Reliability of Machine Components

Mean Time Between Failure - MTB and reliability of machine components and systems

Reliability

Reliability at a given time:

R = e^-^λt (1)

where

R = reliability. Values between 0 - 1 where value 1 indicates 100% live components and value 0 indicates 0% live components.

λ = proportional failure rate - a failure rate expressed as a proportion of initial number of live components - N_o

t = time (hours) - Note! other units can be used as long as the use is consistent through the calculations.

The failure rate can be expressed as

λ = N_F / N_ot

= N_o - N_s / (N_ot) (2)

where

N_F = N_o - N_s = number of failing components at time t

N_s = number of live surviving components at time t

N_o = initial number of live surviving components at time zero

Example - Failure Rate and Reliability

A car manufacturer sells 400000 cars of a certain model in one year. During the the first tree years the owners of 50000 of these cars experience major failures. The failure rate can be calculated as

λ = (50000 cars)/ ((400000 cars) (3 years))

= 0.042 (per year)

The reliability of a model of this car within three years can be calculated as

R = e^-(^{0.042 1/year) (3 year)}

= 0.88

= 88 %

Unreliability - Probability for a Device to Fail

The connection between reliability and unreliability:

R + Q = 1 (3)

where

Q = unreliability. Values between 0 - 1 where value 1 indicates 0% live components and value 0 indicates 100% live components.

(1) and (2) can be used to express unreliability

Q = 1 - e^-^λt (4)

Example - Unreliability

The unreliability of the car model in the example above within three years can be calculated as

Q = 1 - e^-(^{0.042 1/year) (3 year)}

= 0.12

= 12 %

Number of Live Components

The number of live surviving components in a system at a given time:

N_s = N_o e^-^λt (5)

Number of Failure Components

The number of failure dead components in a system at a given time:

N_s = N_o (1 - e^-^λt) (6)

Mean Time Between Failures - MTBF

Mean time between failures - MTBF:

MTBF = 1 / λ (7)

where

MTBF = Mean Time Between Failure (hours)

MTTF - Mean Time To Failure is an alternative to MTBF

mttf vs mtbf

Example - Mean Time Between Failure

From the example above the failure rate is 0.42 per year. MTBF can be calculated as

MTBF = 1 / (0.042 1/year)

= 23.8 years

Mean Time Between Failure (MTBF) can be determined by rating Total Surviving Hours against Number of Failures as

MTBF = t_s / N_F (8)

where

t_s = total surviving hours (hours)

Combining (5) with the formulas for reliability and more

R = e^-^t/m          (1b)

Q = 1 - e^-^t/m       (4b)

N_s = N_o e^-^t/m          (5b)

N_s = N_o (1 - e^-^t/m)       (6b)

Systems Reliability

There are two basic types of reliability systems - series and parallel - and combinations of them.

in a series system - all devices in the system must work for the system to work
in a parallel system - the system works if at least one device in the system works

Reliability systems - series and parallel

Reliability of Systems in Series

Reliability of systems in series can be expressed as

R_s = R₁ R₂ (9)

where

R_s = system reliability

R_1,2 = subsystem reliability

Example - Reliability of Systems in Series

From the example above the reliability of a car over three year is 0.88. If we depends on two cars for a mission - whatever it is - the reliability will be

R_s = 0.88 0.88

= 0.77

= 77 %

Reliability of Systems in Parallel

Reliability of systems in parallel can be expressed as

R_s = R₁ Q₂ + R₂ Q₁ + R₁ R₂ (10)

where

Q_1,2 = (1 - R_1,2) subsystem unreliability

Example - Reliability of Systems in Parallel

From the example above the reliability of a car over three year is 0.88. If we require at least one of our two cars for our mission - the reliability can be calculated as

R_s = 0.88 (1- 0.88) + 0.88 (1- 0.88) + 0.88 0.88

= 0.99

= 99 %

Note! - two systems in parallel with 99 % reliability vs. one system alone with reliability 88 % shows the power of redundancy.