WO2020037995A1 - 一种滚动轴承外圈缺陷二维量化诊断方法 - Google Patents

一种滚动轴承外圈缺陷二维量化诊断方法 Download PDF

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WO2020037995A1
WO2020037995A1 PCT/CN2019/081860 CN2019081860W WO2020037995A1 WO 2020037995 A1 WO2020037995 A1 WO 2020037995A1 CN 2019081860 W CN2019081860 W CN 2019081860W WO 2020037995 A1 WO2020037995 A1 WO 2020037995A1
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outer ring
class
defect
axis
ring defect
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PCT/CN2019/081860
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French (fr)
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崔玲丽
黄金凤
张飞斌
王华庆
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北京工业大学
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    • GPHYSICS
    • G01MEASURING; TESTING
    • G01MTESTING STATIC OR DYNAMIC BALANCE OF MACHINES OR STRUCTURES; TESTING OF STRUCTURES OR APPARATUS, NOT OTHERWISE PROVIDED FOR
    • G01M13/00Testing of machine parts
    • G01M13/04Bearings
    • GPHYSICS
    • G01MEASURING; TESTING
    • G01MTESTING STATIC OR DYNAMIC BALANCE OF MACHINES OR STRUCTURES; TESTING OF STRUCTURES OR APPARATUS, NOT OTHERWISE PROVIDED FOR
    • G01M13/00Testing of machine parts
    • G01M13/04Bearings
    • G01M13/045Acoustic or vibration analysis

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  • the invention belongs to the field of fault diagnosis, and particularly relates to a method for quantitatively diagnosing defects of a rolling bearing outer ring.
  • Quantitative diagnosis of rolling bearing outer ring defects mainly includes two aspects: defect fixed size diagnosis and defect location size diagnosis.
  • the fixed dimension mainly refers to the width of the outer ring defect
  • the positioning dimension refers to the circumferential angular position of the outer ring defect.
  • the theoretical significance and engineering value of fixed-size diagnosis mainly include: in the case of local faults of the bearing that have been diagnosed, comprehensive consideration of economic benefits and production order, allowing the faulty bearing to keep working, but the fixed-size information of the fault must be quantitatively diagnosed In order to grasp the severity of the fault in real time and prevent the fault from spreading to the cause of destructive accidents.
  • the positioning size diagnosis has the following engineering application value: 1.
  • the different locations of bearing outer ring defects may correspond to different causes of failure; for example, defects appearing at the load center may be mainly caused by fatigue damage, and if the defects occur at a location far from the load center, the failure The causes are more likely to be material defects, machining defects, or maintenance defects.
  • the position of the defect is one of the main factors influencing the prediction of the remaining life of the bearing; for example, if the other factors are the same, a defect closer to the load center may have a faster expansion speed, thereby making the remaining life of the bearing shorter.
  • the invention proposes a two-dimensional quantitative diagnosis method for rolling bearings based on the study of the non-linear dynamic mechanism of the bearing system with defects and the vibration acceleration signal processing technology, which can accurately diagnose the width and positioning dimensions of the outer ring defects.
  • a two-dimensional quantitative diagnosis method for a rolling bearing outer ring defect includes the following specific steps:
  • Step 1 Establish a reference coordinate system, take the axial symmetry plane of the rolling bearing as the coordinate system plane, take the output facing the shaft as the front, and take the vertical downward direction as the positive direction of the y-axis; use the positive direction of the y-axis as the reference, and run around the axis Rotate 90 ° to the positive direction of the x-axis; take the positive direction of the x-axis as the angle reference of the reference coordinate system, and set the positive direction of the x-axis to the positive direction of the y-axis as the positive direction of the angular coordinate of the reference coordinate system, that is, the positive direction of the x-axis 0 °, the positive direction of the y-axis is 90 °;
  • Step 2 measure the vibration acceleration signal
  • the acceleration signal measured by the vibration acceleration sensor 1 is named a x ;
  • the acceleration signal measured by the acceleration sensor 2 and the vibration acceleration sensor 2 is named a y ;
  • the sampling frequency F s is set , and the number of sampling points N;
  • the vibration acceleration sensor 1 and the vibration acceleration sensor 2 measure 10 sets of vibration acceleration signals simultaneously, and record them
  • Step 3 Determine the type of fault shock oscillation in the time-domain waveform of the signal.
