Lecture 11 - Time Dependent Failure Models Unit 11...

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Unformatted text preview: Time Dependent Failure Models Unit 11 Quantitative Risk Analysis in Safety Engineering Spring 2014 1 Time Dependent Failure Mode In general, λ (ROCOF) values in an engineering system are not constant, so a ;me dependent model is needed that is versa;le and can be approximated as ;me independent under some condi;ons and periods of ;me. b λ (t) = at , with a > 0 where a linear hazard rate func;on at is generalized by a power func;on with 2 parameters atb, which is linear for b = 1 and λ is constant (exponen;al) for b = 0. Rewrite atb to simplify the associated reliability func;on: β ⎛ t⎞ λ (t) = ⎜ ⎟ θ ⎝θ ⎠ β -1 ; θ > 0, β > 0 ; t ≥ 0 where β is the shape parameter. θ, the scale parameter, is called the characteris)c life, which =1/λ for β = 1 . 2 Reliability Function Using this form for the hazard rate func;on λ(t), the Weibull distribu;on is derived from the basic expression for R(t). t R(t) = e ∫0 − λ (t) dt β ⎛ t⎞ 0θ ⎜θ ⎟ ⎝ ⎠ =e ∫ − t β −1 dt = e ⎛ t⎞ −⎜ ⎟ ⎝θ ⎠ β Simple R(t), which reduces to exponen;al for β = 1, λ = 1/θ Note for t = θ, characteris;c life: R(t = θ ) = e ⎛θ ⎞ –⎜ ⎟ ⎝θ ⎠ β –1 = e = 0.37 So the scale parameter of the Weibull is the ;me t = θ at which 1–0.37 = 63 % of the failures have already occurred for all values of the shape parameter, β. For β = 1, 1/λ = MTTF, − MTTF MTTF e = 0.37 3 Graph of the Reliability Function, R(t) R(t) 1.2 Beta Shape Parameter θ = 2 R(t) 0.5 1 1.5 2.0 0.8 4.0 Same characteris;c life for all β 0.6 ⎛θ ⎞ –⎜ ⎟ ⎝θ ⎠ β θ = t, e = e-­‐1 = 0.37 0.4 Exponen;al for β = 1 0.2 0 0.0 -0.2 1.0 2.0 3.0 4.0 5.0 6.0 Time, t The effect of the shape parameter β for the Weibull reliability func;on. t 4 Weibull, F(t): Effect of β θ = 2 θ = t, 1-­‐ e-­‐1 = 0.63 At t = θ, Reliability has dropped by a factor of 1/e = 2.72 Both shape and scale parameters are required for failure distribution characterization, which is better understood from the parameter values based on observation and measurement of a 5 component during system monitoring or in field tests. Weibull Probability Density Function: Effect ofβon f(t) f(t) 1.0 β < 1, similar to exponen;al β = 1, exponen;al, λ = 1/θ 1 < β < 3: skewed 3 < β, closer to symmetric 0.9 0.8 0.7 0.5 0.6 β 0.5 1.5 2.0 4.0 4.0 0.5 θ = 2 2.0 0.4 1.5 0.3 0.2 0.1 t 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 -0.1 dR(t) β ⎛ t⎞ − = f(t) = ⎜ ⎟ dt θ ⎝θ ⎠ β -1 e ⎛ t⎞ –⎜ ⎟ ⎝θ ⎠ β = λ (t)⋅R(t) 6 The Hazard Rate Function, λ(t) Of special importance is representa;on of the wide ranges of increasing failure rate behavior: nega;ve slope, decreasing posi;ve slope (concave), constant posi;ve slope (Rayleigh), or increasing posi;ve slope (convex). 