% of the larger value. Surface-dry the specimen by removing surface moisture with a towel, and determine the mass. Designate the final surface-dry mass after immersion B.5.2浸泡后饱和质量干燥、冷却试块后确定其质量，再将其浸泡在约21C的水中，不少于48小时，每隔24小时连续测得表面干燥试样两个质量数值，直到数值显示质量的增加小于较大值的0.5%为止。用毛巾擦拭表面水分使其表面干燥，并确定其质量。选取浸泡后测得表面干燥试样的最终质量，定为B。5.3 Saturated Mass After BoilingPlace the specimen, processed as described in 5.2, in a suitable receptacle, covered with tap water, and boil for 5 h. Allow it to cool by natural loss of heat for not less than 14 h to a final temperature of 20 to 25C. Remove the surface moisture with a towel and determine the mass of the specimen. Designate the soaked, boiled, surface-dried mass C.5.3沸腾后饱和质量将按上述5.2处理后的样本试块，放置在一个合适的容器里，用自来水将其淹盖，然后煮5 h。再让它通过自然热损失进行冷却，时间不少于14 h，直到最终温度为20到25C之间。然后用毛巾除去表面的水分并确定样品的质量。选取浸湿和煮沸后测得的表面干燥试样的质量，定为C。5.4 Immersed Apparent MassSuspend the specimen, after immersion and boiling，by a wire and determine the apparent mass in water. Designate this apparent mass D.5.4浸泡后表观质量经过浸泡和煮沸后用一根线使样本悬浮，并且测得样本在水中的表观质量。指定此表观质量为D。6. Calculation6.1 By using the values for mass determined in accordance with the procedures described in Section 5, make the following calculations:6. 计算6.1按上述第5项描述的过程测得质量数据，计算如下:浸泡后吸收率 = (B A)/A100浸泡煮沸后吸收率 = (C A)A100干燥体积密度=A/(C D)=g1浸泡后体积密度=B/(C D)浸泡煮沸后体积密度=C/(C D)表观密度=A/(A D)=g2孔隙率=(g2 g1)/g2100 或者 (C A)/(C D)100式中：A = 烘干质量, 单位gB = 浸泡后表面干燥质量, 单位gC = 浸泡煮沸后表面干燥质量, 单位gD = 浸泡煮沸后的表观质量, 单位gg1 = 干燥体积密度，Mg/m3g2= 表观密度，Mg/m3r = 水密度= 1 Mg/m3 = 1 g/cm3.7. Example7.1 Assume a sample having the following characteristics:7.1.1 Mass of the solid part of the specimen = 1000 g.7.1.2 Total volume of specimen (including solids, “permeable” voids, and “impermeable” voids) = 600 cm37.1.3 Absolute density of solid part of specimen = 2.0 Mg/m37.1.4 Void space in specimen contains initially only air (no water).7.举例7.1假设一个样本有以下特点: 7.1.1样本固体部分的质量= 1000g。 7.1.2样本总体积(包括固体、能渗透孔隙和不透水孔隙)= 600cm3 7.1.3样本固体部分的绝对密度= 2.0 Mg / m3。 7.1.4样本孔隙里最初只包含空气(无水)。7.2 Then, it follows that there are 500 cm3 of solids and 100 cm3 of voids making up the specimen, and the void content is 16 = 16.67 %.7.3 Assume that on immersion 90 mL of water is absorbed.7.4 Assume that after immersion and boiling 95 mL of water is absorbed.7.5 Based on the assumptions given in 7.1-7.4 above, the data that would be developed from the procedures given in Section 5 would be as follows:7.5.1 Oven-dry mass, A = 1000 g.7.5.2 Mass in air after immersion, B = 1090 g.7.5.3 Mass in air after immersion and boiling, C = 1095 g.NOTE 1Since loss of mass in water is equal to mass of displaced water, and volume of specimen = 600cm3, mass of specimen in water after immersion and boiling is 1095 600 = 495 g.7.5.4 Apparent mass in water after immersion and boiling , D = 495 g.7.2 这样一来样本就由500cm3的固体部分和100cm3的孔隙组成的，并且孔隙率为16 = 16.67%7.3假设浸泡中有90mL的水被吸收。7.4假设浸泡煮沸后有95mL的水吸收。7.5基于上述7.17.4的假设，按照第5节给出的过程得出数据如下: 7.5.1烘干的质量，A = 1000g 7.5.2浸泡后的质量，B = 1090g 7.5.3浸泡沸腾后的质量，C = 1095g。注意1因为样本在水中的质量损失等于排开水的质量，而样本体积=600cm3，所以样本在水中浸泡沸腾后的质量为1095600 = 495 g7.6 By using the data given above to perform the calculations described in Section 6, the following results will be obtained (Note 2):通过使用上面给出的数据，进行第6节中描述的计算，将得到以下结果：浸泡后吸收率= (BA)/A100 = (10901000)/1000100 = 9.0浸泡煮沸后吸收率= (CA)/A100 = (10951000)/1000100 = 9.5干燥体积密度= A/(C D) = 1000/(1095495)1 =1.67 Mg/m3 =g1浸泡后体积密度=B / (C D) = 1090/(1095495)1 =1.82 Mg/m3浸泡煮沸后体积密度= C / (C D) = 1095/(1095495)1 =1.83 Mg/m3表观密度= A /(A D) = 1000/(1000495)1 =1.98 Mg/m3 =g2渗透孔隙体积= (g2 g2)/g2100 = (1.981.67)/1.98100 =15.8Or (C A)/(C D)100 = (1095 1000)/(1095495)100 =15.7 NOTE 2This test method does not involve a determination of absolute density. Hence, such pore space as may be present in the specimen that is not emptied during the specified drying or is not filled with water during the specified immersion and boiling or both is considered “impermeable” and is not differentiated from the solid portion of the specimen for the calculations, especially those for percent voids. In the example discussed it was assumed that the absolute density of the solid portion of the specimen was 2.0 Mg/m3, the total void space was 16.67 %, and the impermeable void space was 5 cm3. The operations, if performed, and the calculations, if performed as described, have the effect of assuming that there are 95 cm3 of pore space and 505 cm3 of solids, and indicate that the solid material, therefore, has an apparent density of 1.98 rather than the absolute density of 2.00 Mg/m3 and the specimen has a percentage of voids of 15.8 rather than 16.67.注意2这个测试方法不涉及绝对密度的测定。在样本干燥过程中孔隙内并不是空的，或者在样本浸泡煮沸时孔隙里没有充满水，或者以上两种情况，像这样的孔隙空间可能存在于样本内部，因此认为样本不渗透，并且认为与样本固体部分在计算上相同，特别是孔隙率。在示例中讨论了假设的样本固体部分的绝对密度为2.0Mg/m3，总孔隙空间为16.67%，不透水孔隙空间5 cm3。如果像上述那样，那么操作和计算会受到假设的影响，则会有95cm3的孔隙空间和505cm3的固体体积，因此也表明固体材料的表观密度是1.98而绝对密度不是2.00 Mg/m3，另外样本的孔隙率是15.8而不是16.67。Depending on the pore size distribution and the pore entry radii of the concrete and on the purposes for which the test results are desired, the procedures of this test method may be adequate, or they may be insufficiently rigorous. In the event that it is desired to fill more of the pores than will be filled by immersion and boiling, various techniques involving the use of vacuum treatment or increased