CN105823686A - Method for measuring tensile strength and fracture toughness of cement mortar or concrete - Google Patents
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- 239000011083 cement mortar Substances 0.000 title claims abstract description 50
- 238000000034 method Methods 0.000 title claims abstract description 19
- 238000012360 testing method Methods 0.000 claims abstract description 64
- 238000011068 loading method Methods 0.000 claims abstract description 8
- 238000005520 cutting process Methods 0.000 claims abstract description 5
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- 230000003068 static effect Effects 0.000 claims abstract description 3
- 239000000463 material Substances 0.000 claims description 12
- 239000002245 particle Substances 0.000 claims description 4
- 238000010998 test method Methods 0.000 claims description 4
- 239000000203 mixture Substances 0.000 claims description 3
- 238000013001 point bending Methods 0.000 claims description 3
- 238000004364 calculation method Methods 0.000 claims 1
- 230000005476 size effect Effects 0.000 description 5
- 238000010586 diagram Methods 0.000 description 3
- 238000009826 distribution Methods 0.000 description 3
- 238000007656 fracture toughness test Methods 0.000 description 2
- 238000004519 manufacturing process Methods 0.000 description 2
- 230000009286 beneficial effect Effects 0.000 description 1
- 230000015572 biosynthetic process Effects 0.000 description 1
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- 238000005516 engineering process Methods 0.000 description 1
- 210000003041 ligament Anatomy 0.000 description 1
- 238000012986 modification Methods 0.000 description 1
- 230000004048 modification Effects 0.000 description 1
- 239000004570 mortar (masonry) Substances 0.000 description 1
- 238000012545 processing Methods 0.000 description 1
- 238000009827 uniform distribution Methods 0.000 description 1
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- G—PHYSICS
- G01—MEASURING; TESTING
- G01N—INVESTIGATING OR ANALYSING MATERIALS BY DETERMINING THEIR CHEMICAL OR PHYSICAL PROPERTIES
- G01N1/00—Sampling; Preparing specimens for investigation
- G01N1/28—Preparing specimens for investigation including physical details of (bio-)chemical methods covered elsewhere, e.g. G01N33/50, C12Q
- G01N1/286—Preparing specimens for investigation including physical details of (bio-)chemical methods covered elsewhere, e.g. G01N33/50, C12Q involving mechanical work, e.g. chopping, disintegrating, compacting, homogenising
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Abstract
本发明公开了一种同时测定水泥砂浆或混凝土拉伸强度与断裂韧度的方法,包括浇筑试件、切出裂缝、静力加载和记录峰值荷载、计算名义强度<i>σ</i>n、计算等效裂缝长度<i>a</i>e等步骤,最后经过回归分析同时得出水泥砂浆或混凝土的拉伸强度<i>f</i>t与断裂韧度<i>K</i>IC。本发明方法形式简单,试验易操作,具有足够精度,且易判断结果合理性。只需由小尺寸试件的峰值荷载,即同时确定水泥砂浆及混凝土的拉伸强度<i>f</i>t和断裂韧度<i>K</i>IC。
The invention discloses a method for simultaneously measuring the tensile strength and fracture toughness of cement mortar or concrete, including pouring test pieces, cutting out cracks, static loading and peak load recording, and calculating nominal strength <i>σ</i> n , calculate the equivalent crack length <i>a</i> e and other steps, and finally obtain the tensile strength <i>f</i> t and fracture toughness <i> of cement mortar or concrete at the same time through regression analysis K</i> IC . The method of the invention is simple in form, easy to operate in the test, has sufficient precision, and is easy to judge the rationality of the result. The tensile strength <i>f</i> t and fracture toughness <i>K</i> IC of cement mortar and concrete can be determined at the same time only from the peak load of small-sized specimens.
Description
技术领域technical field
本发明涉及土木与水利工程材料性能技术领域,具体涉及一种同时测定水泥砂浆或混凝土拉伸强度与断裂韧度的方法。The invention relates to the technical field of civil engineering and hydraulic engineering material properties, in particular to a method for simultaneously measuring the tensile strength and fracture toughness of cement mortar or concrete.
