CN105865916A - Method for determining peak load of mortar or concrete members with cracks - Google Patents
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Abstract
本发明属于土木与水利工程领域,特别是涉及一种确定带裂缝砂浆或混凝土构件峰值荷载的方法,该方法通过确定的混凝土材料参数—拉伸强度ft与断裂韧度KIC,可个性化确定带裂缝的不同砂浆或混凝土构件或结构型式的峰值荷载Pmax。本发明方法形式简单,具有足够精度。
The invention belongs to the field of civil engineering and water conservancy engineering, and in particular relates to a method for determining the peak load of cracked mortar or concrete components. The method can be personalized through the determined concrete material parameters—tensile strength f t and fracture toughness K IC Determination of the peak load P max for different mortar or concrete elements or structural types with cracks. The method of the invention is simple in form and has sufficient precision.
Description
技术领域 technical field
本发明属于土木与水利工程领域,特别是涉及一种确定带裂缝砂浆或混凝土构件峰值荷载的方法。 The invention belongs to the field of civil engineering and water conservancy engineering, in particular to a method for determining the peak load of cracked mortar or concrete components.
背景技术 Background technique
混凝土构件或结构的主要组成为骨料、砂与水泥浆等,因此,混凝土从本质上不是均质材料,而是典型的非均质准脆性材料。混凝土骨料分布的随机性、混凝土浇筑条件与养护条件等,都会对混凝土构件或结构的试验结果产生很大影响。因此,采用混凝土制作的构件或结构,其试验结果的离散性是其固有属性。 The main components of concrete components or structures are aggregate, sand and cement slurry, etc. Therefore, concrete is not a homogeneous material in essence, but a typical heterogeneous quasi-brittle material. The randomness of concrete aggregate distribution, concrete pouring conditions and curing conditions will have a great influence on the test results of concrete components or structures. Therefore, for components or structures made of concrete, the discreteness of test results is an inherent property.
对于混凝土构件来说,其骨料粒径与试件尺寸的比值一般在5~20倍。由于混凝土的骨料粒径与构件或结构的比值相对较小,骨料分布对其性能结果影响很大。比如同样条件下浇筑8个混凝土试件,试验结果可能得到8个不同的峰值荷载。目前的研究只能定性说明8个试件结果的差异性或离散性,而定量确定混凝土试验结果的离散性是亟待解决的技术问题。 For concrete members, the ratio of aggregate particle size to specimen size is generally 5 to 20 times. Since the ratio of the aggregate particle size of concrete to the component or structure is relatively small, the aggregate distribution has a great influence on its performance results. For example, pouring 8 concrete specimens under the same conditions, the test results may obtain 8 different peak loads. The current research can only qualitatively explain the difference or dispersion of the results of the eight specimens, but quantitative determination of the dispersion of the concrete test results is a technical problem that needs to be solved urgently.
发明内容 Contents of the invention
针对上述问题,本发明提出一种可个性化确定带裂缝砂浆或混凝土构件峰值荷载的方法,该方法通过确定的混凝土材料参数(拉伸强度ft、断裂韧度KIC)来个性化确定带裂缝的不同砂浆或混凝土构件或结构型式的峰值荷载Pmax,本发明方法形式简单,具有足够精度。 In view of the above problems, the present invention proposes a method for individually determining the peak load of cracked mortar or concrete components. The method uses the determined concrete material parameters (tensile strength f t , fracture toughness K IC ) to individually determine the peak load of cracked mortar or concrete components. The peak load P max of different mortar or concrete elements or structural types of cracks, the method of the invention is simple in form and has sufficient accuracy.
