Reliability Engineering: Failure rate, preventive maintenance, Weibull distribution, Markov differential equations

This article covers topics such as failure rate, preventive maintenance, Weibull distribution, and Markov differential equations in Reliability Engineering. It includes solved problems and calculations. The subject is relevant for students studying courses in engineering, statistics, and operations research. The content is suitable for college and university students.

Introduction to Reliability Engineering
1. The failure rate for a high-speed fan is given by λ (t)=(2.3 ×10−4 +3.2× 10−6 t)/hr, where t is in
hours of operation. The required design-life reliability is 0.96.
a) The required hours of operation for 0.96 reliability is given as:
0.96> 2.3× 10−4 +3.2 ×10−6 t ∴ t< 0.96−2.3 ×10−4
3.2 ×10−6 hours= 2399425
3 =299928.125 hours
b) If, preventive maintenance is taken, the wear contributions to the failure rate can be eliminated.
The hours of operation is given by:
t= 0.96
3.2× 10−6 =300000 hours
c) If the fan is placed in a controlled environment, the contribution of random failure to λ(t)
reduces by a factor of two. Assuming preventive maintenance is negrected. The time of
operation is:
t <0.96−2.3 ×10−4
1.2 ×10−6 hours=2399425
3 =799808.333 hours
d) The extended design life with both preventive maintenance and controlled environment is given
as:
t= 0.96
1.2× 10−6 =800000 hours
2. A water pump on a car engine has a Weibull failure distribution with a shape parameter of 1.89 and a
scale parameter of 475,000 miles.
a) The pump should be replaced for 98 percent reliability is:
.98 designlife :td=475000 × ( ln ( 0.98 ) )
1
1.89 =60267.7416 ≅ 60268
b) Based on the replacement interval in (a) above, the pump reliability at 100,000 miles:
With and without replacement:
R ( 100000 ) =e−( 100000
475000 )
1.89
=0.948752
Rm ( 100000 ) =R ( 60268 ) R ( 100000−60268 ) =( e− ( 60268
475000 )
1.89
)( e−( 100000−60268
475000 )
1.89
) =0.971033

3. A brake assembly has been tested and found to have a capacity that is normally distributed with a
mean of 275 lb. and a standard deviation of 25 lb. If a normally distributed force having a mean of
180 lb and a standard deviation of 30 lb. is applied, what is the reliability? As the variability
(variance) of either the capacity or load increases, what happens to reliability?
The reliability is product of capacity and strength. The static reliability has a distribution which is a
product of normal distribution:
i.e.:
D( x )=∫
−∞
∞
( 1
√2 π σ1
e
−
( ( x−μ1 )2
2 σ1
2 ). 1
√2 π σ2
e
−
( ( x−μ2 )2
2 σ 2
2 ) )dywhere : μ1=275 μ2=180σ 1=25 σ2=30
D ( x ) = 1
2 π (25 ×30) ∫
−∞
∞
( e
−( ( y−275 ) 2
2 × 252 )− ( ( y−180 ) 2
2 ×302 )) dy= 150 √ 122 π
45750 π e
361
122
=2.497076
1500 π ≅ 5.29896014 ×10−4
4. Set up the associated Markov differential equations in matrix form for the following system. Do not
solve.
λi=Failure rates
v 1=Repair rate ¿ State−3 ¿ State−1
v 2=Repair rate ¿ State−3 ¿ State−0
( λ1+ λ3−v2 ) P0− ( λ4 +v1 ) P1− ( λ3 + λ5 ) P2−v2 P3=0 λ1 P0 + ( λ1−λ2 +v1 ) P1− λ3 P2+ v2 P3=0
λ3 P0− ( λ3 + v2 ) P2− ( λ2 + λ3 −λ5 ) P3=0 ( λ4 +λ5 ) P1 + ( λ5−v2 ) P3
Therefore the Markov differential equations in matrix form is given by:

C=¿