EMOSH Reliability.ppt

Reliability
SMJ 4812 Project Mgmt and Maintenance
Eng.

Definition of Reliability
“Probability that a system or product will
perform in a satisfactory manner for a given
period of time when used under specified
operating condition”

Reliability - 4 main elements
1. probability – numerical representation -
number of times that an event occurs
(success) divided by total number trials
2. Satisfactory performance – criteria
established which describe what is
considered to be satisfactory system
operation

3. Specifed time – measure against which degree
of system performance can be related - used to
predict probability of an item surviving without
failure for a designated period of time
4. Specified operating conditions expect a system
to function - environmental factors, humidity,
vibration, shock, temperature cycle, operational
profile, etc.

LIFE CYCLE CURVE
• typical life history curve for infinite no of
items – ‘bathtub curve
• comparison of failure rate with time
• 3 distinct phase – debugging , chance failure
and wear-out phase

Debugging (Infant mortality) Phase
• rapid decrease in failure rate
• Weibull distribution with shape parameter  <
1 is used to describe the occurrences of failure
• Usually covered by warranty period

Chance failure phase
• Constant failure rate – failure occur in random
manner
• Exponential and also Weibull with  =1 can be used
to describe this phase

Wear-out phase
• Sharp rise in failure rate – fatigue, corrosion (old age)
• Normal distribution is one that best describes this
phase
• Also can use Weibull with shape parameter  > 1

Maintainability
• Pertains to the ease, accuracy, safety and
economy in the performance of maintenance
actions
• Ability of an item to be maintained
• Maintainability is a design parameter,
maintenance is a result of design

Measures of Maintainability
MTBM – mean time between maintenance, include
preventive and corrective maintenance
MTBR – mean time between replacement, generate
spare part requirement
- mean active maintenance time
ct – mean corrective maintenance time or mean time
to repair
pt – mean preventive maintenance time
M
M
M

• Frequency of maintenance for a given time is
highly dependent on the reliability of that
item
• Reliability frequency of maintenance
• Unreliable system require extensive
maintenance

Reliability function [R(t)]
• R(t) = 1 – F(t)
• F(t) = probability of a system will fail by time
(t) = failure distribution function
Eg. If probability of failure F(t) is 20%, then
reliability at time t is
R(t) = 1 – 0.20 = 0.80 or 80%

Reliability at time (t)
• R(t) = e-t/
• e = 2.7183
•  = MTBF
 = failure rate
• So,
R(t) = e-t


1


Failure Rate ()
• Rate at which failure occur in a specified time
interval
• Can be expected in terms of failures per hour, % of
failure per 1,000 hours or failures per million hours
= number of failures
total operating hours

Example 1
• 10 components were tested. The components
(not repairable) failed as follows:
Component 1 failed after 75 ours
Component 2 failed after 125 hours

Determine the MTBF
Solution:
Five failures, operating time = 3805 hours
75
125
130
325
525
5 x 525

Solution
 = 5 / 3805 = 0.001314

Example 2
20.2 6.1 7.1 24.4 4.2 35.3 1.8 46.7
Operating time Down time
2.5
a) Determine the MTBF.
Solution:
Total operating time = 20.2 + 6.1 + 24.4 + 4.2 + 35.3 + 46.7
= 136.9 hours
The chart below shows operating time and breakdown time
of a machine.

Solution
 = 4 / 136.9 = 0.02922
Therefore;
 = MTBF = 1/  = 34.22 hours
b) What is the system reliability for a mission time of 20 hours?
R = e-t t = 20 hours
R= e-(0.02922)(20)
R = 55.74%

Reliability Component Relationship
• Application in series network, parallel and
combination of both

Series Network
• Most commonly used and the simplest to
analyze
A B C
Input Output
All components must operate if the system is to function
properly.
R = RA x RB x RC
* The more components in arranged in series network, the lesser the reliability

• If the series is expected to operate for a
specified time period, then
• Rs (t) =
t
n
e )
...
( 3
2
1 


 





Example
• Systems expected to operate for 1000 hours. It
consists of 4 subsystems in series, MTBFA = 6000
hours, MTBFB = 4500 hours, MTBFC = 10,500 hours,
MTBFD = 3200 hours. Determine overall reliability.
A = 1 /MTBFA = 1/6000 = 0.000167
B = 1/MTBFB = 1/4500 = 0.000222
C = 1/MTBFC = 1/10500 = 0.000095
D = 1/MTBFD = 1/3200 = 0.000313
Therefore; R = e-(0.000797)(1000) = 0.4507

Parallel Network
• A number of the same components must fail
order to cause total system failure
A
B
C

Example
• Consider two units A and B in parallel. The
systems fails only when A and B failed.
A
B
Fs(t) = Fa(t) Fb(t)
= [1-Ra(t)][1-Rb(t)]
= 1-Ra(t) Rb(t) + Ra(t) Rb(t)
Rs(t) = 1- Fs(t)
= Ra(t) + Rb(t) – Ra(t) Rb(t)

• If A and B are constant failure rate units, then:
• Ra(t) = Rb(t) =
t
a
e t
b
e 

And Rs(t) =
b
a
b
a
s dt
t
R



 





1
1
1
)
(
0
s = MTBF

Consider 3 components in parallel
• Rs = 1 – Fs
• Fa = 1- Ra Fb = 1- Rb Fc = 1- Rc
• Rs = 1 – (1-Ra)(1-Rb)(1-Rc)
• If components A, B and C are identical, then
the reliability,
Rs = 1 – (1 – R)3
• For a system with n identical components,
Rs=1- (1-R)n
A
B
C

Combined series parallel network
A
B
C
Rs = RA [RB+RC-RBRC]

A
B
C
D
Rs = [1-(1-RA)(1-RB)][1-(1-RC)(1-RD)]

A
B
C
D
E
F
Rs=[1-(1-RA)(1-RB)(1-RC)][RD] x [RE+RF-(RE)(RF)]

• For combined series-parallel network, first
evaluate the parallel elements to obtain unit
reliability
• Overall system reliability is determined by
finding the product of all series reliability

EMOSH Reliability.ppt

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