为解决以上技术问题,本发明通过以下技术方案实现: In order to solve the above technical problems, the present invention is realized through the following technical solutions:
设计一种确定带裂缝砂浆或混凝土构件峰值荷载的方法,包括以下步骤: Devise a method for determining the peak load of a cracked mortar or concrete member comprising the following steps:
一、根据所用试件类型,确定不同带裂缝构件的等效裂缝长度ae;ae由式(1)计算: 1. Determine the equivalent crack length a e of different cracked components according to the type of test piece used; a e is calculated by formula (1):
(1) (1)
根据应力强度因子手册,确定不同构件的几何影响参数Y(α)。 According to the stress intensity factor manual, determine the geometric influence parameter Y(α) of different components.
若采用三点弯曲梁试件,则 If a three-point bending beam specimen is used, then
(2a) (2a)
当L/W=2.5时, When L/W=2.5,
(2b) (2b)
当L/W=4时, When L/W=4,
(2c) (2c)
当L/W=8时, When L/W=8,
(2d) (2d)
若采用紧凑拉伸或楔入劈拉试件,则 If compact tension or wedge split tension specimens are used, then
(3a) (3a)
(3b)。 (3b).
二、将已知确定的试件砂浆或混凝土的材料参数—拉伸强度ft和断裂韧度KIC代入弹塑性理论公式(4) ,即可计算出σn(Pmax)值。 2. Substituting the known and determined material parameters of the specimen mortar or concrete—tensile strength f t and fracture toughness K IC —into the elastoplastic theory formula (4), the value of σ n (P max ) can be calculated.
(4) (4)
三、对于三点弯曲梁试件,基于求得的σn(Pmax),可由弹塑性理论公式(5),通过β取不同的值,解出试件的个性化的峰值荷载Pmax: 3. For three-point bending beam specimens, based on the obtained σ n (P max ), the elastic-plastic theoretical formula (5) can be used to obtain different values of β to obtain the personalized peak load P max of the specimen:
(5); (5);
对于紧凑拉伸或楔入劈拉试件,基于求得的σn(Pmax),可由弹塑性理论公式(6),通过β取不同的值,从而解出试件的个性化的峰值荷载Pmax: For compact tension or wedge split tension specimens, based on the obtained σ n (P max ), the elastic-plastic theory formula (6) can be used to obtain different values of β, so as to solve the personalized peak load of the specimen P max :
(6); (6);
以上各式中,W1=W -a0;W2=W1- (β·dmax);W3=W1 + (β·dmax);W为试件高度,L为试件长度,B为试件厚度;Pmax为峰值荷载; a0为初始裂缝长度,β·dmax为峰值荷载Pmax对应的初始裂缝尖端的裂缝扩展量;dmax为砂浆或混凝土骨料最大粒径。α为缝高比,即初始裂缝长度a0与试件高度W的比值;A(α)为试件形状影响参数,对三点弯曲梁和楔入劈拉试件取不同值;σn为考虑裂缝影响的名义应力;β为考虑试件结果离散性的调整系数,β不是定值,为精确个性化求解每个试件峰值荷载,β可在0.1~2.0之间离散取值,如0.1、0.2、0.3、0.4、0.5、0.6、0.7、0.8…2.0;Y(α)为几何影响参数。 In the above formulas, W 1 =W -a 0 ; W 2 =W 1 - (β·d max ); W 3 =W 1 + (β·d max ); W is the height of the specimen, and L is the length of the specimen , B is the thickness of the specimen; P max is the peak load; a 0 is the initial crack length, β·d max is the crack propagation at the tip of the initial crack corresponding to the peak load P max ; d max is the maximum particle size of mortar or concrete aggregate . α is the fracture height ratio, that is, the ratio of the initial crack length a 0 to the specimen height W; A(α) is a parameter affecting the specimen shape, and different values are taken for three-point bending beams and wedge split tension specimens; σn is Consider the nominal stress affected by cracks; β is an adjustment coefficient considering the discreteness of the test piece results, β is not a fixed value, in order to accurately and individually solve the peak load of each test piece, β can be discretely taken between 0.1 and 2.0, such as 0.1 , 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8…2.0; Y(α) is the geometric influence parameter.