6 θ = 2 β, shape parameter λ(t) f(t) / R(t) 0.5 5 1.5 2.0 Hazard Rate 4 4.0 Convex, β > 2 3 increasing posi;ve slope 2 LFR, Rayleigh distribu;on, β = 2 Concave, 1 < β < 2 1 constant posi;ve slope decreasing posi;ve slope nega;ve slope, β < 1 t 0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 IFR for β > 1; DFR for β < 1; exponen;al, CFR, for β = 1 7 Weibull Shape Parameter Effect onλ(t) •  Value Property •  0 < β < 1 Decreasing Failure Rate (DFR) •  β = 1 Exponen;al Distribu;on (CFR), λ(t) = λ •  1 < β < 2 IFR-­‐Concave failure rate •  β = 2 Rayleigh Distribu;on (LFR, linear failure rate) •  β > 2 IFR – Convex failure rate •  3 < β < 4 IFR – close to Normal Distribu;on -­‐ symmetrical β = 3.4393 Most closely approximates the normal distribu;on β = 3.4394 Mean = median as with normal distribu;on The shape of the distribu;on represents the failure process of the component or system as it develops through ;me. 8 Weibull Characteristic Life: Effect ofθon f(t) Basic Shape is set by β and is unaffected by θ 1.6 f(t) f(t) β = 1.5 1.4 θ 1.2 0.5 0.5 1 1.0 2.0 0.8 1.0 0.6 Variance increases as θ is increased 0.4 2.0 0.2 t 0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 The characteris;c life = θ (scale parameter) value changes the mean, medium, mode, and variance or spread of the 9 distribu;on but not its basic shape that is set by its β. Weibull Characteristic Life: Effect ofθon R(t) R(t) β ~ 2 R(t) broadens and reliability decreases more slowly as θ is increased but it has same basic shape set by β. 10 Weibull Characteristic Life: Effect ofθon λ(t) λ(t) β = 2, Rayleigh distribu;on, linear IFR The slope of λ(t) decreases as θ is increased but the shape set by β remains the same. 11 Weibull MTTF from θ, β, and the Gamma Function ⎛ 1⎞ MTTF = ∫ t f(t)dt = θ Γ ⎜ 1+ ⎟ 0 ⎝ β⎠ ∞ Derived in IRME, App. 4A, p. 87 ∞ Γ(x) = the gamma function = ∫ 0 y x-1e-ydy IRME, Tab A.9, p. 531 Γ(x) = (x - 1)Γ(x - 1) for x >0 Γ(x) = (x - 1)! for x a positive integer 1 lim MTTF = lim θ Γ(1+ ) = θΓ(1) = θ β→∞ β→∞ β The mean approaches the characteris;c life as β increases, because the Weibull f(t) becomes symmetrical. 12 Gamma Function - Selected Values IRME, A.9. p. 531 x 1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.6 1.61 1.62 1.63 1.64 1.67 1.68 1.69 Gamma(x) 0.88757 0.88818 0.88887 0.88964 0.89049 0.89142 0.89243 0.89352 0.89468 0.89592 0.89724 0.89864 0.9033 0.905 0.90678 x 2.23 2.24 2.25 2.26 2.27 2.28 2.29 2.3 2.31 2.32 2.33 2.34 2.35 Gamma(x) 1.12023 1.12657 1.133 1.13954 1.14618 1.15292 1.15976 1.16671 1.17377 1.18093 1.18819 1.19557 1.20305 13 Example Problems Let T = a random variable, the time to failure of a circuit card. T has a Weibull distribution with β = 0.5 and θ = 500 (thousands of hours). Find R(50,000) and the MTTF. 0.5 R(50) = e ⎛ 50 ⎞ −⎜ ⎟ ⎝ 500 ⎠ = 0.729 ( ) MTTF = 500Γ 1+ 2 = 500(2) = 1000 Note DFR because β < 1 Let T = a random variable, the time to failure of a fuse. T has a Weibull distribution with beta = 1.5 and theta = 500 (K hours). Find R(50,000) and the MTTF. ⎛ 50 ⎞ R(50) = e− ⎜ ⎝ 500 ⎟ ⎠ 1.5 = 0.969 ⎛ 2⎞ MTTF = 500Γ ⎜ 1+ ⎟ = 500(0.903) = 451 ⎝ 3⎠ Note IFR because β > 1 14 The Variance and Standard Deviation 2 ⎧ ⎛ 2⎞ ⎡ ⎛ 1⎞ ⎤ 2 2 ⎪ σ = θ ⎨ Γ ⎜ 1+ ⎟ - ⎢ Γ ⎜ 1+ ⎟ ⎥ ⎝ β⎠ ⎣ ⎝ β⎠⎦ ⎪ ⎩ Γ(x) = (x − 1)! ⎫ ⎪ ⎬ ⎪ ⎭ for x = positive integer Variance and σ decrease as β increases, because Weibull f(t) becomes symmetrical as β increases to within (3,4). Derive the variance and σ for exponen;al with β = 1: σ2 = θ2(2 -­‐ 1) = θ2 = (1/λ)2 as is known for exponen;al 15 Example Problem - Standard Deviation θ = 500 θ = 500 β = 1/2 2.33 β = 1.5 σ 2 = 5002 [Γ(5) − 22 ] σ 2 = 5002 [Γ(1+ 4 / 3) − 0.90332 ] = 500 [24 − 4] = 5,000,000 σ = 2236 (thousands of hr.) = 5002 [1.18819 − 0.8160] = 93,048 σ = 305 (thousands of hr.) 2 ⎧ ⎛ 2 ⎞ ⎡ ⎛ 1 ⎞ ⎤2 ⎪ 2 = θ 2 ⎨ Γ ⎜ 1+ ⎟ - ⎢ Γ ⎜ 1+ ⎟ ⎥ σ ⎪ ⎝ β⎠ ⎣ ⎝ β⎠⎦ ⎩ ⎫ ⎪ ⎬ ⎪ ⎭ 16 Design Life and Median set then and R(t) = e t -⎛ θ ⎞ ⎜ ⎟ ⎝ ⎠ β =R tR = θ(- ln R) a targeted value 1 β t 0.50 = t med = θ(-ln0.5) 1 β Note that when β = 1, the expressions for exponen;al appear, showing again that Weibull is an extension of exponen;al! 17 Weibull Mode f(t mode ) = MAX t≥0 1 ⎧ ⎛ 1⎞ β ⎪ θ 1⎪ t mod e = ⎨ ⎜ β⎟ ⎝ ⎠ ⎪ ⎪0 ⎩ f(t) for β > 1 for β ≤ 1 Derived in Appendix 4B exponen;al Recall tmode = 0 for exponen;al, for β = 1 18 Example - Design Life, Median, & Mode θ= 500 1/β β = 0.5 t0.9 = 500 (-ln 0.90)2 = 5.55 tmed = 500 (0.69315)2 = 240 tmode = 0, (β<1) θ= 500 β = 1.5 = 3/2 t0.9 = 500 (-ln 0.90)2/3 = 112 tmed = 500 (0.69315)2/3 = 392 tmode = 500 [1-2/3)]2/3 = 240 Sameθ, but notice the major difference due to different β 19 Weibull Conditional Reliability ⎛ t+T ⎞ −⎜ 0 ⎟ ⎝ θ ⎠ R(T0 + t) e R(t | T0 ) = = β ⎛ T0 ⎞ R(T0 ) −⎜ ⎟ e ⎝θ⎠ β ⎡ ⎢ ⎛ t+T0 ⎞ = exp ⎢ − ⎜ θ ⎟ ⎠ ⎢⎝ ⎣ β ⎛ T0 ⎞ + ⎜ ⎟ ⎝θ⎠ β ⎤ ⎥ ⎥ ⎥ ⎦ Note that when β = 1, the exponen;al R(t) = exp(-­‐λt) appears due to lack or memory of the exponen;al! Find R(50|50) for Weibull with β = 0.5, θ = 500 20 Example - Conditional Reliability θ= 500 β = 0.5; R(50) = 0.7289 θ= 500 β = 1.5; R(50) = 0.969 R(50|50) = R(100)/R(50) = R(50|50) = R(100)/R(50) = exp[-(100/500)0.5] / 0.7289 exp[-(100/500)1.5] / 0.969 = 0.6394 /0.7289 = 0.8772 = 0.9144 / 0.969 = 0.9437 Sameθ, but notice the significant shift due to different β 21 Failure Modes Recall that a system composed of several independent failure modes in series each with a Weibull failure distribu;on has a system hazard rate func;on equal to the sum of the condi;onal failure mode failure rate func;ons. If all failure modes have the iden;cal shape parameter, then the system also will have a Weibull distribu;on. β⎛ t⎞ λS (t) = ∑ ⎜ ⎟ i=1 θ i ⎝ θ i ⎠ n β -1 ⎡ n ⎛ 1⎞β⎤ β β -1 = β t β -1 ⎢ ∑ ⎜ ⎟ ⎥ = t ⎢ i=1 ⎝ θ i ⎠ ⎥ θ S ⎣ ⎦ which is a Weibull hazard rate func;on with a shape parameter β and a characteris;c life for the system, θS ⎡ t = θS = ⎢ ⎢ ⎣ ⎛ 1⎞ ⎤ ∑ ⎜ θi⎟ ⎥ ⎠ ⎥ i=1 ⎝ ⎦ n β − 1 β 22 MTTF and tmed for Jet Engine A jet engine is constructed from