  • the first zero value step of the calibrated fault shock oscillating waveform is A x, 1 , and the first peak point after point A x, 1 is B x, 1 ; when the fault shock oscillates, If there is no second zero-value step point, it is determined that the fault shock oscillation type of the time domain waveform of the signal is Class 1, and the first agitation point in the fault shock oscillation waveform is marked as C x, 1 ; when the fault shock oscillation There is a second zero-value step in the waveform, then it is marked as D x, 1 , and the fault impact type of the time domain waveform of the determined signal is Class 2, and the first peak point after the calibration point D x, 1 is E x, 1 , the end point of the fault shock waveform is calibrated as F x, 1 ; for Class 1 type outer ring defects, perform steps 4 to 6 below; for Class 2 type outer ring defects, perform the following Step 7 to Step 10;
  • Step 4 Diagnose the circumferential dimensions of Class 1 outer ring defects.
  • step 3 If the determination result in step 3 is Class 1, the time interval T c1 between point A x, 1 and point C x, 1 is measured, and the circumferential dimension L c1 of the outer ring defect is diagnosed by the following formula:
  • R r is the ball radius
  • R p is the pitch circle radius
  • f is the inner ring frequency
  • ⁇ 0 is the no-load contact angle
  • Step 5 Determine the center position of the Class 1 outer ring defect.
  • step 3 obtain the ordinate values V Ax, 1 and V Bx, 1 of point A x, 1 and point B x , 1 ; when V Ax, 1 ⁇ V Bx, 1 , determine The defect center is located on the left side of the y-axis of the reference coordinate system; when V Ax, 1 > V Bx, 1 , the defect center is determined to be on the right side of the y-axis of the reference coordinate system;
  • Step 6 Diagnose the size of the center position of the Class 1 outer ring defect.
  • Step 7 Diagnose Class 2 type outer ring defect circumferential dimension L c2 ,
  • step 3 If the judgment result in step 3 is Class 2, the time interval T c2 between point A x, 1 and point D x, 1 is measured, and the circumferential dimension of the outer ring defect is diagnosed by the following formula:
  • Step 8 Determine the left and right edges of the Class 2 outer ring defect.
  • step 3 obtain the ordinate values V Ax, 1 , V Bx, 1 , V of points A x, 1 , B B x, 1 , D D , 1, and E x, 1 . Dx, 1 and V Ex, 1 ; when V Ax, 1 ⁇ V Bx, 1 , the left edge of the defect is determined to be to the left of the y-axis of the reference coordinate system; when V Ax, 1 > V Bx, 1 , the left edge of the defect is determined The edge is located to the right of the y-axis of the reference coordinate system; when V Dx, 1 ⁇ V Ex, 1 , the right edge of the defect is determined to be to the left of the y-axis of the reference coordinate system; when V Dx, 1 > V Ex, 1 , it is determined The right edge of the defect is to the right of the y-axis of the reference coordinate system;
  • Step 9 Calculate the ⁇ vh value of the left and right edges of the Class 2 outer ring defect
  • the ⁇ vh value of the left edge of the Class 2 outer ring defect is calculated by the following formula:
  • N Ax, j , N Dx, j , N Ay, j, and N Dy j represent the number of data points corresponding to points A x, j , D x, j , A y, j , D y, j ; and The following formula calculates the ⁇ vh value of the right edge of the Class 2 outer ring defect:
  • N Fx, j and N Fy, j represent the number of data points corresponding to points F x, j , F y, j , respectively;
  • Step 10 Diagnose the size of the center position of the Class 2 outer ring defect.
