Revision in Fatigue

1. Sample solved problems
1
STATISTICAL ASPECTS OF FATIGUE
End of Chapter Problems
Question 1.
The following data from 43 specimens were obtained from rotating bending fatigue test at one
constant stress level and were grouped as follows:
Specimen failed, n 3 8 6 10 8 5 3
Cycles to failure, Nf 1600 1900 2200 2500 2800 3100 3400
a. Calculate the mean, median, standard deviation, and the coefficient of variation for the
fatigue cycles.
b. Show the data in histograms of fatigue life distribution using both Nf and log Nf
c. Fit the stepped histograms in part b above by frequency distribution curves, assuming
normal and log-normal distributions.
d. Plot the cumulative histograms and distributions
e. Determine B10 and B50 lives for these data, assuming normal distribution.
Cycles to
failure,Nf
Samples
failed, n
Cum.
failure
n*Nf (Nf -X),d d2 n*d n*d2 % C.
Failed
1600 3 3 4800 -872 760384 -2616 2281152 7.0
1900 8 11 15200 -572 327184 -4576 2617472 25.6
2200 6 17 13200 -272 73984 -1632 443904 39.5
2500 10 27 25000 28 784 280 7840 62.8
2800 8 35 22400 328 107584 2624 860672 81.4
3100 5 40 15500 628 394384 3140 1971920 93.0
3400 3 43 10200 928 861184 2784 2583552 100.0
∑n = 43 106300 196 2525488 4 10766512
∑d = 196
2472
Mean X =
(∑n∗Nf)
(∑n)
=
𝟏𝟎𝟔𝟑𝟎𝟎
43
=2472.09
The mean should be a whole number since we cannot have a fraction of a cycle for accuracy
purposes.
Thus the mean is 2472 cycles to failure.
Median =
(43+1)
(2)
(43+1)/2 term = 22nd
item, the cumulative frequency just greater than 22 is 27
and the value of Nf corresponding to 27 is 2500. Hence the median for the fatigue life is 2500
cycles
Standard deviation, S=√[(
∑n∗d2
∑n
) −
∑n∗d)
∑n
)2
]
S = √[(
10766512
43
) − (
4
43
)2
]
S = √250383.991 = 500.38384
=500.384

2. Sample solved problems
2
Coefficient of variation for the fatigue lifes, V =
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝑀𝑒𝑎𝑛
× 100
V =
500.384
2472
× 100= 20.24%
b. /c.
3500
3000
2500
2000
1500
10
8
6
4
2
0
Cycles to Failure, Nf
Frequency,
n
Mean 2472
StDev 506.3
N 43
Normal Histogram of Failed sample - Cycles to failure with fit
4000
3500
3000
2500
2000
1500
10
8
6
4
2
0
Cycles to Failure, Nf
Frequency,
n
Loc 7.792
Scale 0.2110
N 43
Log-Normal Histogram of Failed sample - Cycles to failure with fit

3. Sample solved problems
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d.
3300
3000
2700
2400
2100
1800
40
30
20
10
0
Cycles to Failure, Nf
Frequency
Cumulative Histogram of Failed sample - Cycles to failure
e. B10 = 1642 cycles to failure
B50 = 2300 cycles to failure.

4. Sample solved problems
4
Question 2.
Fatigue testing of Eight, 8 similar components at a given load level has produced the
following fatigue lifes;
14700, 94700, 31700, 27400, 17300, 70100, 21800 and 39800
a. Compute the mean life and standard deviation for this load level.
b. Plot the data on a normal probability paper, log-normal probability paper and Weibull
paper. Which probability distribution do the data fit best?
c. Obtain B50 life from each distribution in b above.
Component
rank, i Fatigue life, Nf (Nf - X),d d2
Plotting
position
1 14700 -24987.5 624375156.3 8
2 17300 -22387.5 501200156.3 20
3 21800 -17887.5 319962656.3 32
4 27400 -12287.5 150982656.3 44
5 31700 -7987.5 63800156.25 56
6 39800 112.5 12656.25 68
7 70100 30412.5 924920156.3 80
8 94700 55012.5 3026375156 92
317500 ∑Nf 0
∑n =8 ∑Nf =317500 (∑Nf – X)2
5611628750
a. Mean X =
∑N𝑓
∑n
=
317500
8
=39687.5
Standard deviation, S =√[(
1
𝑛−1
)∑ (Nf − X)2
]
S = √[(
1
8−1
) ∗ 5611628750]
S =√801661250
Therefore, S = 28313.6
S= 28314
b. Probability plots

5. Sample solved problems
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i.
B50 = 39,688
ii.
B50 = 45244

6. Sample solved problems
6
iii.
100000
10000
1000
99
90
80
70
60
50
40
30
20
10
5
3
2
1
0.1
Fatigue life, Nf (Log scale)
Percent
Failure
Shape 2.012
Scale 59994
N 396
AD 18.848
P-Value <0.010
Weibull Probability distribution plot for the fatigue life of 8 samples
Weibull
B50= 50003
From the probability plots, the Best form of distribution for the data is log-normal
distribution.

