**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...

View
Full Document

- Spring '14
- WillianJ.Roge
- Normal Distribution, Failure rate, Weibull, Weibull distribution, distribu