RELIABILITY IN DESIGN Prof. Dr. Ahmed Farouk Abdul Moneim

GENERAL EXPRESSIONS

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)

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

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

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

_____________________________________________________ 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 = 3.558 KN. Express the Reliability of the column. Find the diameter of the Column that provides a reliability of 0.99 _____________________________________________________ f(L) S L

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

STRENGTH AND LOADS ARE NORMALLY DISTRIBUTED

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

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

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:

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 4360 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

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

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 =

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 = 13080 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

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%

REPEATED LOADS Prof. Dr. Ahmed Farouk Abdul Moneim

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). _____________________________________________________________________________ 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 :

Expected Life time = MTTF ______________________________________________________________________________

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. ______________________________________________________________________________ Smax

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. ___________________________________________________________________________ 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 = (0.9998415)240 = 0.96268 Probability that the die will complete the shift without failure = 0.96268