  • the beneficial effect of the present invention is that a two-dimensional quantitative diagnosis method for rolling bearing outer ring defects is proposed, which classifies the circumferential dimensions of defects based on the type of fault impact, and then uses targeted technology to accurately diagnose their defects based on the characteristics of different types of defects. Circumferential dimensions and position dimensions have important practicality and engineering value.
  • FIG. 1 is a working flowchart of the present invention.
  • Figure 2 Schematic diagram of reference coordinate system and measuring point arrangement of rolling bearing system.
  • FIG. 3 is a time-domain waveform diagram of a vibration acceleration signal of a faulty bearing system to be diagnosed.
  • FIG. 4 is a waveform diagram of a fault shock oscillation of a signal to be diagnosed.
  • Step 1 establish a reference coordinate system
  • Step 2 measure the vibration acceleration signal
  • a vibration acceleration sensor 1 is installed at the intersection of the outer surface of the bearing housing and the negative line of the x-axis, and the acceleration signal measured by the sensor is named a x ;
  • a vibration acceleration sensor 2 is installed at the intersection, and the acceleration signal measured by the sensor is named a y ;
  • the fault characteristic frequency of the loop defect f o 21.5Hz;
  • Step 3 judge the type of fault shock oscillation in the time domain waveform of the signal
  • Figure 3 can be seen after partial enlargement
  • D x, 1 the fault shock oscillation type in this embodiment is Class 2
  • the first peak point is E x, 1
  • the end point of the fault shock oscillation waveform is F x, 1.
  • the waveform after marking is shown in Figure 4.
  • Step 4 Diagnose the circumferential dimensions of Class 2 outer ring defects.
  • Step 5 Determine the left and right edge positions of the Class 2 outer ring defect.
  • V Ax, 1 -0.00073
  • V Bx, 1 0.2649
  • V Dx , 1 -0.012
  • V Ex, 1 0.3733
  • V Ax, 1 ⁇ V Bx, 1 and V Dx, 1 ⁇ V Ex, 1 can be determined, and it can be determined that the left edge of the defect is located on the y-axis of the reference coordinate system. Left and right edges are to the left of the y-axis of the reference coordinate system;
  • Step 6 Calculate the ⁇ vh value of the left and right edges of the Class 2 outer ring defect.
  • N Ax, j , N Dx, j , N Ay, j, and N Dy j represent the number of data points corresponding to points A x, j , D x, j , A y, j , D y, j ; and The following formula calculates the ⁇ vh value of the right edge of the Class 2 outer ring defect:
  • N Fx, j and N Fy, j represent the number of data points corresponding to points F x, j , F y, j , respectively;
  • Step 7 Diagnose the size of the center position of the Class 2 outer ring defect.