five modules each having a Weibull failure distribu;on with β = 1.5 (IFR, concave) and characteris;c life values (opera;ng cycles) of 3,600, 7,200, 5,850, 4,780, and 9,300. Calculate the MTTF, median TTF, the reliability of the engine. ⎡ n ⎛ 1⎞ θS = ⎢ ∑ ⎜ ⎟ ⎢ i=1 ⎝ θ i ⎠ ⎣ β ⎤ ⎥ ⎥ ⎦ − 1 β ⎡⎛ 1 ⎞ 1.5 ⎛ 1 ⎞ 1.5 ⎛ 1 ⎞ 1.5 ⎛ 1 ⎞ 1.5 ⎛ 1 ⎞ 1.5 ⎤ ⎥ θS = ⎢ ⎜ +⎜ +⎜ +⎜ +⎜ 3,600 ⎟ 7,200 ⎟ 5,850 ⎟ 4,780 ⎟ 9,300 ⎟ ⎥ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎢⎝ ⎣ ⎦ ⎛ 1⎞ ⎛ 2⎞ MTTF = θ Γ ⎜ 1+ ⎟ = 1,842.7Γ ⎜ 1+ ⎟ = 1,664.5 cycles ⎝ β⎠ ⎝ 3⎠ − 1 / 1.5 = 1842.7 Note effect of skew to higher t values. 1 β 1/1.5 = 1,443.2 cycles t 0.50 = t med = θ(-ln0.5) = 1,842.7(0.69315) t R(t) = e ∫0 − λ (t) dt ⎛ t⎞ 0θ ⎜θ ⎟ ⎝ ⎠ =e ∫ − tβ β −1 dt = e ⎛ t⎞ −⎜ ⎟ ⎝θ ⎠ β =e ⎛ t ⎞ −⎜ ⎝ 1,842.7 ⎟ ⎠ 1.5 23 Identical Weibull Components If n components have iden;cal hazard rate func;ons with the same scale and shape parameters: then β⎛ t⎞ λ (t) = ⎜ ⎟ θ ⎝θ ⎠ β⎛ t⎞ λS (t) = ∑ ⎜ ⎟ θi ⎝ θi ⎠ i=1 n and t β -1 β -1 = , nβ θ β (t ) β -1 ∫0 λ (t) dt = −n⎛ θt ⎞ ⎜ ⎟ RS (t) = e e ⎝ ⎠ − β 24 Student Exercise - Weibull •  A certain brand lightning arrester has a Weibull failure distribution with a shape parameter of 2.4 and a characteristic life of 10 years. Find: •  a. R(5 yrs) •  b. MTTF •  c. Standard deviation •  d. Median and Mode •  e. 99% (B1) and 95% B(5) design life •  f. R(5|5) 25 Student Exercise - Solution a. R(5) = e ⎛ 5⎞ −⎜ ⎟ ⎝ 10 ⎠ 2.4 = 0.827 b. MTTF = 10 Γ(1+ 1/ 2.4) = 10 Γ(1.42) 10(0.88636) = 8.86 yrs { c. σ 2 = 10 2 Γ(1+ 2 / 2.4) − 0.88636 2 } or σ = 3.93 yr. 26 Student Exercise - Solution (continued) ( ) = 10 (1− 1/ 2.4 ) d. t med = 10 0.69315 t mode t 0.95 = 8.6 yr. 1/2.4 ( ) = 10 ( − ln.95 ) e. t.0.99 = 10 − ln.99 1/2.4 = 8.0 yr. 1/2.4 = 1.5 yr. 1/2.4 = 2.9 yr. 27 Student Exercise - Solution (continued) R(t + T0 ) f. Recall: R(t | T0 ) = R(T0 ) ⎛ 10 ⎞ −⎜ ⎟ ⎝ 10 ⎠ 2.4 R(10) e 0.3679 R(5 | 5) = = = = 0.445 R(5) 0.8274 0.8274 28 Time-­‐Dependent 2-­‐Parameter Reliability Models •  Weibull Distribu;on: A versa;le model with a shape parameter that suggests the progression of a degrada;on or failure process. Defined for posi;ve values of t. •  Normal (Gaussian) Distribu;on for failures that are primarily the result of small (not dominant), addi;ve effects. Defined for posi;ve and nega;ve values of t. •  Lognormal Distribu;on: Defined for posi;ve values of t and especially suitable when failures are the result of mul;plica;ve effects. Similar to the Weibull in versa;lity. •  Gamma Distribu;on: A versa;le model similar to Weibull, including defined for posi;ve values of t. In the same family of distribu;ons as exponen;al and Poisson, so gamma is useful as a prior distribu;on for Bayesian parameter upda;ng. 29 The Normal Probability Density Function(PDF) f(t) = 1 2π σ e 1 (t-µ ) − 2 σ2 2 ,- ∞ < t < ∞ MTTF = µ, location parameter Std Dev = σ, scale or spread parameter There is no shape parameter, because the Normal pdf always has a symmetric shape, but has different scale parameters, σ! 30 Normal Conditional Failure Rate, Effect of σ σ = 0.5 λ(t) = f(t) / R(t) IFR σ = 1 The normal condi;onal failure rate func;on is always increasing, so it represents only the IFR region of component life and cannot be used for modeling defects that are gradually removed in the DFR region. 