本发明具有以下积极有益的技术效果: The present invention has the following positive and beneficial technical effects:
(1)本发明通过混凝土材料参数—拉伸强度与断裂韧度,能够个性化确定不同试件结构的峰值荷载。 (1) The present invention can individually determine the peak load of different specimen structures through concrete material parameters—tensile strength and fracture toughness.
(2)通过参数β·dmax来考虑峰值荷载对应的裂缝扩展量。从统计角度β=1.0,可预测整体构件峰值荷载的平均值;而当β取0.1、0.2、0.3、0.4、0.5、0.6、0.7、0.8…2.0,又可精确个性化求解每个试件峰值荷载。 (2) The crack expansion corresponding to the peak load is considered by the parameter β·d max . From a statistical point of view β=1.0, the average value of the peak load of the overall member can be predicted; and when β is 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8...2.0, the peak value of each specimen can be accurately and personalized load.
附图说明 Description of drawings
图1为外荷载P-位移δ曲线示意图(相同构件对应不同的峰值荷载Pmax); Figure 1 is a schematic diagram of the external load P-displacement δ curve (the same component corresponds to different peak load P max );
图2为平均意义上的峰值荷载时裂缝扩展跨越骨料示意图(考虑试验结果平均值时β=1.0); Figure 2 is a schematic diagram of the crack propagation across the aggregate at the peak load in the average sense (β=1.0 when considering the average value of the test results);
图3为实际情况个性化的峰值荷载时裂缝扩展跨越骨料示意图(个性化确定峰值荷载时β=0.1、0.2、0.3、0.4、0.5、0.6、0.7、0.8…2.0)。 Figure 3 is a schematic diagram of crack propagation across the aggregate under individualized peak loads (individually determined peak loads β=0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8...2.0).
具体实施方式 detailed description
下面结合附图,对本发明实施例中的技术方案进行清楚完整的描述。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,所涉及的材料如无特别说明,均为市售材料。 The technical solutions in the embodiments of the present invention will be clearly and completely described below in conjunction with the accompanying drawings. Some of the steps or methods involved in the following examples are conventional methods in the art unless otherwise specified, and the materials involved are all commercially available materials unless otherwise specified.
实施例1 确定带裂缝砂浆或混凝土构件峰值荷载的方法,包括以下步骤: Embodiment 1 The method for determining the peak load of a cracked mortar or concrete member includes the following steps:
浇筑相同尺寸、相同缝高比的混凝土三点弯曲梁构件4个,其试件尺寸L×B×W=320 × 40 × 80mm,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=4。其初始裂缝长度相同a0 =24mm,即a0/W=0.3都一致。其骨料最大粒径dmax=10 mm。W/dmax = 8。已知该混凝土材料特性—拉伸强度ft=4.7MPa,断裂韧度KIC=1.3 MPa·m1/2。 Four concrete three-point bending beam members with the same size and the same joint height ratio are poured, and the size of the specimen is L×B×W=320×40×80mm, where W is the height of the specimen, B is the thickness of the specimen, and L is the thickness of the specimen. The effective length of the test piece satisfies L/W=4. The initial crack length is the same a 0 =24mm, that is, a 0 /W=0.3 are consistent. The maximum particle size of the aggregate is d max =10 mm. W/d max = 8. The material properties of the concrete are known—tensile strength f t =4.7MPa, fracture toughness K IC =1.3 MPa·m 1/2 .
根据所用试件类型,确定不同带裂缝构件的等效裂缝长度ae。ae可由式 (1) 计算。根据应力强度因子手册,可以确定不同构件或结构型式的几何影响参数Y(α),三点弯曲梁可由式 (2) 计算。 According to the type of specimen used, determine the equivalent crack length a e of different cracked components. a e can be calculated by formula (1). According to the stress intensity factor manual, the geometric influence parameter Y(α) of different components or structural types can be determined, and the three-point bending beam can be calculated by formula (2).
