**Unformatted text preview: **Time Dependent Failure Models Unit 11
Quantitative Risk Analysis
in Safety Engineering
Spring 2016 1 References
• Ebeling, C.E., Introduction to Reliability and Maintainability
Engineering, 2nd ed, Waveland Press, 2010, Chapter 4
(Ebeling, 2010)
• Modarres, M., M. Kaminskiy, V. Krivtsov, Reliability Engineering and Risk
Analysis, 2nd ed, Taylor & Francis, 2010 (Modarres, RERA) • Jordaan, Ian, Decisions Under Uncertainty– Probabilistic Analysis for
Engineering Decisions, Cambridge University Press, 2005 (Jordaan, 2005) • Modarres, M., Risk Analysis in Engineering, Taylor&Francis, 2006
(Modarres, RAE) • Modarres, M., Reliability Engineering and Risk Analysis in Engineering,
Marcel Dekker, 1999 (Modarres, RE) • O’Connor, P.D.T., Practical Reliability Engineering, 4th ed, Wiley, 2002
(O’Connor, PRE) • Rausand, M. and Hoyland, A., System Reliability Theory, 2nd edition,
Wiley, 2004
2 Time Dependent Failure Mode
In general, λ (ROCOF) values in an engineering system are not constant, so a @me dependent failure rate model is needed that is [email protected] and can be approximated as @me independent under some [email protected] and periods of @me. λ (t) = at b , with a > 0 where a [email protected] rate [email protected] λ(t) is a power [email protected] with 2 parameters atb, which is linear for b = 1, and λ is constant ([email protected]) for b = 0. Rewrite the 2-‐parameter atb to simplify the corresponding reliability [email protected]: where β ⎛ t⎞
λ (t) = ⎜ ⎟
θ ⎝θ ⎠ β -1 ; θ > 0, β > 0 ; t ≥ 0 β is the shape parameter. θ, the scale parameter, is called the characteris)c life, which = 1/λ for β = 1 . 3 Reliability Function
Using this form for the hazard rate [email protected] λ(t), the Weibull [email protected] is derived from the basic expression for R(t). Recall! 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 [email protected] for β = 1, λ = 1/θ Note for t = θ, [email protected] 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
We discussed this result for [email protected] in Unit 8. e = 0.37 4 Graph of the Reliability Function, R(t)
R(t) 1.2 Beta Shape Parameter values θ = 2 R(t) 0.5 1 R(t) = e 0.8 ⎛ t⎞
−⎜ ⎟
⎝θ ⎠ β 1.5
2.0
4.0 Same [email protected] life for all β 0.6 ⎛θ ⎞
–⎜ ⎟
⎝θ ⎠ β θ = t, e = e-‐1 = 0.37 0.4 [email protected] for β = 1 0.2 0
0.0 -0.2 1.0 2.0 3.0 4.0 5.0 6.0 Time, t t The eﬀect of the shape parameter β [email protected] in a wide range of shapes for the Weibull reliability [email protected] Recall: [email protected] [email protected] has no shape parameter and only one shape. 5 Weibull, F(t): Eﬀect of β F(t) θ = 2 θ = t, 1-‐ e-‐1 = 0.63 At t = θ, Reliability has dropped by a factor of 1/e = 2.72 t 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
6 component during system monitoring or in field tests. Weibull Probability Density Function:
Effect of Shape Parameter β on f(t) = λ(t)·∙R(t) f(t) 1.0 β < 1, similar to [email protected]: mode at t = 0 β = 1, [email protected], λ = 1/θ 1 < β < 3: skewed 3 < β, closer to symmetric 0.9
0.8
0.7
0.6 0.5 β 0.5
1.5
2.0
4.0 θ = 2 4.0 0.5
0.4 2.0 0.3 1.5 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 ⎠
λ(t) β -1 e ⎛ t⎞
–⎜ ⎟
⎝θ ⎠ β R(t) = λ (t)⋅R(t) λ (t) = f(t)
R(t)
7 Conditional Failure (Hazard) Rate Function, λ(t)
Of special importance is [email protected] of the wide ranges of increasing failure rate behavior: [email protected] slope, decreasing [email protected] slope (concave), constant [email protected] slope (Rayleigh), or increasing [email protected] slope (convex). 6 λ(t) 5 θ = 2 β, shape parameter f(t) / R(t) 0.5
1.5