背景技术Background technique
大量研究表明,水泥砂浆及混凝土的强度与断裂韧度存在明显的尺寸效应。若要得到与试件尺寸无关的真实材料参数,则须浇筑较大尺寸试件。因此,小尺寸水泥砂浆试件的测试成果并不能直接用来评价水泥砂浆和混凝土的真实材料性能。A large number of studies have shown that there is an obvious size effect on the strength and fracture toughness of cement mortar and concrete. To obtain real material parameters independent of the size of the specimen, it is necessary to cast a larger size specimen. Therefore, the test results of small-scale cement mortar specimens cannot be directly used to evaluate the real material properties of cement mortar and concrete.
目前,水泥砂浆和混凝土的拉伸强度一般须采用轴心拉伸试样进行试验,轴心受拉试件的制作和试验都相对困难。由于水泥砂浆及混凝土的非均质性,试件即使能够几何对中,而往往也不能真正的物理对中,从而形成实际的偏心受力状态。At present, the tensile strength of cement mortar and concrete generally needs to be tested with axial tensile specimens, and the fabrication and testing of axial tensile specimens are relatively difficult. Due to the heterogeneity of cement mortar and concrete, even if the specimen can be geometrically centered, it often cannot be truly physically centered, thus forming an actual eccentric stress state.
另一方面,对于水泥砂浆及混凝土断裂韧度的测试,根据美国ASTME399规范,若满足线弹性断裂力学条件,测定无尺寸效应的材料断裂韧度,采用试样的厚度B,裂纹长度a0,韧带尺寸W-a0均需满足下式:On the other hand, for the test of the fracture toughness of cement mortar and concrete, according to the US ASTME399 standard, if the condition of linear elastic fracture mechanics is satisfied, the fracture toughness of the material without size effect is measured, and the thickness B of the sample is used, and the crack length a 0 , The ligament size Wa 0 must satisfy the following formula:
其中,KIC为材料断裂韧度,ft为材料的拉伸强度。则按ASTM的规定确定水泥砂浆的KIC,试件尺寸W一般要达到500mm左右;而混凝土试件尺寸W一般都要超过1000mm。若采用小尺寸试件,则断裂韧度等参数的尺寸效应又不可避免。如何由处于弹塑性断裂条件下的小尺寸试件,比如(W-a0)/dmax=3-10的试件,来确定无尺寸效应的材料真实断裂韧性,是亟待解决的科学难题。Among them, K IC is the fracture toughness of the material, f t is the tensile strength of the material. To determine the K IC of cement mortar according to ASTM regulations, the size W of the test piece should generally reach about 500mm; while the size W of the concrete test piece should generally exceed 1000mm. If small-sized specimens are used, the size effect of parameters such as fracture toughness is unavoidable. How to determine the true fracture toughness of materials without size effect from small-sized specimens under elastic-plastic fracture conditions, such as (Wa 0 )/d max =3-10, is a scientific problem to be solved urgently.
另外,进行断裂韧度测试试验中,对试件型式、加载夹具、加载方式、试验系统等都有严格限制。使得试验试样的制作成本和试验工作量大为增加。In addition, in the fracture toughness test, there are strict restrictions on the type of specimen, loading fixture, loading method, and test system. Make the production cost of the test sample and the test workload increase greatly.
还需强调的是,目前确定水泥砂浆和混凝土的拉伸强度和断裂韧度需采用不同型式的试样和加载方法。而用同一种测试方法和试样,同时确定水泥砂浆和混凝土拉伸强度及断裂韧度的方法还未见报道。It should also be emphasized that different types of specimens and loading methods are currently used to determine the tensile strength and fracture toughness of cement mortars and concrete. However, the method of simultaneously determining the tensile strength and fracture toughness of cement mortar and concrete with the same test method and sample has not been reported yet.