31 Normal Distribution - Applications •  Useful for random stresses over time, such as the additive effects of temperature variation, material wear, and friction leading to increasing conditional failure rate, λ(t), along with Weibull β>1 (concave, linear, or convex behavior cases) •  Tool failures •  Brake lining wear •  Tire tread wear 32 Finding Normal Cumulative Probabilities If T is normally distributed, transform T to Z, where Z is the standard normal deviate T−µ z= σ Then Z has a standard normal distribu;on with a mean of 0 and a standard devia;on of 1. The pdf for Z is given by φ(z) = 1 2π 2 e −z 2 The cdf for failure = F(t), cumula;ve probability of failure to ;me = t, is then given by z -∞ P{Z ≤ z} = Φ(z) = ∫ φ(z) dz 33 Normal Probability Tables •  •  •  •  •  •  •  •  •  •  •  •  •  Z -0.55000 -0.54000 -0.53000 -0.52000 -0.51000 -0.50000 -0.49000 -0.48000 -0.47000 -0.46000 -0.45000 -0.44000 F(Z) 1-F(Z) 0.29116 0.29460 0.29806 0.30153 0.30503 0.30854 0.31207 0.31561 0.31918 0.32276 0.32636 0.32997 0.70884 0.70540 0.70194 0.69847 0.69497 0.69146 0.68793 0.68439 0.68082 0.67724 0.67364 0.67003 P{Z < -­‐ 0.5} = 0.30854 Area under pdf curve up to Z = -­‐ 0.5 P{Z > -­‐ 0.46 = 0.67724 Area under pdf curve > Z = -­‐ 0.46 34 Normal Reliability Function R(t) = ∫ ∞ t 1 2π σ e − (t−µ)2 2σ 2 dt Standardize to Z: Pr of fail for T ≥ t ZT Zt ⎧T − µ t − µ ⎫ R(t) = P{T ≥ t} = P ⎨ ≥ ⎬ = P(Z T ≥ Z t ) σ ⎭ ⎩ σ ⎧ ⎛ t − µ⎞ t−µ⎫ = P ⎨Z T ≥ ⎬ = 1− Φ ⎜ ⎟ = 1− Φ Z t σ ⎭ ⎝ σ ⎠ ⎩ ( ) Tables: A.1, pp. 514-­‐519 35 Example Problem - Normal •  The time to failure of a fan belt is normally distributed with a MTTF = 220 (in hundreds of vehicle miles) and a standard deviation of 40 (in hundreds of vehicle miles). t μ σ •  R(100) = 1 - F[ (100-220)/40] = 1 - F(-3) = 0.9987 •  R(200) = 1 - F[ (200-220)/40] = 1 - F(-0.5) = 0.6915 •  R(300) = 1 - F[ (300-220)/40] = 1 - F(2) = 0.02275 R(t|T0) •  R(100|200) = R(300) / R(200) = 0.02275 / 0.6915 = 0.0329 •  Note: the median = mode = MTTF = 22,000 miles 36 Normal Example problem - Design Life •  A new fan belt is developed from a higher grade of material. It has a time to failure distribution that is normal with a mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles. Find its design life if a 0.97 reliability is desired. •  •  •  •  R(t) = 1 – F[(t - 350)/70] = 0.97; find t ! From the normal CDF table, 1 - F(-1.88) = 0.970 Therefore; (t - 350 ) / 70 = -1.88 = Zt and t0.97 = 350 - 1.88 (70) = 218.4 or 21,840 vehicle miles Design life 37 Student Exercise - Normal •  The operating hours until failure of a halogen headlamp is normally distributed with a mean of 1200 hr. and a standard deviation of 450 hr. •  Find: •  a. The 5 year reliability if normal driving results in the use of the headlamp an average of 0.2 hr a day. •  b. The 0.90 design life in years. 