Zero-Failure(零失效)可靠性测试

原文标题：Zero-Failure Reliability Testing， 转载自：HTTP://TRUE-PROGRESS.COM. 作者：Jeremy Gernand。由iHydrostatics翻译整理，为避免因翻译造成的理解偏差，文末附英文原文。

所以，你想要证明一个新的设计比你自己或其他公司的现有设计更好。以非常小的样本量获得答案的最有效，最快捷的方法是什么？您可能有多种选择，包括加速测试，它们各自都有其优点和缺点。在这里，我建议采用零失效测试，作为一个现实和有用的选择，特别是因为它是我们经常做的事情，但没有数学上的理由。

零失效可靠性测试通常也称为实证性测试。这是为了证实给定设计优于要求或先前设计而进行的测试。如果您知道可能的Weibull形状因子参数（beta），那么您可以轻松计算测试的大小和长度，以证实给定置信水平的设计。特别是某些情况下，你的可用测试单元数量或可用的测试时间非常有限的时候。

首先，在继续之前，确定您所需的置信区间。这通常表示为百分比。对于工程师，通常使用90％或95％。根据设计功能的重要性，数字可能更高或更低。我们不希望出现的情况是，通过选择足够低的置信区间来使测试成功。

其次，您需要确定您的设计的beta或Weibull形状因子。如果您要与之前的类似设计进行比较，则可以根据该设计的失效记录确定beta。如果您正在评估新设计，通常可以从可靠性手册和包含类似零件组件的类似系统或系统的其他已发布信息中获得你需要的数值。小于1.0的beta反映早期失效模式，而1.0代表随机失效模式，高于1.0代表磨损失效模式。

第三，你需要拥有你所需的特征寿命等。这可以从所需的特征寿命或从MTBF（平均故障间隔时间）值导出，或者可以基于先前设计的故障数据来计算。

第四，如果您的测试单位数量或可用测试时间有限，您需要了解这些限制。

现在，您拥有这些数据，您可以根据以下等式计算“k”，即特定测试的特征寿命系数。 Beta是要测试的设计的Weibull形状因子，N是您计划使用的测试样品数，而置信度是您希望的置信度，表示为0到1之间的数字。

然后，您所需的测试时间在没有任何失效的情况下完成并证实该设计优于最后一个或优于所需的测试时间仅仅是所需的特征寿命乘以上面确定的值“k”。

示例：考虑我们有3个机械传动装置，我们要测试它们以90％的置信度展示超过1000小时的特征寿命。从之前的类似设计中我们预计beta（威布尔形状因子）的值为2.2。使用上面的等式，我们计算k为0.8867。然后，我们所需的测试时间为887小时。因此，为了证明符合我们的要求，90％的信心，3个单元必须每个完成887小时的测试而不会失效。

英文原文：

So, you’ve got a design that you want to prove is better than the existing design from your own or another company. What’s the most efficient, fastest way to get to that answer with a very small sample size? Whle there may be several options you have, including accelerated testing, they each can have their benefits and drawbacks. Here, I will advocate for zero-failure testing as a realistic and useful option, especially since it is something we often do anyways, but without the mathematical justification.

Zero-failure reliability testing is also often called substantiation testing. It is a test conducted to substantiate that a given design is better than a requirement or a previous design. If you know the likely Weibull shape factor parameter (beta), then you can easily calculate the size and length of a test to substantiate the design in question for a given confidence level. Most often, either the number of test units available or the test time available is restricted for you.

First, before going any further, determine your required confidence interval. This is usually expressed as a percentage. For engineers, 90% or 95% are generally utilized. The number may be higher or lower depending on the criticality of your design’s function. You don’t want to be in a position after the fact of selecting a low enough confidence interval to make your test count as a success.

Second, you need to determine your design’s beta, or Weibull shape factor. If you are comparing against a similar previous design, you can determine beta from the failure records of that design. If you are evaluating a new design, you can usually gain some kind of insight from reliability handbooks and other published information on similar systems or systems that contain similar piece part components. A beta of less than 1.0 reflects an infant mortality failure mode, while 1.0 represents a random failure mode, and higher than 1.0 represents a wearout failure mode.

Third, you need to have your required characteristic life, eta. This can either be derived from a required characteristic life, or from a MTBF (Mean Time Between Failures) value, or it can be calculated based on the failure data of a previous design.

Fourth, if either your number of test units or your available test time is limited, you need to understand those limits.

Now, that you have those data points, you can calculate ‘k’, the characteristic life multiplier for your particular test from the following equation. Beta is the Weibull shape factor for your design to be tested, N is the number of test units you plan to use, and Confidence is your desired confidence expressed as a number between 0 and 1.

Then, your required test time to complete without any failures and substantiate that this design is better than the last one or better than required is simply the required characteristic life multiplied by the value ‘k’ as determined above.

Example: Consider that we have 3 units of a mechanical transmission that we want to test to demonstrate a better than 1000 hour characteristic life with 90% confidence. From a previous similar design we expect the value of beta (the Weibull shape factor) to be 2.2. Using the equation above, we calculate k to be 0.8867. And, then our required test time is 887 hours. Therefore, to demonstrate compliance with our requirement with 90% confidence, 3 units must each complete 887 hours of testing without failure.