Chapter 5-Fatigue Life Cycle-2223-1

AIRCRAFT STRUCTURES II
Fatigue Life Damage Estimation
Ts. Dr. Haris Ahmad Bin Israr Ahmad

Faculty of Mechanical Engineering
Universiti Teknologi Malaysia
Module outline
➢ PART 1: OVERVIEW OF DAMAGE TOLERANCE CONCEPT
➢ PART 2: FATIGUE DAMAGE INITIATION
➢ PART 3: FATIGUE DAMAGE PROPAGATION
2
5.1 OVERVIEW OF DAMAGE TOLERANCE CONCEPT
Safe Life Approach (Initiation)
• Developed in the 1950s: No damage is allowed in the structure for its whole lifetime, even if
detectable
• A fatigue lifetime N is demonstrated for the structure:
✓ By tests
✓ By calculations for derived versions
• Different sources of scattering are included in the demonstrated lifetime:
✓ Scattering of the material properties,
✓ scattering due to the manufacturing process,
✓ scattering on the in-service loads,
✓ accuracy of the calculation model.
• To take this scattering into account, a safety factor k is applied on the demonstrated lifetime
k = 5 in the general case with a test demonstration 𝑁
k = 3 for a test demonstration on a structure that is 𝑁𝑐𝑒𝑟𝑡𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 =
𝑘
monitored in service
The component is withdrawn from service after this lifetime. 3
Fail Safe Approach (Initiation)
• Developed in the 1960s: The structure must be redundant (multiple load paths) so that in
case of failure of a critical element, the remaining structure is able to sustain the maximal
loads that may be encountered in service (Limit Loads).
• This approach leads to periodic inspections to allow the detection of a crack, the repair and
or the replacement of the component.
Not Fail Safe Fail Safe

4
Damage Tolerance Approach (Propagation):
• Developed in the 1970s: the Damage Tolerance concept
• It is assumed that all the components contain a damage. This damage may exist from
the origin or it may have been created in the service life.
• The Lifetime is therefore defined as the time required to propagate this damage to a
critical size.
Critical size Fracture
crack length
No Crack detected Crack detected
Detectable size
In-service time
First
Inspection Mandatory repair if loss of ultimate
5
loads capability for next interval
PART 2: FATIGUE DAMAGE INITIATION
6
The objectives of fatigue tests are:
• To provide basic data on the methods of analysis used,
• To determine the effect of various factors affecting life of the structure (materials and
processes, fabrication procedures, environmental conditions, etc.)
• To establish design allowable stresses for the materials to be used in the structure
including processing effects,
• To evaluate and compare the fatigue quality of structural design details
• To substantiate that the structure will meet or exceed the design life requirement of the
vehicle (Qualification/Certification testing),
• To locate fatigue critical areas of airplane at ear1iest possible time for production
incorporation,
• To develop inspection and maintenance procedures for customers,
• To provide test data for analytical studies to assist in determining times for inspection
and maintenance.
7
Introduction
Aluminium
alloy coupon
Repeated load test at low stress level

Failure after 650 000 cycles
Circular section
Repeated loading modifies and degrades the material
properties and may lead to the failure even at low stress
level (below yield stress).
This phenomenon is commonly called FATIGUE 8
Introduction
Zoom on the failure test:
When applying the repeated loads, one observes:
1. The apparition of a crack after a certain number
of cycles: this phase is called initiation
2. The increase of crack length: this phase is called
propagation
3. An acceleration of propagation leading to the
final failure
Circular section
9
Behaviour in Fatigue
• Constant Amplitude Loading
A constant amplitude loading is characterised by:
• 𝑆𝑚𝑖𝑛 𝑎𝑛𝑑 𝑆𝑚𝑎𝑥
• 𝑆𝑎 𝑎𝑛𝑑 𝑆𝑚
• 𝑆𝑚𝑎𝑥 = 𝑆𝑎 + 𝑆𝑚 • 𝑆𝑚 = (𝑆𝑚𝑎𝑥 + 𝑆𝑚𝑖𝑛)/2
• 𝑆𝑚𝑖𝑛 = 𝑆𝑚 − 𝑆𝑎 • 𝑆𝑎 = (𝑆𝑚𝑎𝑥 − 𝑆𝑚𝑖𝑛)/2
𝑆𝑚𝑖𝑛 𝑆𝑚 − 𝑆𝑎
The stress ratio, 𝑅 is a characteristic loading parameter 𝑅= =
𝑆𝑚𝑎𝑥 𝑆𝑚 + 𝑆𝑎
𝑅=0 𝑅 = −1
𝑅>0
𝑆𝑎 = 𝑆𝑚 𝑆𝑚 = 0
10
S-N Curve
To quantify the fatigue endurance of a material, Wohler proposed