Abstract

一种滚动轴承外圈缺陷二维量化诊断方法,该方法根据信号波形特征将故障冲击振荡类型划分为Class 1和Class 2。对于Class1型外圈缺陷,则首先基于提出的Class 1型外圈缺陷周向尺寸计算公式量化诊断出缺陷周向尺寸;由提出的Class 1型外圈缺陷中心位置尺寸计算公式完成Class 1型外圈缺陷的二维量化诊断;对于Class 2型外圈缺陷,则首先基于提出的Class 2型外圈缺陷周向尺寸计算公式量化诊断出缺陷周向尺寸;再分别计算出外圈缺陷左、右边缘的σvh值,最后由提出的Class2型外圈缺陷中心位置尺寸计算公式诊断出外圈缺陷中心位置尺寸,完成Class 2型外圈缺陷的二维量化诊断。

Description

一种滚动轴承外圈缺陷二维量化诊断方法 技术领域
本发明属于故障诊断领域,具体涉及一种滚动轴承外圈缺陷量化诊断方法。
背景技术
滚动轴承外圈缺陷的量化诊断主要包括两个方面:缺陷定形尺寸诊断和缺陷定位尺寸诊断。其中定形尺寸主要是指外圈缺陷的宽度,定位尺寸则是指外圈缺陷的周向角位置。定形尺寸诊断的理论意义和工程价值主要包括:在已经诊断出轴承存在局部缺陷故障的情况下,综合考虑经济效益和生产秩序,允许故障轴承保持工作运转,但是必须量化诊断出缺陷的定形尺寸信息以便实时掌握故障的严重程度,防止故障扩展到引起破坏性事故的发生。而定位尺寸诊断则具有以下工程应用价值:1、当缺陷尺寸很微小时,如果能预先诊断出缺陷的位置,将有助于拆机后更快速的排查出故障部位,进而提高工作效率和诊断成功率。2、轴承外圈缺陷出现位置的不同,可能对应于不同的故障原因;比如缺陷出现在承载中心位置,可能主要是疲劳损伤导致的,而如果缺陷出现在距离承载中心较远的位置,则故障原因更可能是材料缺陷、机加工缺陷或维护缺陷等。3、缺陷位置是轴承剩余寿命预测的主要影响因素之一;比如在其他因素相同的情况下,更靠近承载中心的缺陷,可能具有更快的扩展速度,从而使得轴承的剩余寿命也更短。
因此,实现滚动轴承外圈缺陷的量化诊断具有迫切的理论意义和 现实需求。
发明内容
本发明基于含缺陷轴承系统非线性动力学机理研究和振动加速度信号处理技术,提出了一种滚动轴承二维量化诊断方法,可准确地诊断外圈缺陷的宽度尺寸和定位尺寸。
为实现上述目的,本发明的技术方案如下:
一种滚动轴承外圈缺陷二维量化诊断方法,包括以下具体步骤:
步骤1建立参考坐标系,以滚动轴承的轴向对称面为坐标系平面,以面向轴的输出端为正面,以垂直向下为y轴正方向;以y轴正方向为参照,绕轴的运行方向旋转90°为x轴正方向;以x轴正方向为参考坐标系的角度基准,设定x轴正方向旋转到y轴正方向为参考坐标系的角度坐标正方向,即x轴正方向为0°,y轴正方向为90°;
步骤2测取振动加速度信号,
在轴承座外表面与x轴负方向线的交点处安装振动加速度传感器1,振动加速度传感器1测取的加速度信号命名为a x;在轴承座外表面与y轴负方向线的交点处安装振动加速度传感器2,振动加速度传感器2测取的加速度信号命名为a y;设定采样频率F s,采样点数N;由振动加速度传感器1和振动加速度传感器2同时测取10组振动加速度信号,分别记为a x,j和a y,j,j=1,2,3…10,j为振动加速度信号组的序号;根据滚动轴承型号和滚动轴承的内圈转速f,计算出滚动轴承外圈缺陷的故障特征频率f o
步骤3判断信号时域波形中的故障冲击振荡类型,以时间t为横 坐标,a x1为纵坐标绘制振动加速度信号的时域波形图;在时域波形图中找出以1/f o为周期的故障冲击振荡波形,标定故障冲击振荡波形的第一个零值阶跃点为A x,1,点A x,1之后的第一个峰值点为B x,1;当故障冲击振荡波形中不存在第二个零值阶跃点,则判定信号时域波形的故障冲击振荡类型为Class 1,且标记故障冲击振荡波形中的第一个搅动点为C x,1;当故障冲击振荡波形中存在第二个零值阶跃点,则将其标记为D x,1,且判定信号时域波形的故障冲击类型为Class 2,标定点D x,1之后的第一个峰值点为E x,1,故障冲击振荡波形的终止点标定为F x,1;对于Class 1型外圈缺陷,执行下述的步骤4到步骤6;对于Class 2型外圈缺陷,则执行下述的步骤7到步骤10;