2.0 Hazard Rate 4 4.0 constant LFR, Rayleigh [email protected], β = 2 [email protected] slope Convex, β > 2 3 increasing [email protected] slope 2 Concave, 1 < β < 2 1 decreasing [email protected] slope [email protected] 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; [email protected], CFR, for β = 1 8 Weibull Shape Parameter Effect on λ(t)
• Value Property • 0 < β < 1 Decreasing Failure Rate (DFR) • β = 1 [email protected] [email protected] (CFR), λ(t) = λ • 1 < β < 2 IFR-‐Concave failure rate (decreasing + slope) • β = 2 Rayleigh [email protected] (IFR, linear failure rate) • β > 2 IFR – Convex failure rate (increasing + slope) • 3 < β < 4 IFR – close to Normal [email protected] -‐ symmetrical closely approximates the normal [email protected] β = 3.4393: Most The shape of the [email protected] represents the failure process of the component or system as it develops through @me, so it is useful for failure behavior [email protected] 9 Weibull Characteristic Life: Effect of θ on f(t)
1.6 f(t) Basic Shape is set by β, which is unaﬀected by θ 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 t The [email protected] life = θ (scale parameter) value changes the mean, medium, mode, and variance or spread of the 10 [email protected] but not its basic shape, which is set by its β. Weibull Characteristic Life: Effect of θ on 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. 11 Weibull Characteristic Life: Effect of θ on λ(t) λ(t) β = 2, Rayleigh [email protected], linear IFR The [email protected] Life parameter inﬂuences the mean and the dispersion or spread of the Weibull [email protected] As θ is increased, the slope of λ(t) decreases and the [email protected] 12 broadens, but the linear shape set by β remains the same. Weibull MTTF from θ, β, and the Gamma Function MTTF = ∫ ∞ 0 ⎛ 1⎞
t f(t)dt = θ Γ ⎜ 1+ ⎟
⎝ 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; Γ(1) = (0)! = 1 1
lim MTTF = lim θ Γ(1+ ) = θΓ(1) = θ
β→∞
β→∞
β
As β increases, the mean approaches the [email protected] life, because the Weibull f(t) approaches symmetrical. 13 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 14 1.2 1.0 f(t) 0.6
0.4
0.2
0.0 1.0 2.0 R(50) = e 3.0 4.0 ⎛ 50 ⎞
−⎜ ⎟
⎝ 500 ⎠ 5.0 6.0 = 0.729 ( ) MTTF = 500Γ 1+ 2 = 500(2) = 1000
Γ(x) = (x-‐1)! = 2! = 2 Note DFR because β < 1 0.5
1.5
2.0 0.8 R(T) 4.0 0.6 Example Problems t
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
0.0 1 0.5, black 1.5, yellow 0.8 -0.2 1.5 R(t) 0.5 0.4 0.2 0
0.0 1.0 2.0 3.0 4.0 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 ⎟⎠ 5.0 6.0 t
-0.2 1.5 = 0.969 ⎛ 2⎞
MTTF = 500Γ ⎜ 1+ ⎟ = 500(0.903) = 451
⎝ 3⎠ Note IFR because β > 1 15 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 [email protected] with β = 1: σ2 = θ2(2 -‐ 1) = θ2 = (1/λ)2 as is known for [email protected] 16 Example Problem - Standard Deviation
θ = 500 θ = 500 β = 1/2 β = 1.5 2.33 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
2⎪
=
σ θ ⎨ Γ ⎜ 1+ ⎟ - ⎢ Γ ⎜ 1+ ⎟ ⎥
⎪⎩ ⎝ β ⎠ ⎣ ⎝ β ⎠ ⎦ ⎫
⎪
⎬
⎪⎭ 1.0 f(t) 0.9 Compare σ β = ½, black & β = 1.5, yellow 0.8 0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
-0.1 0.0 1.0 2.0 3.0 4.0 5.0 t 6.0 17 Design Life and Median
Set then and R(t) = e - ⎛⎜⎝ θt ⎞⎟⎠ 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 [email protected] appear, showing again that Weibull is an extension of [email protected]! 18 Weibull Mode
f(t mode ) = MAX
t≥0 1
⎧
β
⎛
⎞
1
⎪
⎪ θ 1t mod e = ⎨ ⎜
β⎟
⎝
⎠
⎪
⎪
⎩0 f(t) for β > 1
for β ≤ 1 Derived in Appendix 4B [email protected] Recall tmode = 0 for [email protected], for β = 1 19 Example - Design Life, Median, & Mode
1.2 1.5 R(t)
1 0.5
1.5
