Reliability, maintainability, and safety basics

Introduction

Nowadays, the reliability of engineering systems has become a quite challenging issue during the design process due to the increasing dependence of our daily lives and schedules on the proper functioning of these systems. Some examples of these systems are computers, nuclear powergenerating reactors, automobiles, space satellites, and aircraft.

The alarmingly high operating and support costs of engineering systems/equipment, in part due to failures and subsequent repairs, are the prime reasons for emphasizing maintainability of engineering systems. Some examples of these costs are the expense of maintenance personnel and their training, repair parts, test and support equipment, maintenance instructions and data, training equipment, and maintenance facilities.

Nowadays, engineering systems have become highly complex and sophisticated, and the safety of these systems has become a challenging issue. Needless to say, over the years, various types of methods and approaches have been developed for improving safety, reliability, and maintainability of engineering systems/equipment in the field. This chapter presents various reliability, maintainability, and safety basics considered useful to understand the subsequent chapters of this book.

Bathtub hazard rate curve

This curve is usually used to describe the failure rate of engineering systems/equipment and is shown in Fig. 3.1. The curve is called the bathtub hazard rate curve because it resembles the shape of a bathtub.

As shown in Fig. 3.1, the curve is divided into three sections. These sections are called burn-in period, useful-life period, and wear-out period. During the burn-in period, the system/equipment/item hazard rate decreases with time t. Some of the reasons for the occurrence of failures during this period are substandard materials and workmanship,

Bathtub hazard rate curve

Figure 3.1 Bathtub hazard rate curve.

poor manufacturing methods and processes, inadequate debugging, poor quality control, and human error [1,2]. Other terms used in the published literature from time to time for this region (period) are 'debugging region', 'infant-mortality region', and 'break-in region'.

During the useful life period, the hazard rate remains constant, and some of the causes for the occurrence of failures in this period are as follows [1, 2]:

  • • Higher random stress than expected.
  • • Undetectable defects.
  • • Natural failures.
  • • Abuse.
  • • Low safety factors.
  • • Human errors.

Finally, during the wear-out period, the hazard rate increases with time due to various reasons including poor maintenance practices, wear from aging, incorrect overhaul practices, wear due to friction, corrosion, and creep, and short designed-in life of the system/equipment/item under consideration [1, 2].

Mathematically, the following equation can be used to represent the bathtub hazard rate curve shown in Fig. 3.1 [3]:

where

X(t) is hazard rate (time-dependent failure rate).

t is time.

a is the shape parameter.

P is the scale parameter.

Equation (3.1) at a = 0.5 gives the shape of the bathtub hazard rate curve shown in Fig. 3.1.

General reliability formulas

A number of general formulas are often used for performing various types of reliability analysis. Four of these formulas that are based on the reliability function are as follows:

Failure (or probability) density function

The failure (or probability) density function is defined by [1]

/(<)—^ (3.2)

where

f(t) is the system/item failure (or probability) density function.

R(t) is the system/item reliability at time t.

Example 3.1

Assume that the reliability of an engineering system is expressed by

^(1) = ^' (3.3)

where

(f) is the engineering system reliability at time t. is the engineering system constant failure rate.

Obtain an expression for the failure (probability) density function of the engineering system by using Equation (3.2).

By inserting Equation (3.3) into Equation (3.2), we obtain

de~Xa

dt~ (3.4)

=

Thus, Equation (3.4) is the expression for the failure (probability) density function of the engineering system.

Hazard rate (or time-dependent failure rate) function

This is defined by

M<) = ^ (3.5)

where

X(f) is the system/item hazard rate (or time-dependent failure rate).

By inserting Equation (3.2) into Equation (3.5), we get

dR(t)

M,)—R(i)~ (3'6

Example 3.2

Obtain an expression for the hazard rate of the engineering system by using Equations (3.3) and (3.6).

By substituting Equation (3.3) into Equation (3.6), we get

  • 1 de~*a'
  • (3.7)

Thus, the hazard rate of the engineering system is given by Equation (3.7). It is to be noted that the right-hand side of this equation is not a function of time t. In other words, it is constant. Usually, it is called the constant failure rate of a system/item because it does not depend on time t.

General reliability function

This can be obtained with the aid of Equation (3.6). Thus, with the aid of Equation (3.6), we obtain

~^dt=^dR^

(3.8)

By integrating the both sides of Equation (3.8) over the time interval [0, t], we obtain

i RO)

0 1

Since, at time t = 0, R(t) = 1.

Evaluating the right-hand side of Equation (3.9) and then rearranging it yields

lnR(t) = -j(t)df (3.10)

o

Thus, from Equation (3.10), we obtain

R(t) = e » (3.11)

Thus, Equation (3.11) is the general reliability function. This equation can be used for obtaining the reliability function of a system/item when its times to failure follow any time-continuous probability distribution (e.g., exponential, Weibull, and Rayleigh).

Example 3.3

Assume that the hazard rate of an engineering system is expressed by Equation (3.1). Obtain an expression for the reliability function of the engineering system by using Equation (3.1).

By substituting Equation (3.1) into Equation (3.11), we obtain

  • -HpadSi)“"1«'®1*“ }<(( R(t) = e
  • -'I

<3.12)

Thus, Equation (3.12) is the expression for the reliability function of the engineering system.

Mean time to failure

The mean time to failure of a system/item can be obtained by using any of the three formulas presented below [4, 5]:

co

MTTF = ^R(t)dt

o

(3.13)

or

MTTF = limR(s)

(3.14)

or

00

MTTF = E(t) = Jf/(f)df

  • (3.15)
  • 0

where

MTTF is the mean time to failure of a system/item.

s is the Laplace transform variable.

R(s) is the Laplace transform of the reliability function R(t).

Eft) is the expected value.

Example 3.4

With the aid of Equation (3.3), prove that Equations (3.13) and (3.14) yield the same result for the engineering system mean time to failure.

By substituting Equation (3.3) into Equation (3.13), we obtain

(3.16)

where

MTTF-, is the engineering system mean time to failure.

By taking the Laplace transform of Equation (3.3), we get

R„(s) =

  • (3.17)
  • 0
  • 1
  • (s + Xcs)

where

R,, (s) is the Laplace transform of the engineering system reliability function RK (f).

By inserting Equation (3.17) into Equation (3.14), we obtain

MTTFet = lim —-— s->o (s + k«

As Equations (3.16) and (3.18) are identical, it proves that Equations (3.13) and (3.14) yield the same result for the engineering system mean time to failure.

 
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