发明内容Contents of the invention
目前,测定水泥砂浆或混凝土的拉伸强度对试验试件和试验系统要求较高,另一方面测定无尺寸效应的断裂韧度需要的试件尺寸很大,普通实验室条件下较难完成试样的浇筑和测试工作。为克服现有测定方法的不足,本发明提出一种可同时确定水泥砂浆或混凝土的拉伸强度与断裂韧性的测试方法。At present, the determination of the tensile strength of cement mortar or concrete has high requirements for the test specimen and test system. On the other hand, the test specimen size required for the determination of the fracture toughness without size effect is very large, and it is difficult to complete the test under ordinary laboratory conditions. Such pouring and testing work. In order to overcome the deficiencies of the existing measuring methods, the present invention proposes a testing method that can simultaneously determine the tensile strength and fracture toughness of cement mortar or concrete.
为解决上述技术问题,本发明采用如下技术方案:In order to solve the problems of the technologies described above, the present invention adopts the following technical solutions:
设计一种同时测定水泥砂浆或混凝土拉伸强度与断裂韧度的方法,其特征在于,包括下列步骤:Design a kind of method of measuring cement mortar or concrete tensile strength and fracture toughness simultaneously, it is characterized in that, comprises the following steps:
(1)按给定的水泥砂浆或混凝土配合比要求,浇筑一定数量的尺寸均为W×B×L的水泥砂浆或混凝土三点弯曲梁试件,其中,W为试件高度,B为试件厚度,L为试件有效长度,试件满足L/W=2.5、4或8;(1) According to the given cement mortar or concrete mix ratio requirements, pour a certain number of cement mortar or concrete three-point bending beam specimens with dimensions of W×B×L, where W is the height of the specimen and B is the specimen height. The thickness of the test piece, L is the effective length of the test piece, and the test piece satisfies L/W=2.5, 4 or 8;
(2)将步骤(1)所得试件养护到规定要求的龄期后,采用切割机对所述试件分别切出裂缝,裂缝长度为a0,试件的缝高比α=a0/W在0.02~0.75的范围内取值;从中选取5~8种不同缝高比的试件,每种缝高比对应3~5个试件;其中试件的缝高比α的取值方式为:在0.02~0.75的范围内随机取值、均态分布;(2) After curing the test piece obtained in step (1) to the specified age, use a cutting machine to cut out cracks on the test piece respectively. The length of the crack is a 0 , and the slit height ratio of the test piece is α=a 0 / W ranges from 0.02 to 0.75; select 5 to 8 specimens with different seam height ratios, and each seam height ratio corresponds to 3 to 5 specimens; the value method of the seam height ratio α of the specimens It is: random value within the range of 0.02 to 0.75, uniform distribution;
(3)在普通的压力试验机或者万能试验机上,按静力加载试验方法加载并至试件被破坏,试验过程中记录每个试件的峰值荷载Pmax;(3) On a common pressure testing machine or a universal testing machine, load according to the static loading test method and until the specimen is destroyed, record the peak load P max of each specimen during the test;
(4)基于步骤(3)所得的每个试件的峰值荷载,计算出每个试件的名义强度σn;(4) Calculate the nominal strength σ n of each specimen based on the peak load of each specimen obtained in step (3);
(5)计算出每个试件的等效裂缝长度ae;(5) Calculate the equivalent crack length a e of each specimen;
(6)将步骤(4)、步骤(5)所得的不同的σn与ae值,带入下式(1)进行回归分析,(6) Bring the different σ n and a e values obtained in step (4) and step (5) into the following formula (1) for regression analysis,
即可同时得出水泥砂浆或混凝土的拉伸强度ft与断裂韧度KIC。The tensile strength f t and fracture toughness K IC of cement mortar or concrete can be obtained simultaneously.
优选的,在所述步骤(2)中,所述缝高比α=a0/W的取值依次为0.05,0.1,0.15,0.2,0.25,0.3,0.35,0.4,0.45,0.5,0.55,0.6,0.65,0.7。Preferably, in the step (2), the values of the slit height ratio α=a 0 /W are 0.05, 0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 0.4, 0.45, 0.5, 0.55, 0.6, 0.65, 0.7.
