Time-to-Failure Statistics

Abstract

When nearly identically processed materials/devices are placed under the same set of stress conditions, they will not fail exactly at the same time. An explanation for this occurrence is that slight differences can exist in the materials’ microstructure, even for materials/devices processed nearly identically. This means that not only are we interested in time-to-failure but, more precisely, we are interested in the distribution of times-to-failure. Once the distribution of times-to-failure is established, then one can construct a probability density function f(t) which will permit one to calculate the probability of observing a failure in any arbitrary time interval between t and t + dt, as illustrated in Fig. 7.1.

Problems

• 1. Time-to-rupture data (from a creep study) are shown for steel rods that were held at a fixed level of stress at very high temperatures until rupture occurred. The time-to-failure data are shown in hours. Find the lognormal t₅₀ and σ that describes the data.

44.0	43.3	49.0	36.0	70.8
50.3	45.2	47.4	36.3	68.3
65.6	51.4	43.2	39.4	70.7
54.5	58.3	42.5	52.2	56.5
41.1	60.2	42.7	63.6	40.0

Answers: t₅₀ = 49.8 h, σ = ln(t₅₀/t₁₆) = 0.22

• 2. Given the lognormal distribution (t₅₀ = 49.8 h, σ = 0.22) from Problem 1,

• (a) What is the expected time for 0.1 % of the steel rods to rupture?

• (b) What is the expected time for 99.9 % of the steel rods to rupture?

Answers: (a) t_{0.1 %} = 25.2 h, (b) t_{99.9 %} = 98.3 h

• 3. Given the lognormal distribution (t₅₀ = 49.8 h, σ = 0.22) from Problem 1, what fraction of failures occur between 35 and 55 h?

Answer: 0.620

• 4. Using the time-to-rupture data in problem 1, find the Weibull distribution that gives the best fitting to the data. What are the values for t₆₃ and the Weibull slope β?

Answer: t₆₃ = 55.2 h, β = 5.4

• 5. Given the Weibull distribution (t₆₃ = 55.2 h, β = 5.4) from Problem 4,

• (a) What is the expected time for 0.1 % of the steel rods to rupture?

• (b) What is the expected time for 99.9 % of the steel rods to rupture?

Answers: (a) t_{0.1 %} = 15.4 h, (b) t_{99.9 %} = 79.0 h

• 6. Given the Weibull distribution (t₆₃ = 55.2 h, β = 5.4) from Problem 4, what fraction of the failures occurred between 35 and 55 h?

Answer: 0.543

• 7. Using the normal distribution in Chap. 5, fit the data shown in Table 7.1.

• (a) What are the values of t₅₀ and sigma for the normal distribution?

• (b) Compare your normal fit to the lognormal-fit shown in Fig. 7.3. Which distribution gives the better fitting, normal or lognormal?

Answers:

• (a) t₅₀ = 1,573 h, σ = 576 h,

• (b) Lognormal distribution gives a better fitting to this data set.

• 8. The following time-to failure data was collected and found to have two failure mechanisms in the time-to-failure data.

Cum fraction F	Time-to-failure (h)
0.05	16
0.09	20
0.15	25
0.22	30
0.3	35
0.38	42
0.45	104
0.51	110
0.63	118
0.68	127
0.75	135
0.82	143
0.87	150
0.9	155

• (a) Perform a lognormal plot of the above data.

• (b) Find the point of inflection which separates the two mechanisms.

• (c) Replot the data for the two mechanisms.

• (d) What are the (t₅₀, σ) values for the two mechanisms?

Answers:

(b) Point of inflection: F = 0.42,

(d) Mechanism A: t₅₀ = 27.3 h, σ = 0.4,

Mechanism B: t₅₀ = 131.1 h, σ = 0.17.

9. Using the time-to failure data (shown in the table in Problem 8):

• (a) Perform a Weibull plot of the above data.

• (b) Find the point of inflection which separates the two mechanisms.

• (c) Replot the data for the two mechanisms.

• (d) What are the (t₆₃, β) values for the two mechanisms?

Answers:

(b) Point of inflection: F = 0.42,

(d) Mechanism A: t₆₃ = 32.3 h, β = 2.99 Mechanism B: t₆₃ = 139.6 h, β = 6.29.