将已知确定的该试件混凝土的材料参数—拉伸强度ft和断裂韧度KIC代入弹式(4),推出σn(Pmax)值。 Substitute the known and determined material parameters of the specimen concrete—tensile strength f t and fracture toughness K IC into elastic formula (4) to derive the value of σ n (P max ).
由式(5),通过β取不同的值,从而反解出不同试件的个性化的峰值荷载Pmax。参见图1,由于混凝土类准脆性材料自身特性决定,即使浇筑相同尺寸相同型式的构件,每个构件的峰值荷载也不尽相同。 According to formula (5), different values of β are used to inversely solve the personalized peak load P max of different specimens. Referring to Figure 1, due to the inherent characteristics of concrete-like quasi-brittle materials, the peak load of each component is different even if components of the same size and type are poured.
参见图2,从统计角度考虑,β=1.0可预测整体构件峰值荷载的平均值。 Referring to Figure 2, from a statistical point of view, β = 1.0 predicts the average value of the peak load of the overall member.
参见图3,考虑到试验结果的离散性,对于每个试件峰值荷载试验值的预测,则β=0.1、0.2、0.3、0.4、0.5、0.6、0.7、0.8…2.0,β值不是定值,取值范围大小与材料性能离散性相关。 See Figure 3, considering the discreteness of the test results, for the prediction of the peak load test value of each specimen, β=0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8...2.0, the β value is not a fixed value , the value range is related to the discreteness of material properties.
参见表1,对于本实施例的4个相同尺寸相同缝高比的试件,其实测峰值荷载都不相同。采用本发明方法,4个试件的β分别取不同值0.9、1.1、0.8、0.7,预测值就会分别与4个峰值荷载试验值很好吻合。β=1.0能够预测整体4个构件的平均值。 Referring to Table 1, for the four specimens of the same size and the same slit-height ratio in this embodiment, the measured peak loads are different. By adopting the method of the present invention, the β of the four test pieces take different values of 0.9, 1.1, 0.8, and 0.7 respectively, and the predicted values are in good agreement with the test values of the four peak loads respectively. β=1.0 can predict the mean of the overall 4 components.
表1 个性化确定四个带裂缝构件的峰值荷载 Table 1 Individually determined peak loads of four members with cracks
。 .
实施例2确定带裂缝砂浆或混凝土构件峰值荷载的方法,包括以下步骤: Embodiment 2 The method for determining the peak load of a cracked mortar or a concrete component comprises the following steps:
浇筑相同尺寸、相同缝高比的混凝土三点弯曲梁构件4个,其试件尺寸L×B×W=320 × 40 × 80mm,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=4。其初始裂缝长度相同a0 =32mm,即a0/W=0.4都一致。其骨料最大粒径dmax=10 mm。W/dmax = 8。已知该混凝土材料特性—拉伸强度ft=4.7MPa,断裂韧度KIC=1.3 MPa·m1/2。 Four concrete three-point bending beam members with the same size and the same joint height ratio are poured, and the size of the specimen is L×B×W=320×40×80mm, where W is the height of the specimen, B is the thickness of the specimen, and L is the thickness of the specimen. The effective length of the test piece satisfies L/W=4. The initial crack length is the same a 0 =32mm, that is a 0 /W=0.4 are consistent. The maximum particle size of the aggregate is d max =10 mm. W/d max = 8. The material properties of the concrete are known—tensile strength f t =4.7MPa, fracture toughness K IC =1.3 MPa·m 1/2 .
根据所用试件类型,确定不同带裂缝构件的等效裂缝长度ae。ae可由式 (1) 计算。根据应力强度因子手册,可以确定不同构件或结构型式的几何影响参数Y(α),三点弯曲梁可由式 (2) 计算。 According to the type of specimen used, determine the equivalent crack length a e of different cracked components. a e can be calculated by formula (1). According to the stress intensity factor manual, the geometric influence parameter Y(α) of different components or structural types can be determined, and the three-point bending beam can be calculated by formula (2).