优选的,在所述步骤(4)中,由下式(2)计算出每个试件的名义强度σn,Preferably, in said step (4), the nominal strength σ n of each test piece is calculated by the following formula (2),
式(2)中,W1=W-a0;W2=W1-(γ·dmax);W3=W1+(γ·dmax);Pmax为各试件实测峰值荷载;B为试件厚度;a0为初始裂缝长度,γ·dmax为峰值荷载Pmax对应的初始裂缝扩展量;dmax为骨料最大粒径。In formula (2), W 1 =Wa 0 ; W 2 =W 1 -(γ·d max ); W 3 =W 1 +(γ·d max ); P max is the measured peak load of each specimen; B is The thickness of the specimen; a 0 is the initial crack length, γ·d max is the initial crack expansion corresponding to the peak load P max ; d max is the maximum particle size of the aggregate.
优选的,在所述式(2)中,当γ=1.0时,从统计角度计算得出水泥砂浆或混凝土强度与韧度的平均值;Preferably, in said formula (2), when γ=1.0, the average value of cement mortar or concrete strength and toughness is calculated from a statistical point of view;
当γ=0.2,0.4,0.6,0.8…2.0时,从水泥砂浆或混凝土的材料性能的离散性角度,计算得出水泥砂浆或混凝土强度与韧度的精确值。When γ = 0.2, 0.4, 0.6, 0.8...2.0, from the discrete point of view of the material properties of cement mortar or concrete, the exact value of the strength and toughness of cement mortar or concrete can be calculated.
优选的,在所述步骤(5)中,每个试件的等效裂缝长度ae由下式(3a)~(3d)计算:Preferably, in the step (5), the equivalent crack length a e of each test piece is calculated by the following formulas (3a) to (3d):
当L/W=2.5时, When L/W=2.5,
当L/W=4时, When L/W=4,
当L/W=8时,Y(α)=1.106-1.552α+7.71α2-13.53α3+14.23α4 When L/W=8, Y(α)=1.106-1.552α+7.71α 2 -13.53α 3 +14.23α 4
——式(3d)。- Formula (3d).
本发明的有益效果在于:The beneficial effects of the present invention are:
1.本发明的方法中采用小尺寸试样,试件尺寸不需满足现有规范测定线弹性断裂条件下,试件的高度、厚度、初始裂缝长度等都要大于一定数值的严格要求。1. In the method of the present invention, small-sized samples are adopted, and the size of the test piece does not need to meet the strict requirement that the height, thickness, initial crack length, etc. of the test piece should be greater than a certain value under the condition of linear elastic fracture of existing norms.
2.本发明采用一种试样和测试方法,即可同时确定水泥砂浆或混凝土的拉伸强度及断裂韧度。克服了目前确定水泥砂浆和混凝土的拉伸强度和断裂韧度需采用不同型式的试样和加载方法。2. The present invention adopts a sample and a test method to simultaneously determine the tensile strength and fracture toughness of cement mortar or concrete. It overcomes the need to use different types of samples and loading methods to determine the tensile strength and fracture toughness of cement mortar and concrete.
3.对于断裂韧性试验的初始裂纹的形成,只需采用切割机切缝,而不需进行试件浇筑时的预制裂缝。大大降低了加工试样裂缝的难度,减轻了相应的工作量。3. For the formation of the initial cracks in the fracture toughness test, it is only necessary to use a cutting machine to cut the cracks, instead of prefabricating the cracks when pouring the specimen. The difficulty of processing the cracks of the sample is greatly reduced, and the corresponding workload is reduced.
4.通过参数γ·dmax来考虑峰值荷载对应的裂缝扩展量,从而可准确估计整体试验结果的离散性,计算单个试件的拉伸强度和断裂韧度,即,从统计角度取值γ=1.0,可计算整体试验的平均值;精确求解每个试件的拉伸强度和断裂韧度时,则γ可取值0.2、0.4、0.6、0.8…2.0。4. Considering the amount of crack growth corresponding to the peak load through the parameter γ·d max , so that the dispersion of the overall test results can be accurately estimated, and the tensile strength and fracture toughness of a single specimen can be calculated, that is, the value γ is taken from a statistical point of view =1.0, the average value of the overall test can be calculated; when the tensile strength and fracture toughness of each specimen are accurately calculated, then γ can take values of 0.2, 0.4, 0.6, 0.8...2.0.