2.0 0.8 R(T) 4.0 0.6 0.5 0.4 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 0.2 0 0.0 1.0 2.0 3.0 4.0 5.0 1.0 Compare mode β = ½, black & β = 1.5, yellow f(t) 0.8 0.6
0.4
0.2
0.0
-0.2 0.0 1.0 2.0 3.0 4.0 5.0 t 6.0 6.0 t -0.2 20 Weibull Conditional Reliability
⎛ t+T ⎞
−⎜ 0 ⎟
⎝ θ ⎠ R(T0 + t) e
R(t | T0 ) =
=
β
⎛
⎞
R(T0 )
− ⎜ T0 ⎟
e ⎝θ⎠ β ⎡
⎢ ⎛ t+T0 ⎞
= exp ⎢ −
⎜⎝ θ ⎟⎠
⎢⎣ β ⎛
T
0⎞
+
⎜⎝ ⎟⎠
θ β ⎤
⎥
⎥
⎥⎦ Note that when β = 1, the [email protected] R(t) = exp(-‐λt) appears due to lack of memory of the [email protected]! Find R(50|50) for Weibull with β = 0.5, θ = 500 21 Example - Conditional Reliability 1.2 R(t) 0.8 Diﬀerent failure behavior shown by change in shape, β 0.5 1.5 1 1.5
2.0
4.0 0.6 0.5 0.4 0.2 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
increased 0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 t -0.2 = 0.9144 / 0.969 = 0.9437
decreased Same θ, but notice the significant shift due to different β 22 Failure Modes
Recall that a system composed of several independent failure modes in series each with a Weibull failure [email protected] has a system [email protected] failure rate [email protected] equal to the sum of the [email protected] failure mode failure rate [email protected] If all failure modes have the [email protected] shape parameter, then the system also has a Weibull [email protected] 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 [email protected] failure rate [email protected] with a shape parameter β and a [email protected] life for the 1
−
system, θS ⎡ n ⎛ 1⎞β ⎤ β
System θS t = θS = ⎢ ∑ ⎜ ⎟ ⎥
⎢⎣ i=1 ⎝ θ i ⎠ ⎥⎦
23 MTTF and tmed for Jet Engine
A jet engine is constructed from 5 modules each [email protected] Weibull [email protected] failure behavior with β = 1.5 (IFR, concave). Is the engine system also Weibull. Why or why not? The [email protected] life values ([email protected] cycles) of 3,600, 7,200, 5,850, 4,780, and 9,300. Calculate the, θS , MTTFS, median TTF, and 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
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠ ⎥
⎢⎣⎝
⎦ ⎛
⎛ 2⎞
1⎞
MTTFS = θ S 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⎞
0 θ ⎜⎝ θ ⎟⎠ =e ∫
− tβ β −1 dt = e ⎛ t⎞
−⎜ ⎟
⎝θ ⎠ β =e ⎛ t ⎞
−⎜
⎝ 1,842.7 ⎟⎠ 1.5 24 Identical Weibull Components
If n independent components in series have iden8cal [email protected] failure rate [email protected] with the same scale and β -1
shape parameters: β⎛ t⎞
λ (t) = ⎜ ⎟
θ ⎝θ ⎠ then β⎛ t⎞
λS (t) = ∑ ⎜ ⎟
θi ⎝ θi ⎠
i=1
n and ⎡
θS = ⎢
⎢⎣ t β -1 = ,
nβ θ β (t ) β -1 ⎛ t⎞
λ (t) dt
∫
−n
RS (t) = e 0
= e ⎜⎝ θ ⎟⎠
− ⎛ 1⎞ ⎤
∑ ⎜⎝ θ i ⎟⎠ ⎥⎥
i=1
⎦
n β − 1
β θ
= 1/ β
n β ⎛
1⎞
MTTFS = θ S Γ ⎜ 1+ ⎟
⎝ β⎠ 25 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) 26 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. 27 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. 28 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 29 Time-‐Dependent 2-‐Parameter Reliability Models • Weibull [email protected]: A [email protected] reliability model with a shape parameter that represents the progression of a [email protected] or failure process. Deﬁned for [email protected] values of t. • Normal (Gaussian) [email protected] for failures that are primarily the result of small (not dominant), [email protected] eﬀects. Deﬁned for [email protected] and [email protected] values of t. Represents only the IFR region where the [email protected] failure rate increases with @me. • Lognormal [email protected]: Deﬁned for [email protected] values of t and especially suitable when failures are the result of [email protected]@ve eﬀects. Similar to the Weibull in ve[email protected] Because of a skewed tail to larger t, ideal for modeling repair frequency and other skewed [email