Mechanical Engineering Reliability Case Study

1
Mechanical Engineering - Reliability Case Study

• Problem Definition
• How reliable is a product or system, how likely
  is it to fail?

2
The Bathtub Curve
Number of failures
Infant deaths
Old age deaths
Time
3
Number of Failures

• Expressed
• as a ratio - number per thousand of items
  manufactured - 80/1000
• or as a percentage - 8
• or as a decimal - 0.08 (probability),
  failuressuccesses1 pf0.08, ps0.92
• successesreliable products psr

4
Failure is Binary

• Although we will be expressing overall failure
  rates in one of these ways, remember that failure
  is binary. A product either fails or works. So
  each product is either a 100 success or a 100
  failure. These numbers talk of the experience of
  a large number of products.

5
Series Layout

• A typical system has a series layout
• All must work to drive.

Fuel
Ignition
Wheels
In
Out
Transmission
6
Parallel Layout

• Sometimes a system has the better configuration
  of a parallel layout
• Only one has to work to eat.

Stove
Microwave
In
Out
Barbeque
7
Estimates of Product Reliability

• MAT 2371/2377
• formal analysis to determine the overall
  reliability
• for 3 elements in series each having the
  reliability r, the total reliability rtotr3
• for 3 elements in parallel, rtotr3-3r23r.
• If r0.8, then for series rtot 0.512
• and for parallel rtot 0.992.

8
Computed Estimate

• We can simulate a production line and build
  systems and test them.
• We will use the random number generator to
  determine if a particular element in the system
  works or fails.

9
Revisiting rand()

• Remember (Lect07) that each call to the function
  rand, returns a different number in the random
  number sequence. The values from rand are in the
  range 0 to RAND_MAX
• We saw that we could generate a number in the
  range 0 to 1
• float p
• p ((float) rand()/RAND_MAX)

10
Deciding on Failure

• We can decide on the failure of a single part by
  saying
• float p, r0.8
• .
• .
• p ((float) rand()/RAND_MAX)
• if(pltr) //Success
• else //failure

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Procedure

• Lets look at 3 elements in series, if any one
  fails, the whole system has failed
• define OK 1
• define FAIL -1
• int i, flagOK
• for(i0 ilt3 i) //3 in series
• p ((float) rand()/RAND_MAX)
• if(pgtr)
• flagFAIL //failure

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Calculation Strategy

• 1. Use the loop above as the heart of the system.
• 2. Run the loop many times to generate the
  statistics - say 1000 times and count the number
  of successes, then you will have the success rate
  per 1000
• rtotcount/1000

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Procedure - cont.

• int j, count0 float rseries
• for(j0 jlt1000 j)
• flagOK
• //Above loop
• if(flagOK) count
• rseries(float) count/1000.0
• printf("Series reliability based on an element
  reliability f is f\n", r, rseries)

14
Testing

• Consider the effect of different numbers of
  trials.
• Consider different values of the elemental
  reliability r.
• Consider the use of srand().

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Experimental Extension

• After you have confidence in working with this
  Monte Carlo analysis
• Write for a parallel system.
• Write for a compound system.