5.本发明的方法形式简单,试验易操作,具有足够精度,且易判断结果合理性。该方法只需由小尺寸试件的峰值荷载,即同时确定水泥砂浆的拉伸强度ft和断裂韧度KIC。不需要满足现行国内外规范对试验试样尺寸、型式,加载条件等的严格规定。5. The method of the present invention is simple in form, easy to operate in the test, has sufficient precision, and is easy to judge the rationality of the result. This method only needs to determine the tensile strength f t and fracture toughness K IC of cement mortar simultaneously from the peak load of a small-sized specimen. It is not necessary to meet the strict regulations on the size, type and loading conditions of test specimens in current domestic and foreign standards.
附图说明Description of drawings
图1为本发明试件形状及峰值荷载时的应力分布图;Fig. 1 is the stress distribution figure when the specimen shape of the present invention and peak load;
图2为λ·dmax的取值示意图:其中,图2a表示统计意义下γ=1.0时的取值;图2b表示γ=0.2,0.4,0.6,0.8…2.0时的精确取值;Figure 2 is a schematic diagram of the value of λ·d max : among them, Figure 2a shows the value when γ=1.0 in a statistical sense; Figure 2b shows the exact value when γ=0.2, 0.4, 0.6, 0.8...2.0;
图3为由实施例1试验数据回归确定的水泥砂浆拉伸强度与断裂韧度示意图:其中,图3a为第一系列试件L×B×W=400×50×100mm的拉伸强度与断裂韧度;图3b为第二系列试件L×B×W=240×50×60mm的拉伸强度与断裂韧度;Fig. 3 is the schematic diagram of tensile strength and fracture toughness of cement mortar determined by the regression of test data in Example 1: wherein, Fig. 3a is the tensile strength and fracture toughness of the first series of test pieces L * B * W = 400 * 50 * 100mm Toughness; Figure 3b is the tensile strength and fracture toughness of the second series of test pieces L×B×W=240×50×60mm;
图4为由实施例2试验数据回归确定的混凝土拉伸强度与断裂韧度示意图。Fig. 4 is a schematic diagram of concrete tensile strength and fracture toughness determined by regression of test data in Example 2.
具体实施方式detailed description
下面结合附图和实施例来说明本发明的具体实施方式,但以下实施例只是用来详细说明本发明,并不以任何方式限制本发明的范围。以下实施例中所涉及的一些步骤或方法,如无特殊说明,均为本领域的常规方法,所涉及的材料、仪器设备,如无特别说明,均为常规材料和仪器设备。The specific implementation of the present invention will be described below in conjunction with the accompanying drawings and examples, but the following examples are only used to describe the present invention in detail, and do not limit the scope of the present invention in any way. Some of the steps or methods involved in the following examples are conventional methods in the art unless otherwise specified, and the materials and equipment involved are conventional materials and equipment unless otherwise specified.
首先,按给定的水泥砂浆配合比要求,浇筑尺寸均为W×B×L的水泥砂浆或混凝土三点弯曲梁试件15~40个,其中,W为试件高度,B为试件厚度,L为试件有效长度,试件满足L/W=2.5、4或8。在满足试验机测试精度要求和其他配套测量仪器精度要求的前提下,试件的尺寸W可足够小。First, according to the given cement mortar mix ratio, 15 to 40 concrete mortar or concrete three-point bending beam specimens with dimensions of W×B×L are poured, where W is the height of the specimen and B is the thickness of the specimen , L is the effective length of the specimen, and the specimen satisfies L/W=2.5, 4 or 8. Under the premise of meeting the test accuracy requirements of the testing machine and other supporting measuring instruments, the size W of the test piece can be small enough.