38 Student Exercise - Solution •  a. t = 0.2 hr./da. x 365 day/yr x 5 yr = 365 hr. •  R(365) = 1 – F[(365 - 1200)/450] = 1 – F[-1.86] = 0.969 •  b. R(t0.90) = 0.90 or 1 – F[(t0.90 - 1200)/450] = 0.90 •  (t0.90 - 1200) / 450 = -1.28 •  t0.90 = 1200 - 1.28 (450) = 624 hr. or t0.90 = 624 / (0.2 x 365) = 8.5 yr. 39 some normal logs The Lognormal Failure Process Let T = a random variable, the time to failure. If T has a lognormal distribution, then the logarithm of T has a normal distribution. 40 40 Lognormal Density Function^, Effect of Shape Parameter, s f(t) = 1 2π s t e 1 ⎛ t ⎞ − 2 ⎜ ln ⎟ 2 s ⎝ tMED ⎠ 2 ;t ≥ 0 Defined for only posi;ve value of t. More symmetrical for smaller s values tmed = median ;me to failure, loca;on parameter s = shape parameter s = 0.1 s = 1 s = 0.5 ^Lognormal pdf is some;mes represented in terms of the mean and std. dev. of the underlying normal distribu;on. Here the parameters are median (more representa;ve than mean for skewed) and shape parameter s (~ σ). 41 Lognormal F(t), Effect of Shape Parameter, s s = 0.1 s = 0.5 s = 1.0 As s increases, the variance increases. Data represented by Weibull will owen be represented sa;sfactorily by lognormal. 42 Lognormal for Repair Frequency Majority of repair ;mes near Mode t med e Short repair ;mes s2 / 2 Long repair ;mes NASA, Lesson 0840 The lognormal distribu;on is widely used to describe the frequencies of system repair, because it reflects normal dura;on repair-­‐;mes, a large number of repairs closely grouped about a modal value, and long repair ;me data points with decreasing frequency in the tail. 43 Lognormal/Normal Relationship •  Given T is a lognormal random variable, then Log T space T space •  Distribu;on Lognormal Normal •  Mean •  Variance •  Mode s2 /2 t med e 2 s2 s2 med t e [e −1] ln tmed s2, σ = S t = t med ln tmed mode s2 e 44 Lognormal Failure and Reliability Distribution Pr of failure up to t F(t) = P {T ≤ t} = P {lnT ≤ lnt} loca;on parameter logic expression Standardize: ⎧ lnT - ln tMED lnt - ln tMED ⎫ = P⎨ ≤ ⎬ s s ⎩ ⎭ ZT Zt ⎧ ⎛1 1 t ⎫ t ⎞ = P ⎨z ≤ ln ⎬ = Φ ⎜ ln s tMED ⎭ ⎝ s tMED ⎟ ⎠ ⎩ Values in Std. Normal tables Pr of failure up to t ⎛1 t ⎞ R(t) = 1− Φ ⎜ ln ⎝ s tMED ⎟ ⎠ 45 Lognormal Design Life and Median Specify R Find Z in the normal table Calculate tR ⎛1 t ⎞ 1- Φ ⎜ ln R ⎟ = R ⎝ s t med ⎠ ⎛1 t ⎞ Φ ⎜ ln R ⎟ = 1- R ⎝ s t med ⎠ Find z1-­‐R = such that: Φ(z 1−R ) = 1- R 1 tR ln = Z1−R s t med Solve for tR t R = t med e s z1−R 46 Lognormal Conditional Failure Rate Function, λ(t) Effect of Shape Parameter, s 0.30 0.25 S = 0.4 tmed = 10 λ(t) HAZARD RATE 0.20 S = 0.6 0.15 S = 