将已知确定的该试件混凝土的材料参数—拉伸强度ft和断裂韧度KIC代入弹式(4),推出σn(Pmax)值。 Substitute the known and determined material parameters of the specimen concrete—tensile strength f t and fracture toughness K IC into elastic formula (4) to derive the value of σ n (P max ).
由式(5),通过β取不同的值,从而反解出不同试件的个性化的峰值荷载Pmax。参见图1,由于混凝土类准脆性材料自身特性决定,即使浇筑相同尺寸相同型式的构件,每个构件的峰值荷载也不尽相同。 According to formula (5), different values of β are used to inversely solve the personalized peak load P max of different specimens. Referring to Figure 1, due to the inherent characteristics of concrete-like quasi-brittle materials, the peak load of each component is different even if components of the same size and type are poured.
参见图2,从统计角度考虑,β=1.0可预测整体构件峰值荷载的平均值。 Referring to Figure 2, from a statistical point of view, β = 1.0 predicts the average value of the peak load of the overall member.
参见图3,考虑到试验结果的离散性,对于每个试件峰值荷载试验值的预测,则β=0.1、0.2、0.3、0.4、0.5、0.6、0.7、0.8…2.0,β值不是定值,取值范围大小与材料性能离散性相关。 Referring to Figure 3, considering the discreteness of the test results, for the prediction of the peak load test value of each specimen, β=0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8...2.0, the value of β is not a fixed value , the value range is related to the discreteness of material properties.
参见表2,对于本实施例的4个相同尺寸相同缝高比的试件,其实测峰值荷载都不相同。采用本发明方法,4个试件的β分别取不同值1.6、0.9、1.1、1.3,预测值就会分别与4个峰值荷载试验值很好吻合。β=1.0能够预测整体4个构件的平均值。 Referring to Table 2, for the four specimens of the same size and the same slit-height ratio in this embodiment, the measured peak loads are different. By adopting the method of the present invention, the β of the four test pieces take different values of 1.6, 0.9, 1.1, and 1.3 respectively, and the predicted values are in good agreement with the test values of the four peak loads respectively. β=1.0 can predict the mean of the overall 4 components.
表2 个性化确定四个带裂缝构件的峰值荷载 Table 2 Individually determined peak loads of four members with cracks
。 .
实施例3确定带裂缝砂浆或混凝土构件峰值荷载的方法,包括以下步骤: Embodiment 3 The method for determining the peak load of a cracked mortar or concrete member comprises the following steps:
浇筑相同尺寸、相同缝高比的混凝土三点弯曲梁构件4个,其试件尺寸L×B×W =400 × 100 × 100 mm,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=4。其初始裂缝长度相同a0 =40mm,即a0/W=0.4都一致。其骨料最大粒径dmax=20 mm。W/dmax = 5。已知该混凝土材料特性—拉伸强度ft=4.7MPa,断裂韧度KIC=2.2 MPa·m1/2。 Four concrete three-point bending beam members with the same size and the same joint height ratio were poured, and the specimen size L×B×W =400×100×100 mm, where W is the height of the specimen, B is the thickness of the specimen, and L is The effective length of the test piece, the test piece satisfies L/W=4. The initial crack length is the same a 0 =40mm, that is a 0 /W=0.4 are consistent. The maximum particle size of the aggregate is d max =20 mm. W/d max = 5. The material properties of the concrete are known—tensile strength f t =4.7MPa, fracture toughness K IC =2.2 MPa·m 1/2 .
根据所用试件类型,确定不同带裂缝构件的等效裂缝长度ae。ae可由式 (1) 计算。根据应力强度因子手册,可以确定不同构件或结构型式的几何影响参数Y(α),三点弯曲梁可由式 (2) 计算。 According to the type of specimen used, determine the equivalent crack length a e of different cracked components. a e can be calculated by formula (1). According to the stress intensity factor manual, the geometric influence parameter Y(α) of different components or structural types can be determined, and the three-point bending beam can be calculated by formula (2).