to perform constant amplitude tests.
One obtains an endurance curve, also called Wohler curve or S-N
curve. (usually used to predict the initiation)
Number of
cycles to
failure
11
• S-N Curve
The classical 𝑆𝑚𝑎𝑥 − 𝑁 diagram does not A 𝑆𝑎 − 𝑁 representation better shows
directly illustrates the effect of mean stress 𝑆𝑚 the effect of mean stress 𝑆𝑚
c
cv c
c v
v
v
𝑅1 < 𝑅2 < 𝑅1 𝑆𝑚1 < 𝑆𝑚2 < 𝑆𝑚3
The higher is the mean stress, the lower is 12

the fatigue life.
• Complex Spectrum
Typical aircraft spectrum:

In real life, constant amplitude spectrum
almost never exists
Method:
Step 1 : From complex spectrum to Block constant Step 2 : Damage calculated block per block then
amplitude spectrum Cumulative law to estimate the total damage
Total
Damage
D
13
Complex Spectrum - Miner's rule basis
𝑆𝑡𝑟𝑒𝑠𝑠
𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
Here, the material is
𝑆𝑚𝑎𝑥
broken, D=1
𝑆𝑎 Palmgren-Miner Linear
𝑆𝑎 Damage Rule:
D=0 for a safe material
D=1 for a broken material
Here, the material is
safe D=0 Number of
cycles to
𝑁𝑓 failure
14
• Complex Spectrum - Miner's rule applied to block constant amplitude spectrum:
𝑛1
𝐷𝑎𝑚𝑎𝑔𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑛1 𝑐𝑦𝑐𝑙𝑒𝑠 𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝑆𝑎1 𝑑1 = 𝑇𝑜𝑡𝑎𝑙 𝑑𝑎𝑚𝑎𝑔𝑒
𝑁𝑓1
𝐷 = 𝑑1 + 𝑑2 + 𝑑3
𝑛2
𝐷𝑎𝑚𝑎𝑔𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑛2 𝑐𝑦𝑐𝑙𝑒𝑠 𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝑆𝑎2 𝑑2 = 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑤ℎ𝑒𝑛 𝐷 = 1
𝑁𝑓2 1
𝑛3 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑦𝑐𝑙𝑒𝑠 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 =
𝐷𝑎𝑚𝑎𝑔𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑛3 𝑐𝑦𝑐𝑙𝑒𝑠 𝑎𝑡 𝑙𝑒𝑣𝑒𝑙 𝑆𝑎3 𝑑2 = 𝐷
16
𝑁𝑓3
Rainflow method
• The Rainflow method is usually used to analyze a complex spectrum
• The basis of the Rain-Flow method is to prevent the 'small cycle' from hiding the 'big cycle'.
• The method consists in letting a drop flow out along the spectrum.
➢ The spectrum must be re-arranged so that the first peak is the extremum (max or min)
of the sequence;
➢ The 1st flow starts at the first peak, the 2nd starts at the other extremum till the end is
reached;
➢ Each flow is stopped according to 1 of the 3 following rules (see pictures). It leads to a
semi-cycle which is then associated with its complement to create a full cycle;
➢ The so defined cycle whole is the spectrum to consider for the damage computation.
17
Rainflow method
18
Rainflow method: Example
19
Fatigue Cycles Estimation Classical S-N curve approach of Endurance Fatigue
20
Damage Initiation Estimation
• Classical S-N curve approach
A classical method to estimate the fatigue life of a structural component using Wohler curve
BUT such an approach requires a huge number of tests associated with the unwanted cost.
Therefore, more generic theoretical approaches have been developed such as Modified Goodman,
Gerber Parabola, Soderberg Relation etc… to be used to estimate the damage initiation of a
structural component
For OEM like Airbus or Boeing, they used their own engineering approaches developed by their
design office and the data used generally are confidential.
21
Modified Goodman, Gerber Parabola & Soderberg Estimation
𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
𝑆𝑢 = 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠
𝑆𝑒 𝑆𝑦 = 𝑌𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠
𝑆𝑒 = 𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 𝑙𝑖𝑚𝑖𝑡 𝑜𝑟 𝐹𝑎𝑡𝑖𝑔𝑢𝑒 𝑙𝑖𝑚𝑖𝑡
𝑆𝑦 𝑆𝑢 𝑀𝑒𝑎𝑛 𝑆𝑡𝑟𝑒𝑠𝑠
22
Modified Goodman, Gerber Parabola & Soderberg Estimation
𝑛
𝑆𝑎 𝑆𝑚
+ =1
𝑆𝑎𝑟 𝑆𝑢
𝑛 = 1 𝑚𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝐺𝑜𝑜𝑑𝑚𝑎𝑛
𝑛 = 2 𝐺𝑒𝑟𝑏𝑒𝑟 𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑎
𝐹𝑜𝑟 𝑆𝑜𝑑𝑒𝑟𝑏𝑒𝑟𝑔 𝑅𝑒𝑙𝑎𝑡𝑖𝑜𝑛,

