1. RELIABILITY IN DESIGN
Prof. Dr. Ahmed Farouk Abdul Moneim

2. GENERAL EXPRESSIONS

3. A Structure or an element of a structure is said to be RELIABLE If its Strength, Capacity or Resistance is greater than applied External Loads
S is the Strength of the member
L is the applied load
The Strength is assumed Uncertain and considered as a Random Variable with
Pdf S = f(S). The Applied load is assumed also Uncertain with Pdf L = f(L)
Pdf S
Pdf L L
……(1)

4. Alternatively, Reliability may given in the following form:
Pdf L
Pdf S S
….(2)

5. R Important Special Cases Case 1: The LOAD is CONSTANT
The STRENGTH is RANDOM with Pdf S = f(S).
From equation (1) we have, R
PDF S L

6. R Case 2: The STRENGTH is CONSTANT
The LOAD is RANDOM with Pdf L= f(L).
From equation (2) we have, R
PDF L S

7. _____________________________________________________
Example 1
A steel column has the critical buckling Strength F C
Modulus of Elasticity E = 200 G Pa
Length of column l = 3 meters
One Pascal=One Newton/square meter
The Applied Load is Random distributed according to Weibull with β = 1.76 and
η = KN. Express the Reliability of the column.
Find the diameter of the Column that provides a reliability of 0.99
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f(L) S L

8. Example 2
The Strength of a structure is Random with Pdf:
The applied load is Random with Pdf:
Evaluate the Reliability of the structure
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9. STRENGTH AND LOADS ARE NORMALLY DISTRIBUTED

10. Then
Since S and L are Normal, then S – L = W is also Normal with the following parameters
Example Given:
Then

11. FACTOR OF SAFETY AS A FUNCTION OF RELIABILITY
Prof. Dr. Ahmed Farouk Abdul Moneim

12. The Factor of Safety is defined as the Ratio of Mean Strength to Mean Load:
The Coefficient of Variation of Strength and Load CVS and CVL
Therefore,
….(1)
Given : R, CV S and CV L
Required FS
Equation (1) may be rewritten as follows:

13. Determination of Factor of Safety that provides
A Cantilever beam of length 3 m. and of circular cross section with diameter D is subjected
To a vertical load L. If the target reliability should not be less than 0.9, what should be
the diameter of the beam if it will be manufactured from a material of yield point given below:
Load L
Yield Point
Modulus of Elasticity
Mean N.
270 MN / m2
205 G Pa
Standard Deviation
162 MN / m2
12.3 G Pa
Coeff. Of Var. (CV) 1
0.6
L = 4360 3
Determination of Factor of Safety that provides
Target Reliability of 0.9.
It is very important to notice that it is IMPOSSIBLE to design a structure with target reliability equals
UPPER LIMIT since Factor of Safety in this case will tends to INFINITY
2) Put Limit State Equation
Stress Generated due to Load <= Maximum Allowable Stress
Maximum Allowable Stress = Mean Yield / FS

14. Stress Generated due to Load <= Maximum Allowable Stress
3) Satisfy the Limit State Equation by selecting the appropriate Cross section
of the Beam
The Beam is under bending with the following Bending Moment Diagram
L = 4360 D 3
M max = L * length = 13080
Maximum Stress generated = M max / Section Modulus = M max / Z
Stress Generated due to Load <= Maximum Allowable Stress
Maximum Allowable Stress = Mean Yield / FS

15. DETERMINISTIC Loads and Strength s given: P L
Example 2
For the cantilever beam given in the previous example, what should be the diameter
If the deflection at its end should not exceed 3 mm.
From theory of structures, equation of deflection of cantilever beam in case of
DETERMINISTIC Loads and Strength s given: P L X δ
… (1)
… (2) y D Z
I ZZ is the Area Moment of Inertia of the beam
cross section about Z-axis
The effect of Randomness of Load and strength is accounted by Multiplying
the Load P by the Factor of Safety FS. Therefore (1) will become:
… (3)
Considering (2) in (3) , we find
Cross sectional area =

16. Stress Generated due to Load <= Maximum Allowable Stress
Repeat the previous exercise for a beam with square cross section with side W
Stress Generated due to Load <= Maximum Allowable Stress
Maximum Alowable Stress = Mean Yield / FS
…(1)
BM max = N.m
…(2) y Z
Therefore, from(1) and (2)
Remember for circular
As far as Bending strength is concerned the square cross section is more ECONOMIC
Reduction in weight =
10.5% reduction

17. The reduction in weight in case of square cross section amounts to
Concerning Rigidity, which cross section is more Economic?
Remember for circular cross section
The reduction in weight in case of square cross section amounts to
Reduction in weight =
The reduction in weight is 2.5%

18. REPEATED LOADS
Prof. Dr. Ahmed Farouk Abdul Moneim

19. The Applied Wind speed during a Gust is Weibull Variable:
Example 1
A building is subject to random (Poisson distribution) wind gusts at an average of TWO
Gusts per year. The building is so designed to withstand winds with speeds up to
100 mph. Wind speed during a gust is Random with Weibull distribution with
β = 2 and η = 40 mph. Determine the Building Reliability as function of time.
Find the Expected Life of the building (Mean Number of Years till Failure).
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The Applied Wind speed during a Gust is Weibull Variable:
Reliability in one Gust = Probability of Having applied Wind speed less than 100 mph
UDWS
The number of Gusts X in a specified period of time t is random and distributed
According to POISSON’s with mean λ t (λ = 2 Gusts / year)
The probability that the building could survive (Reliability RS) a specified period of time t :

20. Expected Life time = MTTF
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21. Example 2
An Oil Rig is subjected to severe sea waves thrusts. The number of these severe thrusts
Per year is a Poissonian variable with mean of THREE Thrusts per year. These waves
Propagates at random speed distributed according to Weibull with β = 3 and η = 10 m / sec
What is maximum wave speed that the rig may withstand in order to have at least
30 years of expected life.
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Smax

22. of completing an 8-hours shift without die failure.
Example 3
A die is designed to withstand a force of 35 KN. A hydraulic forge is exerting a force that
Is random and distributed according to exponential distribution with a mean of 4 KN.
If the forgings are made at the rate of one every 2 minutes, what is the reliability
of completing an 8-hours shift without die failure.
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The force exerted by the forging hammer in one stroke is Random and distributed
Exponentially with λ =1/mean = 1/4.
λ = 1/4
Failure probability in a single strike =
Then Reliability in a single Strike = 35
Load L
The probability that the die will not fail (will survive) after N Strikes will be = RN
Number of Strikes in an 8 hours shift = N = 8*60/2 = 240 Strikes
Probability of Survival (Reliability) after 240 Strikes = RN = ( )240 =
Probability that the die will complete the shift without failure =