protected] [email protected] • Gamma [email protected]: A [email protected] reliability model similar to Weibull, including deﬁned for [email protected] values of t. In the same family of [email protected] as [email protected] and Poisson, so gamma is useful also as a prior [email protected] for Bayesian parameter [email protected] with [email protected] or Poisson as the likelihood [email protected] 30 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 with a wide range of scale parameter, σ. 31 Normal Conditional Failure Rate, Effect of σ σ = 0.5 λ(t) = f(t) / R(t) IFR σ = 1 The Normal [email protected] failure rate [email protected] 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. 32 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
33 Finding Normal Cumulative Probabilities
If T is normally distributed, transform T to Z, which is the standard normal variable T−µ
z=
σ Then Z has a standard normal [email protected] with a mean of 0 and a standard [email protected] of 1. The pdf for Z is given by φ(z) = 1
2π 2 e −z
2 The cdf for failure = F(t), [email protected] probability of failure to @me = t, is then given by z
-∞ P{Z ≤ z} = Φ(z) = ∫ φ(z) dz 34 Standard 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 35 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 Z T Z t ⎧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 Z t
⎬ = 1− Φ ⎜
⎟
σ ⎭
⎝ σ ⎠
⎩ ( ) Tables: A.1, pp. 514-‐519 36 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
37 Normal Example problem - Design Life
• A new fan belt is developed from a higher grade of
material. The belt 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 38 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. 39 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.
40 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. 41 41 Lognormal Density Function^, Effect
of Shape Parameter, s
f(t) = 1
2π s t s2 = ln(1+ δ 2t ), δ t = e 1 ⎛
t ⎞
− 2 ⎜ ln
⎟
2 s ⎝ tMED ⎠ σt
= cov
µt 2 ;t ≥ 0 Deﬁned for only [email protected] value of t. More symmetrical for smaller s values s2 ~ δ 2t , δ t ≤ 0.3 tmed = median @me to failure, [email protected] parameter s = shape parameter s = 0.1 s = 1 s = 0.5 ^Lognormal pdf is [email protected] represented in terms of the mean and std. dev. of the underlying normal [email protected] Here the parameters are median (more [email protected] than mean for skewed) and shape parameter s (~ σ). 42 Lognormal F(t), Eﬀect of Shape Parameter, s F(t) s = 0.1 s = 0.5 s = 1.0 As s increases, the variance increases and the [email protected] is more asymmetric. Data represented by Weibull will o~en be represented [email protected] by lognormal. 43 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 [email protected] is widely used to describe the frequencies of system repair, because it reﬂects normal [email protected] 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. 44 Lognormal/Normal Relationship
• Given T is a lognormal random variable, then Log T space T space • [email protected] 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 2 ln tmed mode
es
45 Lognormal Failure and Reliability Distribution
Pr of failure up to t F(t) = P {T ≤ t} = P {lnT ≤ lnt} logic expression [email protected] parameter ⎧ lnT - ln tMED lnt - ln tMED ⎫
= P⎨
≤
⎬
s
⎩ Z s
⎭
Z Standardize: t T Zt ⎧
⎛1
1
t ⎫
t ⎞
= P ⎨z T ≤ ln
⎬ = Φ ⎜ ln
⎟⎠
s
s
⎝
tMED ⎭
tMED
⎩ Values in Std. Normal tables Pr of failure ≤ t ⎛1
t ⎞
R(t) = 1− Φ ⎜ ln
⎝ s tMED ⎟⎠
Pr of failure > t 46 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
47 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...

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