𝑆𝑎 𝑆𝑚
+ =1
𝑆𝑎𝑟 𝑆𝑦
23
Basquin’s Equation
• The S-N curve in the high cycle fatigue is sometime described by the Basquin equation
• Basquin proposed a mathematical equation to represent the part of the S-N curve for the region
𝑁 < 106
𝑆𝑡𝑟𝑒𝑠𝑠 𝐿𝑒𝑣𝑒𝑙
𝐿𝐶𝐹
𝐻𝐶𝐹
𝑆𝑒 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑎𝑡𝑖𝑔𝑢𝑒 𝑙𝑖𝑚𝑖𝑡 𝑜𝑟 𝑆𝑒
𝑘𝑛𝑜𝑤𝑛 𝑎𝑠 𝑒𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 𝑙𝑖𝑚𝑖𝑡
Number of
cycles to failure
24
𝐵𝑎𝑠𝑞𝑢𝑖𝑛’𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑆𝑎𝑟 = 𝐴𝑁 𝐵 (1)
𝑆𝑎𝑟 is a reversing stress, will give a fatigue life of N cycles (𝑁 < 106 )and A & B are two
material constants.
𝑇𝑎𝑘𝑖𝑛𝑔 log 𝑜𝑓 1 , log(𝑆𝑎𝑟 ) = log 𝐴 + 𝐵 log(𝑁) (2)
For a small number of stress reversals, N = 1000, the range stress, 𝑆𝑎𝑟 can be approximated
to 90% of the ultimate strength of the material:
log(0.9𝑆𝑢 ) = log 𝐴 + 𝐵 log 1000 = log 𝐴 + 3𝐵 (3)
We also know that 𝑆𝑎𝑟 = 𝑆𝑒 , for 𝑁 = 106 . Putting values in (2) we get
log(𝑆𝑒 ) = log 𝐴 + 𝐵 log 1000,000 = log 𝐴 + 6𝐵 (4)
25
log 𝑆𝑒 − log 0.9𝑆𝑢

Subtracting (4)-(3) and simplifying: 𝐵=
3
𝑆𝑒
𝐹𝑟𝑜𝑚 4 , 𝐴 = 6𝐵
10
Thus, from the known values of ultimate strength (𝑆𝑢 ) and the endurance strength (𝑆𝑒 ) of the
part material, we can determine the material constants A and B of Basquin’s equation
Once the constants A, B and fully reversal stress (𝑆𝑎𝑟 ) are known, then we can determine the
expected number of cycles to failure for a given mean stress (𝑆𝑚 ) and alternate stress (𝑆𝑎 ) .
1
𝑆𝑎𝑟 𝐵
𝑁=
𝐴
26
Example 1
Suppose that a material with 𝑆𝑢 = 620 𝑀𝑃𝑎 and 𝑆𝑒 = 275 𝑀𝑃𝑎, is to be subjected to a
completely reversing stress of 𝑆𝑎𝑟 = 310 𝑀𝑃𝑎. Use Basquin’s equation to determine the
expected number of cycles to failure for this stress level.
Solution:
(i) Find A & B of Basquin’s Equation (ii) Apply Basquin’s Equation to calculate N
1
log 𝑆𝑒 − log 0.9𝑆𝑢
𝐵= 𝑆𝑎𝑟 𝐵
3 𝑁=
𝐴
log 275 × 106 − log 0.9 × 620 × 106 1
𝐵=
3 310 × 106 −0.1024
𝑁=
𝐵 = −0.1024 1132 × 106
𝑁 = 311,182 𝑐𝑦𝑐𝑙𝑒𝑠
𝑆𝑒 275 × 106
𝐴 = 6𝐵 = (6×−0.1024) = 1132 × 106
10 10
Example 2
Let consider the following stress spectrum which contains 1 flight. The fatigue properties of the
component is as follows:
𝑆𝑢 = 480 𝑀𝑃𝑎 𝐴 = 1365 𝑀𝑃𝑎 𝐵 = −0.166
1. Apply the rain flow method to this stress spectrum
2. Determine the allowable number of flights using Modified Goodman and Basquin’s equations
𝑆𝑡𝑟𝑒𝑠𝑠 (𝑀𝑃𝑎)
450 𝑀𝑃𝑎
430 𝑀𝑃𝑎
425 𝑀𝑃𝑎
400 𝑀𝑃𝑎
200 𝑀𝑃𝑎
100 𝑀𝑃𝑎
50 𝑀𝑃𝑎
0 𝑀𝑃𝑎 0 𝑀𝑃𝑎
Example 2
Solution:
450 𝑀𝑃𝑎 1
430 𝑀𝑃𝑎 2
425 𝑀𝑃𝑎
3 3 3 400 𝑀𝑃𝑎
4
200 𝑀𝑃𝑎
100 𝑀𝑃𝑎
50 𝑀𝑃𝑎
0 𝑀𝑃𝑎 0 𝑀𝑃𝑎
Example 2
Solution:
𝐺𝑟𝑜𝑢𝑝 𝑆𝑚𝑖𝑛 𝑆𝑚𝑎𝑥 𝑆𝑚 𝑆𝑎 𝑆𝑎𝑟 𝑛𝑖 𝑁𝑖 𝑑 = 𝑛𝑖 /𝑁𝑖