步骤4诊断Class 1型外圈缺陷周向尺寸,
如果步骤3中的判定结果为Class 1,则量取点A x,1到点C x,1之间的时间间隔T c1,并由以下公式诊断出外圈缺陷的周向尺寸L c1
Figure PCTCN2019081860-appb-000001
其中,R r是滚珠半径,R p是节圆半径,f为内圈转频,α 0是空载接触角;
步骤5判断Class 1型外圈缺陷中心方位,
根据步骤3中绘制的时域波形图,取得点A x,1和点B x,1的纵坐标值V Ax,1和V Bx,1;当V Ax,1<V Bx,1,则判定缺陷中心位于参考坐标系y轴的左侧;当V Ax,1>V Bx,1,则判定缺陷中心位于参考坐标系y轴的右侧;
步骤6诊断Class 1型外圈缺陷中心位置尺寸,
利用步骤2中测取的振动加速度信号,由以下公式计算外圈缺陷 中心位置尺寸θ c1
Figure PCTCN2019081860-appb-000002
其中,
Figure PCTCN2019081860-appb-000003
Figure PCTCN2019081860-appb-000004
分别表示第j组振动加速度信号a x,j和a y,j的平均值,σ 0是振动加速度信号的故障基准标准差,N是数据点数,i是循环控制变量;至此完成Class 1型外圈缺陷的周向尺寸和位置尺寸的二维量化诊断。
步骤7诊断Class 2型外圈缺陷周向尺寸L c2
如果步骤3中的判定结果为Class 2,则量取点A x,1到点D x,1之间的时间间隔T c2,并由以下公式诊断出外圈缺陷的周向尺寸:
Figure PCTCN2019081860-appb-000005
步骤8判断Class 2型外圈缺陷左、右边缘方位,
根据步骤3中绘制的时域波形图,取得点A x,1、点B x,1、点D x,1和点E x,1的纵坐标值V Ax,1、V Bx,1、V Dx,1和V Ex,1;当V Ax,1<V Bx,1,则判定缺陷左边缘位于参考坐标系y轴的左侧;当V Ax,1>V Bx,1,则判定缺陷左边缘位于参考坐标系y轴的右侧;当V Dx,1<V Ex,1,则判定缺陷右边缘位于参考坐标系y轴的左侧;当V Dx,1>V Ex,1,则判定缺陷右边缘位于参考坐标系y轴的右侧;
步骤9计算Class 2型外圈缺陷左、右边缘的σ vh
按照步骤3中的规则标定信号a x,j的故障冲击振荡波形中的点A、B、D、E和F,并分别记为A x,j、B x,j、D x,j、E x,j和F x,j;按照步骤3中的规则标定信号a y,j的故障冲击振荡波形中的点A、B、D、E和F,并分别记为A y,j、B y,j、D y,j、E y,j和F y,j;由以下公式计算Class 2型外圈缺陷左边缘的σ vh值:
Figure PCTCN2019081860-appb-000006
其中N Ax,j、N Dx,j、N Ay,j和N Dy,j分别表示点A x,j、D x,j、A y,j、D y,j所对应的数据点数;再由以下公式计算Class 2型外圈缺陷右边缘的σ vh值:
Figure PCTCN2019081860-appb-000007
其中N Fx,j、N Fy,j分别表示点F x,j、F y,j所对应的数据点数;
步骤10诊断Class 2型外圈缺陷中心位置尺寸,
对步骤9中求解出的σ vh,l和σ vh,r进行比较,当σ vh,lvh,r,则由以下公式计算Class 2型外圈缺陷左边缘的位置尺寸θ l
Figure PCTCN2019081860-appb-000008
再结合公式(3)计算出Class 2型外圈缺陷中心的位置尺寸θ c2
Figure PCTCN2019081860-appb-000009