将上述试件养护到规定要求的龄期后,采用切割机对所述试件分别切出裂缝,裂缝长度为a0,试件的缝高比α=a0/W在0.02~0.75的范围内取值;从中选取5~8种不同缝高比的试件,每种缝高比对应3~5个试件。After curing the above test pieces to the required age, use a cutting machine to cut out cracks on the test pieces respectively. The length of the cracks is a 0 , and the slit height ratio of the test pieces α=a 0 /W is in the range of 0.02 to 0.75 Inner value; select 5 to 8 specimens with different seam height ratios, and each seam height ratio corresponds to 3 to 5 specimens.
实施例1:本实施例1试验所用水泥砂浆骨料最大粒径dmax=2.5mm。分两个系列:第一系列L×B×W=400×50×100mm,初始缝高比α=a0/W=0.03~0.72,W/dmax=40。第二系列L×B×W=240×50×60mm,初始缝高比α=a0/W=0.04~0.68,W/dmax=24。Example 1: The maximum particle size of the cement mortar aggregate used in the test of Example 1 is d max =2.5mm. There are two series: the first series L×B×W=400×50×100mm, initial seam height ratio α=a 0 /W=0.03~0.72, W/d max =40. The second series L×B×W=240×50×60mm, initial seam height ratio α=a 0 /W=0.04~0.68, W/d max =24.
参见图1,考虑水泥砂浆为准脆性材料,峰值荷载时的应力分布如图1所示,对应的裂缝扩展量Δafic=γ·dmax。基于下式(2),由实测的峰值荷载Pmax,可计算出每个试件的名义强度σn:Referring to Fig. 1, considering that cement mortar is a quasi-brittle material, the stress distribution at peak load is shown in Fig. 1, and the corresponding crack extension Δa fic = γ·d max . Based on the following formula (2), the nominal strength σ n of each specimen can be calculated from the measured peak load P max :
从统计角度考虑γ=1.0,可计算得出水泥砂浆拉伸强度与断裂韧度的平均值,参见图2a。考虑水泥砂浆试验结果的离散性问题而精确求解,可使γ=0.2,0.4,0.6,0.8…2.0,计算得出水泥砂浆拉伸强度与断裂韧度的精确值,参见图2b。Considering γ=1.0 from a statistical point of view, the average value of tensile strength and fracture toughness of cement mortar can be calculated, see Figure 2a. Considering the discretization of the cement mortar test results and solving it accurately, γ=0.2, 0.4, 0.6, 0.8...2.0 can be used to calculate the exact values of the tensile strength and fracture toughness of the cement mortar, see Figure 2b.
每个试件的等效裂缝长度ae可由下式(3a)~(3d)计算:The equivalent crack length a e of each specimen can be calculated by the following formulas (3a)~(3d):
当L/W=2.5时, When L/W=2.5,
当L/W=4时, When L/W=4,
当L/W=8时,Y(α)=1.106-1.552α+7.71α2-13.53α3+14.23α4 When L/W=8, Y(α)=1.106-1.552α+7.71α 2 -13.53α 3 +14.23α 4
——式(3d)。- Formula (3d).
得到不同的σn与ae后,带入式(1)进行回归分析,即可同时得出水泥砂浆的拉伸强度ft与断裂韧度KIC:After obtaining different σ n and a e , put them into the formula (1) for regression analysis, and the tensile strength f t and fracture toughness K IC of cement mortar can be obtained at the same time:
实施例1试验数据回归确定的水泥砂浆拉伸强度与断裂韧度结果参见图3,其中,图3a为第一系列试件L×B×W=400×50×100mm的拉伸强度与断裂韧度;图3b为第二系列试件L×B×W=240×50×60mm的拉伸强度与断裂韧度。确定的水泥砂浆拉伸强度与断裂韧度数据见下表1。The results of the cement mortar tensile strength and fracture toughness determined by the regression of test data in Example 1 are shown in Fig. 3, wherein Fig. 3a shows the tensile strength and fracture toughness of the first series of test pieces L×B×W=400×50×100mm Degree; Figure 3b shows the tensile strength and fracture toughness of the second series of test pieces L×B×W=240×50×60mm. The determined tensile strength and fracture toughness data of cement mortar are shown in Table 1 below.