将已知确定的该试件混凝土的材料参数—拉伸强度ft和断裂韧度KIC代入弹式(4),推出σn(Pmax)值。 Substitute the known and determined material parameters of the specimen concrete—tensile strength f t and fracture toughness K IC into elastic formula (4) to derive the value of σ n (P max ).
由式(5),通过β取不同的值,从而反解出不同试件的个性化的峰值荷载Pmax。参见图1,由于混凝土类准脆性材料自身特性决定,即使浇筑相同尺寸相同型式的构件,每个构件的峰值荷载也不尽相同。 According to formula (5), different values of β are used to inversely solve the personalized peak load P max of different specimens. Referring to Figure 1, due to the inherent characteristics of concrete-like quasi-brittle materials, the peak load of each component is different even if components of the same size and type are poured.
参见图2,从统计角度考虑,β=1.0可预测整体构件峰值荷载的平均值。 Referring to Figure 2, from a statistical point of view, β = 1.0 predicts the average value of the peak load of the overall member.
参见图3,考虑到试验结果的离散性,对于每个试件峰值荷载试验值的预测,则β=0.1、0.2、0.3、0.4、0.5、0.6、0.7、0.8…2.0,β值不是定值,取值范围大小与材料性能离散性相关。 See Figure 3, considering the discreteness of the test results, for the prediction of the peak load test value of each specimen, β=0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8...2.0, the β value is not a fixed value , the value range is related to the discreteness of material properties.
参见表3,对于本实施例的4个相同尺寸相同缝高比的试件,其实测峰值荷载都不相同。采用本发明方法,4个试件的β分别取不同值0.9、1.2、0.95、1.0,预测值就会分别与4个峰值荷载试验值很好吻合。β=1.0能够预测整体4个构件的平均值。 Referring to Table 3, for the four specimens of the same size and the same slit-height ratio in this embodiment, the measured peak loads are different. By adopting the method of the present invention, the β of the four test pieces take different values of 0.9, 1.2, 0.95, and 1.0 respectively, and the predicted values are in good agreement with the test values of the four peak loads respectively. β=1.0 can predict the mean of the overall 4 components.
表3 个性化确定四个带裂缝构件的峰值荷载 Table 3 Individually determined peak loads of four members with cracks
。 .
实施例4确定带裂缝砂浆或混凝土构件峰值荷载的方法,包括以下步骤: Embodiment 4 The method for determining the peak load of a cracked mortar or concrete member may include the following steps:
浇筑相同尺寸、相同缝高比的混凝土三点弯曲梁构件5个,其试件尺寸L×B×W =400 × 100 × 100 mm,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=4。其初始裂缝长度相同a0 =50mm,即a0/W=0.5都一致。其骨料最大粒径dmax=20 mm。W/dmax = 5。已知该混凝土材料特性—拉伸强度ft=4.7MPa,断裂韧度KIC=2.2 MPa·m1/2。 Five concrete three-point bending beam members with the same size and the same joint height ratio were poured, and the specimen size L×B×W =400×100×100 mm, where W is the height of the specimen, B is the thickness of the specimen, and L is The effective length of the test piece, the test piece satisfies L/W=4. The initial crack length is the same a 0 =50mm, that is a 0 /W=0.5 are consistent. The maximum particle size of the aggregate is d max =20 mm. W/d max = 5. The material properties of the concrete are known—tensile strength f t =4.7MPa, fracture toughness K IC =2.2 MPa·m 1/2 .
根据所用试件类型,确定不同带裂缝构件的等效裂缝长度ae。ae可由式 (1) 计算。根据应力强度因子手册,可以确定不同构件或结构型式的几何影响参数Y(α),三点弯曲梁可由式 (2) 计算。 According to the type of specimen used, determine the equivalent crack length a e of different cracked components. a e can be calculated by formula (1). According to the stress intensity factor manual, the geometric influence parameter Y(α) of different components or structural types can be determined, and the three-point bending beam can be calculated by formula (2).