7. Sample solved problems
7
Question 3
A fatigue life test program on a new product consisted of 10 units subjected to the same load
spectrum. The units failed in the following number of hours in ascending order: 75, 100, 130,
150, 185, 200, 210, 240, 265, and 300. Obtain the percent median rank values for the 10 failures.
Plot the percent median rank values versus hours to failure on Weibull paper. Draw the best
line of fit curve through the data and find
a. 10%, B10 expected failure life and
b. 50%. B50 expected failure life
c. What percent of population will have failed in 230 hours?
d. Consider 1 or 0.1% expected life would be.
Component
rank, i
Fatigue life, Nf Percent
Median rank
1 75 6.7
2 100 16.3
3 130 26.0
4 150 35.6
5 185 45.2
6 200 54.8
7 210 64.4
8 240 74.0
9 265 83.7
10 300 93.3
Mean X =
∑N𝑓
∑n
＝
1855
10
= 185.5
Percent median ranks for each are obtained from the formula;
% median rank =
( i − 0.3)
(n+0.4)
× 100
Since n= 10, For i=1, the % median rank will be =
( 1− 0.3)
(n+0.4)
× 100 =
( 0.7)
(10.4)
× 100 =6.7%
All other values of the percent median ranks are calculated and tabulated as shown in the table
above

8. Sample solved problems
8
a. B10 = 131 hours
b. B50 = 213 hours
c. the % that will have failed after 230 hours is 61%
d. 1% life or B1 =71 hours.

9. Sample solved problems
9
Question 4
Plane strain fracture toughness tests of six identical samples from a metal produced the
following values of KIc : 57.2, 54.0, 59.7, 55.5, 62.4, and 58.6 MPa √m. Assuming a normal
distribution, estimate the plane strain fracture toughness of the metal with 90% reliability and
90% confidence level.
S. No.
Plane strain Fracture
toughness, KIc Sorted (Kic - X), d d2
1 54 54 -3.90 15.21
2 55.5 55.5 -2.40 5.76
3 57.2 57.2 -0.70 0.49
4 58.6 58.6 0.70 0.49
5 59.7 59.7 1.80 3.24
6 62.4 62.4 4.50 20.25
∑Kic =347.4 ∑ d = 0.00 ∑ d2 = 45.44
Mean X ∑Kic /n 57.9
1/(n-1)*∑d2
9.088
StDev, S = √ 9.088 3.0146
The mean plane strain fracture toughness = 57.9 MPa √m.
The standard deviation = 3.015
Assuming normal distribution, with p = 90% and CL = 90 %, and n = 6 the value of k = 2.49
Therefore the estimated plane strain fracture toughness is obtained using the equation;
X= x –k S
X= 57.9 - 2.49 (3.015) = 50.392
= 50.4 MPa √m.

10. Sample solved problems
10
Question 5
Fatigue strength Sf , at 106
cycles versus Brinell hardness (HB) for seven grades of SAE 1141
steel are given as follows
Fatigue Strength,
MPa
Brinell Hardness,
HB
276 199
287 217
286 223
296 229
342 241
332 252
433 277
a. Using linear regression, find the slope, intercept, and correlation coefficient for these
data.
b. Is there a good correlation between the hardness and the fatigue strength of these steel
grades
c. What fatigue strength corresponds to a Brinell hardness of 225 =304 MPa
From the graph,
The slope = 2.0155,
The intercept = - 149.9
R= √0.8696 = 0.932
This indicates that there is a strong correlation between Fatigue strength and Brinell hardness
values for the steel grades.
Fatigue strength corresponding to Brinell hardness of 225 HB =304 MPa
y = 2.0155x - 149.9
R² = 0.8696
240
290
340
390
440
490
190 200 210 220 230 240 250 260 270 280 290
FATIGUE
STRENGTH,
MPA
BRINELL HARDNESS, HB
Scatterplot of Fatigue Strength vs Brinell Hardness

11. Sample solved problems
11
Question 6
Constant amplitude crack growth rate data in the threshold region for a cast aluminium alloy
are listed as a function off stress intensity factor range, ∆𝐾:
da/dN(×10-10
m/cycle) 8.6 5.9 6.8 3.9 4.5 1.6 2.2
∆𝐾(MPa √m) 1.91 1.84 1.89 1.79 1.84 1.67 1.73
Fit these data with a linear regression and determine the threshold stress intensity factor
range, ∆𝐾th, defined at a 10-10
m/cycle fatigue crack growth rate.
The threshold stress intensity factor is stress found in region I of the crack growth rate versus
change in stress intensity graph. It indicates the threshold value ∆𝐾 th, below which no
observable crack growth. It corresponds to the value of ∆𝐾 for which the crack growth is of the
order of 1×10-10
m/cycle and below. From the liear regression graph, the value of
∆𝐾th = 1.675 (MPa √m)
y = 27.758x - 45.456
R² = 0.9152
0
1
2
3
4
5
6
7
8
9
10
1.4 1.5 1.6 1.7 1.8 1.9 2
da/dn(×10
-10
m/cycle)
∆𝐾(Mpa √m)
Graph of da/dN-∆𝐾(MPa √m)