当σ vh,lvh,r,则先对由以下公式计算Class 2型外圈缺陷右边缘的位置 尺寸θ r
Figure PCTCN2019081860-appb-000010
再结合公式(3)计算出Class 2型外圈缺陷中心的位置尺寸θ c2
Figure PCTCN2019081860-appb-000011
至此完成Class 2型外圈缺陷的周向尺寸和位置尺寸的二维量化诊断。
本发明的有益效果是:提出了一种滚动轴承外圈缺陷二维量化诊断方法,基于故障冲击类型对缺陷周向尺寸进行分类,再根据不同类型缺陷的特点采用针对性的技术精确诊断出它们的周向尺寸和位置尺寸,有重要的实用性和工程价值。
附图说明
图1本发明的工作流程图。
图2滚动轴承系统参考坐标系和测点布置示意图。
图3待诊断故障轴承系统的振动加速度信号时域波形图。
图4待诊断信号的故障冲击振荡波形图。
具体实施方式
下面具体结合附图与实施例对本发明的诊断方法进行详细说明,但是本发明的保护范围不局限于所述实施例。
步骤1建立参考坐标系,
以滚动轴承的轴向对称面为坐标系平面,以面向轴的输出端为正面,以垂直向下为y轴正方向;以y轴正方向为参照,绕轴的运行方向旋转90°为x轴正方向;以x轴正方向为参考坐标系的角度基准, 设定x轴正方向旋转到y轴正方向为参考坐标系的角度坐标正方向,即x轴正方向为0°,y轴正方向为90°,建立完成的参考坐标系如图2所示;
步骤2测取振动加速度信号,
如图2所示,在轴承座外表面与x轴负方向线的交点处安装振动加速度传感器1,该传感器测取的加速度信号命名为a x;在轴承座外表面与y轴负方向线的交点处安装振动加速度传感器2,该传感器测取的加速度信号命名为a y;设定采样频率F s=131072Hz,采样点数N=131072;由传感器1和传感器2同时测取10组振动加速度信号,分别记为a x,j和a y,j,j=1,2,3…10;本实施例中选用的滚动轴承型号为6308,滚动轴承的内圈转速为f=7Hz,则可计算出滚动轴承外圈缺陷的故障特征频率f o=21.5Hz;
步骤3判断信号时域波形中的故障冲击振荡类型,
以时间t为横坐标,a x1为纵坐标绘制本实施例的振动加速度信号时域波形图,如图3所示;在图3中找出以1/f o=0.0465为周期的故障冲击振荡波形,标定故障冲击振荡波形的第一个零值阶跃点为A x,1,点A x,1之后的第一个峰值点标定为B x,1;将图3进行局部放大后可看出故障冲击振荡波形中存在第二个零值阶跃点,将其标记为D x,1,并由此判定本实施例中的故障冲击振荡类型为Class 2,标定点D x,1之后的第一个峰值点为E x,1,以及故障冲击振荡波形的终止点为F x,1,标记完成后的波形如图4所示。
步骤4诊断Class 2型外圈缺陷周向尺寸,
由图4量取点A x,1到点D x,1之间的时间间隔T c2=0.0051,并由以下公式诊断出外圈缺陷的周向尺寸L c2
Figure PCTCN2019081860-appb-000012
根据公式(1)可计算出本实施例中的外圈缺陷周向尺寸为L c2=0.0034;
步骤5判断Class 2型外圈缺陷左、右边缘方位,
由图4取得点A x,1、点B x,1、点D x,1和点E x,1的纵坐标值分别为V Ax,1=-0.00073,V Bx,1=0.2649,V Dx,1=-0.012,V Ex,1=0.3733;因此可得出V Ax,1<V Bx,1和V Dx,1<V Ex,1,则可判定缺陷左边缘位于参考坐标系y轴的左侧,右边缘位于参考坐标系y轴的左侧;
步骤6计算Class 2型外圈缺陷左、右边缘的σ vh值,
按照步骤3中的规则标定信号a x,j的故障冲击振荡波形中的点A、B、D、E和F,并分别记为A x,j、B x,j、D x,j、E x,j和F x,j;按照步骤3中的规则标定信号a y,j的故障冲击振荡波形中的点A、B、D、E和F,并分别记为A y,j、B y,j、D y,j、E y,j和F y,j;本实施例的振动加速度信号的故障基准标准差为σ 0=0.29,由以下公式计算Class 2型外圈缺陷左边缘的σ vh值:
Figure PCTCN2019081860-appb-000013
其中N Ax,j、N Dx,j、N Ay,j和N Dy,j分别表示点A x,j、D x,j、A y,j、D y,j所对应的数据点数;再由以下公式计算Class 2型外圈缺陷右边缘的σ vh值:
Figure PCTCN2019081860-appb-000014
其中N Fx,j、N Fy,j分别表示点F x,j、F y,j所对应的数据点数;
步骤7诊断Class 2型外圈缺陷中心位置尺寸,
对步骤6中求解出的σ vh,l和σ vh,r进行比较,可得出σ vh,lvh,r,又由步骤5可知V Dx,1<V Ex,1,则由以下公式计算Class 2型外圈缺陷右边缘的位置尺寸θ r
θ r=180°+arccot(σ hv,r)=180°+89.84°=269.84°         (4)
再结合公式(1)计算出Class 2型外圈缺陷中心的位置尺寸θ c2
Figure PCTCN2019081860-appb-000015
至此完成本实施例的外圈缺陷周向尺寸和位置尺寸的二维量化诊断。

Claims (1)

  1. 一种滚动轴承外圈缺陷二维量化诊断方法,其特征在于:包括以下具体步骤:
    步骤1建立参考坐标系,以滚动轴承的轴向对称面为坐标系平面,以面向轴的输出端为正面,以垂直向下为y轴正方向;以y轴正方向为参照,绕轴的运行方向旋转90°为x轴正方向;以x轴正方向为参考坐标系的角度基准,设定x轴正方向旋转到y轴正方向为参考坐标系的角度坐标正方向,即x轴正方向为0°,y轴正方向为90°;
    步骤2测取振动加速度信号,
    在轴承座外表面与x轴负方向线的交点处安装振动加速度传感器1,振动加速度传感器1测取的加速度信号命名为a x;在轴承座外表面与y轴负方向线的交点处安装振动加速度传感器2,振动加速度传感器2测取的加速度信号命名为a y;设定采样频率F s,采样点数N;由振动加速度传感器1和振动加速度传感器2同时测取10组振动加速度信号,分别记为a x,j和a y,j,j=1,2,3…10,j为振动加速度信号组的序号;根据滚动轴承型号和滚动轴承的内圈转速f,计算出滚动轴承外圈缺陷的故障特征频率f o
    步骤3判断信号时域波形中的故障冲击振荡类型,以时间t为横坐标,a x1为纵坐标绘制振动加速度信号的时域波形图;在时域波形图中找出以1/f o为周期的故障冲击振荡波形,标定故障冲击振荡波形的第一个零值阶跃点为A x,1,点A x,1之后的第一个峰值点为B x,1;当故障冲击振荡波形中不存在第二个零值阶跃点,则判定信号时域波形的故障冲击振荡类型为Class 1,且标记故障冲击振荡波形中的第一个 搅动点为C x,1;当故障冲击振荡波形中存在第二个零值阶跃点,则将其标记为D x,1,且判定信号时域波形的故障冲击类型为Class 2,标定点D x,1之后的第一个峰值点为E x,1,故障冲击振荡波形的终止点标定为F x,1;对于Class 1型外圈缺陷,执行下述的步骤4到步骤6;对于Class 2型外圈缺陷,则执行下述的步骤7到步骤10;
    步骤4诊断Class 1型外圈缺陷周向尺寸,
    如果步骤3中的判定结果为Class 1,则量取点A x,1到点C x,1之间的时间间隔T c1,并由以下公式诊断出外圈缺陷的周向尺寸L c1
    Figure PCTCN2019081860-appb-100001
    其中,R r是滚珠半径,R p是节圆半径,f为内圈转频,α 0是空载接触角;
    步骤5判断Class 1型外圈缺陷中心方位,
    根据步骤3中绘制的时域波形图,取得点A x,1和点B x,1的纵坐标值V Ax,1和V Bx,1;当V Ax,1<V Bx,1,则判定缺陷中心位于参考坐标系y轴的左侧;当V Ax,1>V Bx,1,则判定缺陷中心位于参考坐标系y轴的右侧;