表1实施例1确定的水泥砂浆拉伸强度与断裂韧度The cement mortar tensile strength and fracture toughness that table 1 embodiment 1 determines
可见,从统计意义上看,当γ=1.0时,水泥砂浆拉伸强度与断裂韧度都可与试验实测平均值相吻合。若考虑试验中每个不同试件的离散性,则γ=0.6,0.8…2.0。It can be seen that, from a statistical point of view, when γ=1.0, the tensile strength and fracture toughness of cement mortar can be consistent with the average value measured in the test. If the discreteness of each different specimen in the test is considered, then γ=0.6,0.8...2.0.
实施例2:本实施例2试验所用混凝土骨料最大粒径dmax=19mm。试件尺寸L×B×W=1143×76.2×305mm,初始缝高比α=a0/W=0.36~0.68,W/dmax=16。Example 2: The maximum particle size of the concrete aggregate used in the test of Example 2 is d max =19mm. The size of the test piece is L×B×W=1143×76.2×305mm, the initial seam height ratio α=a 0 /W=0.36~0.68, W/d max =16.
试件峰值荷载时的应力分布如图1所示,对应的裂缝扩展量Δafic=γ·dmax。基于式(2),由实测的峰值荷载,可计算出每个试件的名义强度σn。The stress distribution of the specimen at the peak load is shown in Fig. 1, and the corresponding crack growth Δa fic = γ·d max . Based on formula (2), the nominal strength σ n of each specimen can be calculated from the measured peak load.
从统计角度考虑γ=1.0,可计算得出混凝土拉伸强度与断裂韧度的平均值。考虑混凝土试验结果的离散性问题而精确求解,则使γ=0.6,0.8…2.0,计算得出混凝土拉伸强度与断裂韧度的精确值。Considering γ=1.0 from a statistical point of view, the average value of concrete tensile strength and fracture toughness can be calculated. Considering the discretization of the concrete test results and solving it accurately, then make γ=0.6,0.8...2.0, and calculate the exact values of concrete tensile strength and fracture toughness.
每个试件的等效裂缝长度ae可由式(3a)~(3d)计算。The equivalent crack length a e of each specimen can be calculated by formulas (3a)~(3d).
得到不同的σn与ae后,带入式(1)进行回归分析,即可同时得出混凝土的拉伸强度ft与断裂韧度KIC。实施例(2)中所指的式(1)、式(2)、式(3a)~(3d)与实施例1中相同。After obtaining different σ n and a e , put them into formula (1) for regression analysis, and then obtain the tensile strength f t and fracture toughness K IC of concrete at the same time. The formulas (1), formula (2), and formulas (3a) to (3d) referred to in the embodiment (2) are the same as those in the embodiment 1.
参见图4,为实施例2试验数据回归确定的混凝土拉伸强度与断裂韧度结果。确定的混凝土拉伸强度与断裂韧度具体数据见下表2。Referring to Fig. 4, it is the results of concrete tensile strength and fracture toughness determined by regression of test data in Example 2. The specific data of the determined concrete tensile strength and fracture toughness are shown in Table 2 below.
表2实施例2确定的混凝土拉伸强度与断裂韧度The concrete tensile strength and fracture toughness that table 2 embodiment 2 determines
对所公开实施例的上述说明,使本领域技术人员能够实现或使用本发明。对这些实施例的多处修改对本领域技术人员来说是显而易见的,本文中所定义的一般原理可以在不脱离发明的精神或范围的前提下,在其它实施例中实现。因此,本发明将不限制于本文所显示的这些实施例,而是要符合与本文公开原理和新颖特点相一致的最宽范围。The above description of the disclosed embodiments is provided to enable any person skilled in the art to make or use the invention. Many modifications to these embodiments will be readily apparent to those skilled in the art, and the general principles defined herein may be implemented in other embodiments without departing from the spirit or scope of the invention. Therefore, the present invention is not to be limited to the embodiments shown herein, but is to be accorded the widest scope consistent with the principles and novel features disclosed herein.
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