将已知确定的该试件混凝土的材料参数—拉伸强度ft和断裂韧度KIC代入弹式(4),推出σn(Pmax)值。 Substitute the known and determined material parameters of the specimen concrete—tensile strength f t and fracture toughness K IC into elastic formula (4) to derive the value of σ n (P max ).
由式(5),通过β取不同的值,从而反解出不同试件的个性化的峰值荷载Pmax。参见图1,由于混凝土类准脆性材料自身特性决定,即使浇筑相同尺寸相同型式的构件,每个构件的峰值荷载也不尽相同。 According to formula (5), different values of β are used to inversely solve the personalized peak load P max of different specimens. Referring to Figure 1, due to the inherent characteristics of concrete-like quasi-brittle materials, the peak load of each component is different even if components of the same size and type are poured.
参见图2,从统计角度考虑,β=1.0可预测整体构件峰值荷载的平均值。 Referring to Figure 2, from a statistical point of view, β = 1.0 predicts the average value of the peak load of the overall member.
参见图3,考虑到试验结果的离散性,对于每个试件峰值荷载试验值的预测,则β=0.1、0.2、0.3、0.4、0.5、0.6、0.7、0.8…2.0,β值不是定值,取值范围大小与材料性能离散性相关。 See Figure 3, considering the discreteness of the test results, for the prediction of the peak load test value of each specimen, β=0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8...2.0, the β value is not a fixed value , the value range is related to the discreteness of material properties.
参见表4,对于本实施例的5个相同尺寸相同缝高比的试件,其实测峰值荷载都不相同。采用本发明方法,5个试件的β分别取不同值0.55、0.85、1.0、1.1、1.4,预测值就会分别与5个峰值荷载试验值很好吻合。β=1.0能够预测整体5个构件的平均值。 Referring to Table 4, for the five specimens of the same size and the same slit-to-height ratio in this embodiment, the measured peak loads are different. By adopting the method of the present invention, if the β of the five test pieces take different values of 0.55, 0.85, 1.0, 1.1, and 1.4 respectively, the predicted values will be in good agreement with the five peak load test values respectively. β = 1.0 is able to predict the mean of the overall 5 components.
表4 个性化确定四个带裂缝构件的峰值荷载 Table 4 Individually determined peak loads of four members with cracks
。 .
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,但显然所描述的实施例仅为本发明示意性的部分具体实施方式,并非用以限定本发明的范围,任何本领域的技术人员在不脱离本发明构思和原则的前提下所做出的等同变化与修改,均应属于本发明保护的范围。 The above descriptions of the disclosed embodiments enable those skilled in the art to implement or use the present invention, but it is obvious that the described embodiments are only illustrative partial implementations of the present invention, and are not intended to limit the scope of the present invention. Equivalent changes and modifications made by those skilled in the art without departing from the concepts and principles of the present invention shall fall within the protection scope of the present invention.
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CN114577564A (en) * | 2022-02-11 | 2022-06-03 | 中国电建集团西北勘测设计研究院有限公司 | Method for loading cracks of prefabricated mortar test piece by three-point bending |
CN114970093A (en) * | 2022-04-15 | 2022-08-30 | 华北水利水电大学 | Construction and Application of Compatibility Control Model of Concrete Material Strength and Fracture Toughness |
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CN114577564A (en) * | 2022-02-11 | 2022-06-03 | 中国电建集团西北勘测设计研究院有限公司 | Method for loading cracks of prefabricated mortar test piece by three-point bending |
CN114577564B (en) * | 2022-02-11 | 2023-02-10 | 中国电建集团西北勘测设计研究院有限公司 | Method for prefabricating cracks of mortar test piece by three-point bending loading |
CN114970093A (en) * | 2022-04-15 | 2022-08-30 | 华北水利水电大学 | Construction and Application of Compatibility Control Model of Concrete Material Strength and Fracture Toughness |
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