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

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) L Pdf S Pdf L ……(1) Prof. Dr. Ahmed Farouk Abdul Moneim

4. Alternatively, Reliability may given in the following form: Pdf S Pdf L S ….(2) Prof. Dr. Ahmed Farouk Abdul Moneim

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

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

7. Example 1A steel column has the critical buckling Strength F C Modulus of Elasticity E = 200 G Pa Length of column l = 3 meters The Applied Load is Random distributed according to Weibull with β = 1.76 and η = 3.558 MN. Express the Reliability of the column. Find the diameter of the Column that provides a reliability of 0.99 _____________________________________________________ S One Pascal=One Newton/square meter Prof. Dr. Ahmed Farouk Abdul Moneim

8. Example 2The Strength of a structure is Random with Pdf: The applied load is Random with Pdf: Evaluate the Reliability of the structure _____________________________________________________________________

9. STRENGTH AND LOADS ARE NORMALLY DISTRIBUTED Prof. Dr. Ahmed Farouk Abdul Moneim

10. Then Since S and L are Normal, then S – L = W is also Normal with the following parameters 0 Example Given: Then Prof. Dr. Ahmed Farouk Abdul Moneim

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

12. Therefore, Given : R, CV S and CV L Required FS ….(1) Equation (1) may be rewritten as follows: The Factor of Safety is defined as the Ratio of Mean Strength to Mean Load: The Coefficient of Variation of Strength and Load CV S and CV L Provided that!!!! Prof. Dr. Ahmed Farouk Abdul Moneim

13. 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 LYield PointModulus of Elasticity Mean4360 N.270 MN / m 2 205 G Pa Standard Deviation4360 N.162 MN / m 2 12.3 G Pa Coeff. Of Var. (CV)10.6 L = 4360 3 1)Determination of Factor of Safety that provides Target Reliability of 0.9. 2) Put Limit State Condition Stress Generated due to Load <= Maximum Allowable Stress Maximum Alowable Stress = Mean Yield / FS Prof. Dr. Ahmed Farouk Abdul Moneim

14. 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 3 M max = L * length = 13080 Maximum Stress generated = M max / Section Modulus = M max / Z D Stress Generated due to Load <= Maximum Allowable Stress Maximum Alowable Stress = Mean Yield / FS Prof. Dr. Ahmed Farouk Abdul Moneim

15. 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 δ I ZZ is the Area Moment of Inertia of the beam cross section about Z-axis D y Z X … (1)… (2) Considering (2) in (3), we find 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) Cross sectional area = Prof. Dr. Ahmed Farouk Abdul Moneim

16. 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 M max = 13080 N.m y Z Therefore, from(1) and (2) …(2) …(1) Remember for circular As far as Bending strength is concerned the square cross section is more ECONOMIC Reduction in weight = About 10% reduction Prof. Dr. Ahmed Farouk Abdul Moneim

17. 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 The reduction in weight is more than 2% Reduction in weight = Prof. Dr. Ahmed Farouk Abdul Moneim

19. Example 1 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. ___________________________________________________________________________ The force exerted by the forging hammer in one stroke is Random and distributed Exponentially with λ =1/mean = 1/4. λ = 1/4 35 Load L Failure probability in a single strike = Then Reliability in a single Strike = The probability that the die will not fail (will survive) after N Strikes will be = R N Number of Strikes in an 8 hours shift = N = 8*60/2 = 240 Strikes Probability of Survival (Reliability) after 240 Strikes = R N = (0.9998415) 240 = 0.96268 Probability that the die will complete the shift without failure = 0.96268 Prof. Dr. Ahmed Farouk Abdul Moneim

20. Example 2 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 = 50 mph. Determine the Building Reliability as function of time. Find the Expected Life of the building (Mean Number of Years till Failure). _____________________________________________________________________________ The Wind speed during a Gust is Weibull Variable: 100 mph Reliability in one Gust = Probability of Wind speed is less 100 mph 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 R S ) a specified period of time t : Prof. Dr. Ahmed Farouk Abdul Moneim

21. Expected Life time = MTTF ______________________________________________________________________________ Example 3 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 = 40 m / sec What is maximum wave speed that the rig may withstand in order to have at least 30 years of expected life. ______________________________________________________________________________ Smax Prof. Dr. Ahmed Farouk Abdul Moneim