𝑀𝑃𝑎 𝑀𝑃𝑎 𝑀𝑃𝑎 𝑀𝑃𝑎 𝑀𝑃𝑎 (𝑐𝑦𝑐𝑙𝑒) (𝑐𝑦𝑐𝑙𝑒)
1 0 450 225 225 423.5 1 1152.8 8.675 × 10−4
2 100 430 265 165 368.4 1 2671.7 3.743 × 10−4
3 200 425 312.5 112.5 322.4 3 5965.2 5.029 × 10−4
4 50 400 225 175 329.4 1 5238.9 1.91 × 10−4
σ 1.936 × 10−3
𝐵𝑎𝑠𝑞𝑢𝑖𝑛′ 𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛,
1 1
𝑆𝑎 𝑆𝑚 𝑆𝑎𝑟 𝐵
+ =1 𝑁=
𝑆𝑎𝑟 𝑆𝑢 𝐴
Example 2
Solution:
4
𝑛𝑖
𝐷𝑓𝑙𝑖𝑔ℎ𝑡 = ෍ = 1.936 × 10−3
𝑁𝑖
𝑛=1
1 1
𝑁𝑓𝑙𝑖𝑔ℎ𝑡 = = = 516.5 𝑓𝑙𝑖𝑔ℎ𝑡𝑠 (𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑)
𝐷 1.936 × 10−3
𝑁 516.5
𝑁𝑓𝑙𝑖𝑔ℎ𝑡 = = = 103 𝑓𝑙𝑖𝑔ℎ𝑡𝑠 (𝑐𝑒𝑟𝑡𝑖𝑓𝑖𝑒𝑑)
𝐾 5
Example 3
A landing gear component of an aircraft made from Titanium Ti-6AI-AV, experiences a stress history as
given below in its normal operation
𝑆𝑢 = 900𝑀𝑃𝑎 𝐴 = 2625𝑀𝑃𝑎 𝐵 = −0.1702
1. Apply the rain flow method to this stress spectrum
2. Determine the expected number of cycles to failure using Modified Goodman and Basquin’s equations
600
350
200
0
Example 3
Solution:
1
2 2 2 1
2 2 2 2 2
3 3 3
Example 3
Solution:

𝑀𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝐺𝑜𝑜𝑑𝑚𝑎𝑛, 𝐵𝑎𝑠𝑞𝑢𝑖𝑛′ 𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛,

1 1
𝑆𝑎 𝑆𝑚 𝑆𝑎𝑟 𝐵
+ =1 𝑁=
𝑆𝑎𝑟 𝑆𝑢 𝐴
Example 4
1. Let consider the following stress spectrum which contains 1 flight. The material properties of the component
is as follows:
𝑆𝑢 = 480 𝑀𝑃𝑎 𝑆𝑒 = 150 𝑀𝑃𝑎 𝑆𝑦 = 345 𝑀𝑃𝑎
420
370
300
220
170
80
50 50
Example 4
Solution:

1 0 420 210 210 373.3 1
2 50 170 110 70 90.8 2
4 80 220 150 70 101.8 3
σ 4.525 × 10−4
𝑀𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝐺𝑜𝑜𝑑𝑚𝑎𝑛, 𝐵𝑎𝑠𝑞𝑢𝑖𝑛′ 𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛,
1 1
𝑆𝑎 𝑆𝑚 𝐴 = 1243.6 × 106 𝑆𝑎𝑟 𝐵
+ =1 𝑁=
𝐴
𝑆𝑎𝑟 𝑆𝑢 𝐵 = −0.1531
Assignment 5i
1. Let consider the following stress spectrum which contains 1 flight. The material properties of the
component is as follows:
𝑆𝑢 = 480 𝑀𝑃𝑎 𝑆𝑒 = 150 𝑀𝑃𝑎 𝑆𝑦 = 345 𝑀𝑃𝑎
a) Apply the rain flow method to this flight
b) Determine the allowable number of flights using Modified Goodman and Basquin’s equations
300
275
250 250
200 200
100
80 80
50 20
Assignment 5ii
2. Reconsider Question 1, determine the allowable number of flights using Soderberg Relation. Please
comment on the result obtained as compared to question 1.
300
275
250 250
200 200
100
80 80
50 20
Engineering/Industrial Approach
The standard sizing is driven by the Total Life and uses knock down factors in function of
technological criteria such as:
• Stress concentration
• Surface treatment,
• Mechanical treatment,
• Type of fastener
The industrial methods are based on general principles which limit the number of required
tests.
➢ For example, the Indice de Qualite de Fatigue, IQF (Fatigue Quality Index) approach is often
found in the aeronautical industry (Airbus, Boeing, ...), it is based on a simplified
description of the Wohler curves.
➢ Nevertheless, it is limited to certain type of structural configurations instead of generic and
applicable for all type structural components and configurations.
44
Engineering/Industrial Approach - General Equations
𝑇𝑦𝑝𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 𝑓𝑜𝑟 𝑃:

𝑝 = 4.5 𝑓𝑜𝑟 𝑎𝑙𝑢𝑚𝑖𝑛𝑖𝑢𝑚 𝑎𝑙𝑙𝑜𝑦𝑠
𝑝 = 6.5 𝑓𝑜𝑟 𝑡𝑖𝑡𝑎𝑛𝑖𝑢𝑚 𝑎𝑙𝑙𝑜𝑦𝑠
𝑝 = 7.5 𝑓𝑜𝑟 𝑠𝑡𝑒𝑒𝑙𝑠
45
Engineering/Industrial Approach - IQF general law

The IQF is determined via en empirical law resulting from the
exploitation of thousands attests. This analysis lead to an expression
of IQF in function of main parameters assumed independent:
46
Engineering/Industrial Approach - IOF general law:
47
Engineering/Industrial Approach - IOF general law:
49
PART 3: FATIGUE DAMAGE PROPAGATION
Introduction
• The presence of a crack can significantly reduce the life of a component or structure.
• To make life estimations via Fracture mechanics approach for fatigue crack growth and
damage tolerant design, the following information are often needed:
➢ The stress intensity factor, K.
➢ The fracture toughness, Kc.
➢ The applicable fatigue crack growth rate expression.
➢ The initial crack size, ai (ao).
➢ The final or critical crack size, af (ac).
50
Introduction
• Fatigue life can often be split into initiation phase and propagation phase.
• Cracks can form
✓ due to fretting damages
✓ due to corrosion damages
✓ as a consequence of manufacturing processes (i.e. deep machining marks or voids in welds),
✓ and metallurgical discontinuities (i.e. inclusions)
• Therefore, fatigue life with pre-existing flaws must be considered as these flaws may
grow under cyclic loading
𝑁𝑇𝑜𝑡𝑎𝑙 = 𝑁𝐼𝑛𝑖𝑡𝑖𝑎𝑡𝑖𝑜𝑛 + 𝑁𝑃𝑟𝑜𝑝𝑎𝑔𝑎𝑡𝑖𝑜𝑛

51
Introduction
Typical questions answered by Fracture Mechanics
• What is the residual strength of a structure with a given crack length ?
• Given the in-service loads, what is the maximum allowable crack size for as structure?
• How long will it take for an initial undetected defect or crack to reach the maximum
allowable crack size?
• What is the adequate inspection interval to control a propagating crack and avoid the
rupture of the structure?
52
Analysis of Crack Bodies
• Cracks grow in a variety of ways and the stress fields that develop around the crack
tips may be divided into three basic modes:
(Opening) (Shearing) (Tearing)
• Mode I is the mode that occurs most commonly, it is also the most dangerous mode.
53
Analysis of Crack Bodies - Stress Intensity Factor (SIF)

𝑅𝑒𝑚𝑜𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑆∞
𝐾𝐼 = 𝑆∞ 𝜋. 𝑎
𝑆𝑡𝑟𝑒𝑠𝑠 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟
The SIF, is a key parameter of Fracture

Mechanics. Its unit is ′𝑴𝑷𝒂. √𝒎′
2𝑎
𝐾𝐼 = 𝑆∞ 𝜋. 𝑎 . 𝑓 𝑎, 𝑤 …
𝑤
𝑓 𝑎, 𝑤, 𝑙 𝑖𝑠 𝑎 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑛𝑔 𝑜𝑛 𝑐𝑟𝑎𝑐𝑘 𝑙𝑒𝑛𝑔𝑡ℎ 𝑎𝑛𝑑
𝑜𝑛 𝑡𝑦𝑝𝑖𝑐𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ𝑠/𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑦
𝑆∞ 54
Analysis of Crack Bodies - Stress Intensity Factor (SIF)
Application: Finite width plate with a centre crack (thick plate / Plane strain)
𝑆∞ 𝜋. 𝑎
𝑲𝑰 =
𝑎
cos 𝜋. 𝑤 Corrective factor
For brittle materials and thick plate, fracture occurs when the SIF reaches a
critical value, called plain strain fracture toughness, 𝑲𝑰𝒄
𝑲𝑰 = 𝑲𝑰𝒄
The fracture is thus defined by a relationship between the applied load

