雨流循环计数

使用循环计数从随机载荷序列中提取离散的简单的“等效”恒幅循环。

理解“循环计数”的一种方法是将其视为应力-应变随时间变化的信号。循环计数计算应力-应变滞后回线的数量,并跟踪它们的范围/平均值或最大/最小值。

雨流循环计数是应用最广泛的循环计数方法。它要求应力时间历史仅转换为峰和谷,并重新排列,以便从最高峰或最低谷(以绝对值较大者为准)开始。三个连续的应力点(1、2 和 3)将定义两个连续的范围为 Δ S 12 = S 1 S 2 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIXaGaaGOmaaqabaGccqGH9aqpdaabdaqaaiaa dofadaWgaaWcbaGaaGymaaqabaGccqGHsislcaWGtbWaaSbaaSqaai aaikdaaeqaaaGccaGLhWUaayjcSdaaaa@4287@ Δ S 23 = S 2 S 3 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIYaGaaG4maaqabaGccqGH9aqpdaabdaqaaiaa dofadaWgaaWcbaGaaGOmaaqabaGccqGHsislcaWGtbWaaSbaaSqaai aaiodaaeqaaaGccaGLhWUaayjcSdaaaa@428C@ 。只有当 Δ S 12 Δ S 23 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIXaGaaGOmaaqabaGccqGHKjYOcqGHuoarcaWG tbWaaSbaaSqaaiaaikdacaaIZaaabeaaaaa@3F79@ ,才能提取从点 1 到点 2 的循环。提取循环后,形成循环的两个点将被丢弃,其余点将相互连接。重复此过程,直到耗尽剩余的数据点。以下示例对 SimSolid 中的雨流计数流程进行了说明。

使用简单载荷历史的示例

考虑以下载荷历史:


Figure 1.
由于它是连续的,此载荷历史将转换为仅具有峰值和谷值的载荷历史:


Figure 2.
1、2、3 和 4 是四个峰值和谷值。点 4 是载荷历史中的峰值应力,并在重新排列期间移到前面,如 Figure 3 所示。重新排列后,峰值和谷值被重新编号。
Figure 3.

接下来,选择前三个应力值(1、2 和 3),并确定是否存在循环。如果 S i MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa aaleaacaWGPbaabeaaaaa@37E6@ 表示应力值,点 1,那么

Δ S 12 = S 1 S 2 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIXaGaaGOmaaqabaGccqGH9aqpdaabdaqaaiaa dofadaWgaaWcbaGaaGymaaqabaGccqGHsislcaWGtbWaaSbaaSqaai aaikdaaeqaaaGccaGLhWUaayjcSdaaaa@4287@
Δ S 23 = S 2 S 3 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIYaGaaG4maaqabaGccqGH9aqpdaabdaqaaiaa dofadaWgaaWcbaGaaGOmaaqabaGccqGHsislcaWGtbWaaSbaaSqaai aaiodaaeqaaaGccaGLhWUaayjcSdaaaa@428C@

由于 Δ S 12 Δ S 23 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIXaGaaGOmaaqabaGccqGHLjYScqGHuoarcaWG tbWaaSbaaSqaaiaaikdacaaIZaaabeaaaaa@3F8A@ ,无法提取点 1 至点 2 的循环。现在考虑下面三个点,如下:

Δ S 23 = S 2 S 3 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIYaGaaG4maaqabaGccqGH9aqpdaabdaqaaiaa dofadaWgaaWcbaGaaGOmaaqabaGccqGHsislcaWGtbWaaSbaaSqaai aaiodaaeqaaaGccaGLhWUaayjcSdaaaa@428C@
Δ S 34 = S 3 S 4 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIZaGaaGinaaqabaGccqGH9aqpdaabdaqaaiaa dofadaWgaaWcbaGaaG4maaqabaGccqGHsislcaWGtbWaaSbaaSqaai aaisdaaeqaaaGccaGLhWUaayjcSdaaaa@4290@

Δ S 23 Δ S 34 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIYaGaaG4maaqabaGccqGHKjYOcqGHuoarcaWG tbWaaSbaaSqaaiaaiodacaaI0aaabeaaaaa@3F7D@ ,因此提取了点 2 至点 3 的循环。由于已提取循环,因此从图形中删除这两个点。


Figure 4.

对其余的点执行相同的过程:

Δ S 14 = S 1 S 4 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIXaGaaGinaaqabaGccqGH9aqpdaabdaqaaiaa dofadaWgaaWcbaGaaGymaaqabaGccqGHsislcaWGtbWaaSbaaSqaai aaisdaaeqaaaGccaGLhWUaayjcSdaaaa@428C@
Δ S 45 = S 4 S 5 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaI0aGaaGynaaqabaGccqGH9aqpdaabdaqaaiaa dofadaWgaaWcbaGaaGinaaqabaGccqGHsislcaWGtbWaaSbaaSqaai aaiwdaaeqaaaGccaGLhWUaayjcSdaaaa@4294@
在这种情况下, Δ S 14 = Δ S 45 MathType@MTEF@5@5@+= feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam 4uamaaBaaaleaacaaIXaGaaGinaaqabaGccqGH9aqpcqGHuoarcaWG tbWaaSbaaSqaaiaaisdacaaI1aaabeaaaaa@3ED0@ ,所以又提取了点 1 至点 4 的循环。这两点将被丢弃。还剩下 5 个点,因此雨流计数流程已完成。

从该载荷历史中提取了两个循环(2→3 和 1→4)。选择最高的峰值/谷值并重新排列载荷历史的一个主要原因是为了确保始终提取最大的循环(在本例中为 1→4)。如果在重新排列之前观察载荷历史,并对其执行相同的雨流计数流程,则无法提取 1→4 循环。

使用更复杂载荷历史的示例

无论有多少载荷历史点,雨流计数流程都是相同的。然而,根据用于重排的最高峰/谷的位置,重排过程如何进行可能并不明显。下面的示例仅显示了具有更复杂载荷历史的重排过程。随后的雨量计数只是上述简单示例中提到的流程的外推,这里不再重复。

请考虑下面的载荷历史:


Figure 5.
由于此载荷历史是连续的,它将转换为仅具有峰值和谷值的载荷历史,如下所示:


Figure 6.
载荷点 11 是最高值载荷,所以现对载荷历史进行重新排列和重新编号,如下所示:


Figure 7.
载荷历史被重新排列,以便将包括最高载荷和最高载荷之后的所有点移动到载荷历史的开头,并从载荷历史的末尾删除。