0.8 0.10 0.05 S = 1.0 0.00 1 6 11 16 21 S=.4 26 S=.6 31 S=.8 36 41 46 TIME S=1 λ(t) increases until it reaches a max. and then it gradually decreases. Smaller the s, the smaller the variance and longer time until the max. 47 Lognormal Conditional Failure Rate Function, λ(t) for tmed = 10 in previous slide Lognormal Mode, MTTF, λ(t) with several shape parameters: s t mode = t med / e 1.0 0.8 0.6 5.3 13.8 10 7.0 12.0 16 0.4 s2 /2 Mode 3.7 f(t): MTTF 16.5 Max λ(t) 7 8.5 10.8 20 beyond MTTF If for small values of s, the maximum point of λ(t) occurs at a large value of t rela;ve to the MTTF and mode, then the lognormal will represent an increasing λ(t) beyond MTTF, as shown above for s = 0.6, 0.4 48 Lognormal Example •  The failure distribution of an exhaust system is lognormal with tmed = 50,000 vehicle miles and s = 0.8. Therefore: •  a. MTTF = 50,000 S2 e0.64/2 = 68,856 mi. •  b. tmode = 50,000 / e0.64 = 26,640 mi. •  c. Variance = 50,0002 e0.64 [e0.64 - 1] and the standard deviation = 65,195 mi. 49 Lognormal Example (continued) Zt ⎛ 1 10,000 ⎞ R(10,000) = 1- Φ ⎜ ln ⎟ = 1− Φ −2.01 = 0.9770 ⎝ 0.8 50,000 ⎠ ( S ) tmed ⎛ 1 20,000 ⎞ R(20,000) = 1- Φ ⎜ ln ⎟ = 1− Φ −1.15 = 0.8749 ⎝ 0.8 50,000 ⎠ ( ) Calculate the condi;onal reliability: R(10,000 | 10,000) = R(20,000) / R(10,000) = 0.8749 / 0.9770 = 0.8955 50 Example (continued)   Find the design life corresponding to a 90 percent reliability. ⎛ 1 t 0.9 ⎞ R(t 0.9 ) = 1- Φ ⎜ ln ⎟ = 0.90 ⎝ 0.8 50,000 ⎠ From the standard normal probability tables: ⎛ 1 t 0.9 ⎞ ⎜ 0.8 ln 50,000 ⎟ = −1.285 ⎝ ⎠ Zt = -­‐1.285 corresponds to Pr = 0.90 or t 0.9 = 50,000 e−1.285(0.8) = 17,886 mi. 51 Class Exercise - Lognormal •  Reliability testing of a new 1.6 liter automotive engine has resulted in a time to failure distribution that is lognormal with tmed = 100,000 mi. and s = 0.70. Find: •  a. R(36,000 mi.) •  b. MTTF and Std. Dev. •  c. R(100,000|36,000) •  d. t0.95 52 Lognormal Class Exercise - Solution •  a. R(36,000) = 1 – F[(1/0.7)ln(36,000/100,000] •  = 1 - F[ -1.46] = 0.92786 •  b. MTTF = 100,000 e0.49/2 = 127,762 mi. •  Var = 100,0002 e0.49 [ e0.49 -1] = 1.032 x 1010 •  Std Dev = 101,594 mi. •  c. R(64,000|36,000) •  = R(100,000)/R(36,000) •  = 5.0 / 0.92786 = 0.54 53 Class Exercise - Solution •  d. R(t0.95) = 0.95 •  1 - F[ (1/0.7) ln(t0.95 /100,000) ] = 0.95 •  (1/.7) ln(t0.95 /100,000) = -1.645 •  t0.95 = 100,000 e-1.645 x .7 = 31,616 mi. general approach: tR = t med e s z1−R 54 The Gamma as a Failure Distribu;on •  Density func;on, pdf with 2 parameters (and rela;on to previous nota;on for cases of k a posi;ve integer): •  (= k), shape parameter; α (= 1/λ), scale parameter t γ −1e − t/α f(t) = γ α Γ(γ ) γ, α > 0 a...
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  • Spring '14
  • WillianJ.Roge
  • Normal Distribution, Failure rate, Weibull, Weibull distribution, distribu

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