𝑆∞ and the crack length 𝑎 :
𝑎
𝐾𝐼𝑐 cos 𝜋. 𝑤
𝑺∞ =
𝜋. 𝑎 55
Analysis of Crack Bodies - Typical values of KIc
Generally, for a given alloy, when 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 increases, 𝐾1𝑐 decreases
56
Analysis of Crack Bodies - Effect of thickness
• The general relationship between fracture toughness,
Kc, and thickness is shown.
• Thin parts have a high value of Kc accompanied by
appreciable “shear lips".
• As the thickness is increased, the percentage of
"shear lips" decreases, as does Kc. This type of
fracture appearance is called mixed-mode implying
both slant and flat fracture.
• For thick parts, the entire fracture surface is flat, and
Kc approaches an asymptotic minimum value, KIc.
• Plastic zone sizes at fracture are much larger in thin

parts as compared to thick parts.
57
Analysis of Crack Bodies - Effect of thickness
• Generally, the fracture behaviour is dependent on the thickness of the material. 𝑉𝑟𝑜𝑚𝑎𝑛
defines a fracture toughness Kc as a function of thickness, 𝑡 .
2
2𝑡
− 2
𝐾1𝑐
൘𝑆
𝑦
𝐾𝑐 = 𝐾1𝑐 1 + 𝑒
𝑤ℎ𝑒𝑟𝑒:
𝑡 𝑖𝑠 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
𝑆𝑦 𝑖𝑠 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙
58
Analysis of Crack Bodies – Other Effects
• Fracture toughness 𝑲𝒄 of metals is also dependent on:

✓ temperature,
✓ strain rate, and
✓ corrosive environment.
• Besides that, 𝑲𝒄 can be very sensitive to metallurgical conditions such as:

✓ grain orientation,
✓ chemistry, and
✓ microstructure.
59
Example 5
A 10 mm crack has been discovered in a wing skin (thickness = 4 mm)
Let consider the following fracture mechanics data:
𝑲𝟏𝒄 = 𝟏𝟏𝟎𝟎 𝑴𝑷𝒂. 𝒎𝒎𝟎.𝟓
𝑺𝒚 = 𝟑𝟖𝟎 𝑴𝑷𝒂
𝑺𝒍𝒊𝒎𝒊𝒕 = 𝟐𝟓𝟎 𝑴𝑷𝒂
(a) Determine the Kc

(b) What is the strength of the structure with the present crack (maximum stress one can
apply before failure)
(c) What is the critical crack length?
Example 5: Solution
2
2𝑡
(a) Determine the Kc − 2
𝐾1𝑐
൘𝑆
𝑦
𝑲𝒄 ≠ 𝑲𝟏𝒄 = 𝟏𝟏𝟎𝟎 𝑴𝑷𝒂. 𝒎𝒎𝟎.𝟓 𝐾𝑐 = 𝐾1𝑐 1 + 𝑒
𝐾𝑐 = 1542 𝑀𝑃𝑎 𝑚𝑚0.5
(b) What is the strength of the structure with the present crack (maximum stress one can apply before failure)
𝟐𝒂 = 𝟏𝟎 𝒎𝒎 𝒂 = 𝟓 𝒎𝒎
𝐾𝐼 = 𝐾𝐶 = 𝑆∞ 𝜋. 𝑎
1542 = 𝑆∞ 𝜋. (5)
1542
𝑆∞ = = 389 𝑀𝑃𝑎
𝜋. (5)
Example 5: Solution
(c) What is the critical crack length?
𝐾𝑙𝑖𝑚𝑖𝑡 = 𝑆𝑙𝑖𝑚𝑖𝑡 𝜋𝑎𝑐𝑟 = 𝐾𝐶
2
𝐾𝐶
𝑎𝑐𝑟 =
𝑆𝑙𝑖𝑚𝑖𝑡 𝜋
𝑎𝑐𝑟 = 12.11 𝑚𝑚
Fatigue Crack Growth
Toughness 𝐾𝐼 = 𝑓 𝜎, 𝑎 = 𝐾𝑐
𝑎𝑐 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑐𝑟𝑎𝑐𝑘 𝑠𝑖𝑧𝑒
𝑎𝑜 𝑁𝑜 𝑜𝑓
????
𝑝𝑟𝑜𝑝𝑎𝑔𝑎𝑡𝑖𝑜𝑛
𝑐𝑦𝑐𝑙𝑒𝑠
• How long will it take for an initial undetected defect or crack to reach this maximum
allowable crack size?
• What is the adequate inspection interval to control a propagating crack and avoid the
rupture of the structure? 63
Fatigue Crack Growth - Crack growth rate
𝑼𝒏𝒔𝒕𝒂𝒃𝒍𝒆 𝒑𝒓𝒐𝒑𝒂𝒈𝒂𝒕𝒊𝒐𝒏
𝐹𝑟𝑎𝑐𝑡𝑢𝑟𝑒
𝑲𝑰 = 𝒇 (𝝈, 𝒂) = 𝑲𝒄
𝐶𝑟𝑎𝑐𝑘 𝑝𝑟𝑜𝑝𝑎𝑔𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒

𝑑𝑎
𝑑𝑁
64
• Zone 1
✓ discontinuous phenomena
✓ Influence of microstructure and environment
✓ Existence of a threshold ∆𝐾𝑡ℎ under which
there is few or no propagation.
✓ Generally associated to a propagation speed of
10−7 mm/cycle
• Zone 2
✓ Continuous phenomena
✓ Low influence of microstructure and environment
✓ Linear behaviour in a log-log plane
• Zone 3
✓ Static phenomena
✓ Close to unstable propagation
✓ Behaviour driven by 𝐾𝐶
65
𝐶𝑜𝑚𝑚𝑜𝑛𝑙𝑦 𝑢𝑠𝑒𝑑 𝑙𝑎𝑤: 𝑷𝒂𝒓𝒊𝒔′ 𝒍𝒂𝒘

(𝒐𝒏𝒍𝒚 𝒗𝒂𝒍𝒊𝒅 𝒇𝒐𝒓 𝒍𝒊𝒏𝒆𝒂𝒓 𝒃𝒆𝒉𝒂𝒗𝒊𝒐𝒖𝒓, 𝒛𝒐𝒏𝒆 𝑰𝑰)
𝑑𝑎
= 𝐶. ∆𝐾 𝑛
𝑑𝑁
∗ 𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑧𝑜𝑛𝑒 𝐼𝐼
66
Example 6
Let consider the following crack growth rate curve. Using the Paris Law formula, determine the
parameters C and n
𝑑𝑎/𝑑𝑁
10−3
10−4
∆𝐾
350 800 1
𝑀𝑃𝑎. 𝑚𝑚2
67
Example 3: Solution
log(10−3 ) = log 𝐶 + 𝑛 log 800 … … . (1)
log(10−4 ) = log 𝐶 + 𝑛 log 350 … … . (2)
log(10−3 ) − log(10−4 )
𝑛=
log 800 − log 350
𝑛 = 2.785
𝑑𝑎
= 𝐶. ∆𝐾 𝑛
𝑑𝑁
log(10−3 ) = log 𝐶 + 2.785 log 800
𝑑𝑎
log = log 𝐶 + 𝑛 log ∆𝐾 𝐶 = 8.22 × 10−12
𝑑𝑁
68
𝜎𝑚𝑖𝑛 𝐾1𝑚𝑖𝑛
𝑅= =
𝜎𝑚𝑎𝑥 𝐾1𝑚𝑎𝑥
𝐹𝑜𝑟 𝑎 𝑔𝑖𝑣𝑒𝑛 ∆𝐾:
𝑅 ↑ → 𝑑𝑎/𝑑𝑁 ↑
𝑇ℎ𝑒 𝑖𝑛𝑓𝑙𝑢𝑒𝑛𝑐𝑒 𝑜𝑓 𝑅 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑡ℎ𝑒
𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙, 𝑒𝑠𝑝𝑒𝑐𝑖𝑎𝑙𝑙𝑦 𝑜𝑛 𝑖𝑡𝑠 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑒
𝑎𝑛𝑑 𝑖𝑡𝑠 𝑒𝑙𝑎𝑠𝑡𝑜𝑝𝑙𝑎𝑠𝑡𝑖𝑐 𝑏𝑒ℎ𝑎𝑣𝑖𝑜𝑢𝑟
𝐹𝑟𝑎𝑐𝑡𝑢𝑟𝑒 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤ℎ𝑒𝑛 𝐾𝑚𝑎𝑥 𝑟𝑒𝑎𝑐ℎ𝑒𝑠 𝐾𝑐 :