    步骤6诊断Class 1型外圈缺陷中心位置尺寸,
    利用步骤2中测取的振动加速度信号,由以下公式计算外圈缺陷中心位置尺寸θ c1
    Figure PCTCN2019081860-appb-100002
    其中,
    Figure PCTCN2019081860-appb-100003
    Figure PCTCN2019081860-appb-100004
    分别表示第j组振动加速度信号a x,j和a y,j的平均值,σ 0是振动加速度信号的故障基准标准差,N是数据点数,i是循环控制变量;至此完成Class 1型外圈缺陷的周向尺寸和位置尺寸的二维量化诊断;
    步骤7诊断Class 2型外圈缺陷周向尺寸L c2
    如果步骤3中的判定结果为Class 2,则量取点A x,1到点D x,1之间的时间间隔T c2,并由以下公式诊断出外圈缺陷的周向尺寸:
    Figure PCTCN2019081860-appb-100005
    步骤8判断Class 2型外圈缺陷左、右边缘方位;
    根据步骤3中绘制的时域波形图,取得点A x,1、点B x,1、点D x,1和点E x,1的纵坐标值V Ax,1、V Bx,1、V Dx,1和V Ex,1;当V Ax,1<V Bx,1,则判定缺陷左边缘位于参考坐标系y轴的左侧;当V Ax,1>V Bx,1,则判定缺陷左边缘位于参考坐标系y轴的右侧;当V Dx,1<V Ex,1,则判定缺陷右边缘位于参考坐标系y轴的左侧;当V Dx,1>V Ex,1,则判定缺陷右边缘位于参考坐标系y轴的右侧;
    步骤9计算Class 2型外圈缺陷左、右边缘的σ vh
    按照步骤3中的规则标定信号a x,j的故障冲击振荡波形中的点A、 B、D、E和F,并分别记为A x,j、B x,j、D x,j、E x,j和F x,j;按照步骤3中的规则标定信号a y,j的故障冲击振荡波形中的点A、B、D、E和F,并分别记为A y,j、B y,j、D y,j、E y,j和F y,j;由以下公式计算Class 2型外圈缺陷左边缘的σ vh值:
    Figure PCTCN2019081860-appb-100006
    其中N Ax,j、N Dx,j、N Ay,j和N Dy,j分别表示点A x,j、D x,j、A y,j、D y,j所对应的数据点数;再由以下公式计算Class 2型外圈缺陷右边缘的σ vh值:
    Figure PCTCN2019081860-appb-100007
    其中N Fx,j、N Fy,j分别表示点F x,j、F y,j所对应的数据点数;
    步骤10诊断Class 2型外圈缺陷中心位置尺寸,
    对步骤9中求解出的σ vh,l和σ vh,r进行比较,当σ vh,lvh,r,则由以下公式计算Class 2型外圈缺陷左边缘的位置尺寸θ l
    Figure PCTCN2019081860-appb-100008
    再结合公式(3)计算出Class 2型外圈缺陷中心的位置尺寸θ c2
    Figure PCTCN2019081860-appb-100009
    当σ vh,lvh,r,则先对由以下公式计算Class 2型外圈缺陷右边缘的位置尺寸θ r
    Figure PCTCN2019081860-appb-100010
    再结合公式(3)计算出Class 2型外圈缺陷中心的位置尺寸θ c2
    Figure PCTCN2019081860-appb-100011
    至此完成Class 2型外圈缺陷的周向尺寸和位置尺寸的二维量化诊断。
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