# Reliability of Machine Components

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

Reliability characterizes components or system of components by the probability they will perform the desired functions for a given time.

In general -

- more components and/or more complicated systems reduces reliability
- simpler systems with few components increases reliability

Reliability equations:

### 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*

_{o }t_{ }* = N_{o} - N_{s} / (N_{o }t) (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

#### 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 of Systems in Series

Reliability of systems in series can be expressed as

*R _{s} = R_{1} R_{2} (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} = R_{1} Q_{2} + R_{2} Q_{1} + R_{1} R_{2} (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.

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