∆𝐾𝐶 = 𝐾𝑚𝑎𝑥 − 𝐾𝑚𝑖𝑛 𝐶 = 1 − 𝑅 𝐾𝑚𝑎𝑥
69
Fatigue Crack Growth – Influence of stress ratio
• Elber model explain the effect of stress ratio R on the crack propagation by the crack tip plastic zone
• The plastic zone is constrained by the

elastic zone to a zero-strain state
• Compression stresses arise in the

vicinity of the crack
Fatigue Crack Growth – Influence of stress ratio
• Compression stresses close the crack before complete unloading
𝑆𝑡𝑟𝑒𝑠𝑠 • Elber proposed that the crack can propagate only when it
is fully open and modifies Paris’s Law
𝐾𝑚𝑎𝑥 𝑑𝑎
𝑑𝑎 2
= 𝐶. ∆𝐾 𝑛 = 𝐶𝑒𝑓𝑓 . ∆𝐾𝑒𝑓𝑓
𝑑𝑁 𝑑𝑁
∆𝐾𝑒𝑓𝑓
𝑤𝑖𝑡ℎ ∆𝐾𝑒𝑓𝑓 = 𝐾𝑚𝑎𝑥 − 𝐾𝑜𝑝
∆𝐾
• As the opening level is not easy to determine by experiment,
𝐾𝑜𝑝 Elber proposed to derive it from the complete ∆𝐾 as follow:
𝐾𝑚𝑖𝑛 = 0 ∆𝐾𝑒𝑓𝑓 = 𝐾𝑚𝑎𝑥 − 𝐾𝑜𝑝 = 𝑈 𝑅 . ∆𝐾
• Where U(R) is a function of the stress ratio R for which Elber proposed:
𝑈 𝑅 = 𝐴. 𝑅 + 𝐵
Fatigue Crack Growth – Influence of stress ratio, Elber Model
𝑑𝑎 2
= 𝐶𝑒𝑓𝑓 . ∆𝐾𝑒𝑓𝑓
𝑑𝑁
𝑪𝒆𝒇𝒇 𝒊𝒔 𝒂 𝒕𝒓𝒖𝒆 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝐸𝑙𝑏𝑒𝑟 𝑀𝑜𝑑𝑒𝑙:
𝑑𝑎
= 𝐶𝑒𝑙𝑏𝑒𝑟 𝐴𝑅 + 𝐵 𝑛 . ∆𝐾 𝑛
𝑑𝑁
𝑪 𝒅𝒆𝒑𝒆𝒏𝒅𝒔 𝒐𝒏 𝑹 → 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒄𝒖𝒓𝒗𝒆𝒔

*For a typical aeronautical aluminium alloy 2024, one finds A=0.39 and B=0.61
72
Example 5
Crack in infinite plate (2024-T351, 2 mm thick)
(a) Calculate the critical crack size, 𝑎𝑐𝑟

(b) Calculate the number of cycles to grow from 𝑎𝑜 to 𝑎𝑐𝑟
73
Example 5: Solution
(a) Calculate the critical crack size, 𝑎𝑐𝑟
𝐹𝑎𝑖𝑙𝑢𝑟𝑒 ≈ 𝐾𝑙𝑖𝑚𝑖𝑡 = 𝑆𝑙𝑖𝑚𝑖𝑡 𝜋𝑎𝑐𝑟 = 𝐾𝐶

2
𝐾𝐶
𝑎𝑐𝑟 =
𝑆𝑙𝑖𝑚𝑖𝑡 𝜋
2
2150
𝑎𝑐𝑟 =
300 𝜋
𝑎𝑐𝑟 = 16.34 𝑚𝑚
Example 5: Solution (b) Calculate the number of cycles to grow from 𝑎𝑜 to 𝑎𝑐𝑟
∆𝐾 = 1 − 𝑅 𝐾𝑚𝑎𝑥 = (1 − 𝑅)𝑆𝑚𝑎𝑥 𝜋𝑎
𝑑𝑎
= 𝐶𝑒𝑙𝑏𝑒𝑟 𝐴𝑅 + 𝐵 𝑛 . ∆𝐾 𝑛
𝑑𝑁
𝑑𝑎 𝑛 2
𝑛
= 𝐶𝑒𝑙𝑏𝑒𝑟 𝐴𝑅 + 𝐵 1 − 𝑅 𝑆𝑚𝑎𝑥 𝜋 𝑎
𝑑𝑁
𝑎𝑐𝑟
1 −𝑛ൗ
∆𝑁 = 𝑛
න 𝑎 2 𝑑𝑎
𝐶𝑒𝑙𝑏𝑒𝑟 𝐴𝑅 + 𝐵 1 − 𝑅 𝑆𝑚𝑎𝑥 𝜋 𝑎𝑜
1−𝑛/2 1−𝑛/2
1 𝑎𝑐𝑟 − 𝑎𝑜
∆𝑁 =
𝐶𝑒𝑙𝑏𝑒𝑟 𝐴𝑅 + 𝐵 1 − 𝑅 𝑆𝑚𝑎𝑥 𝜋 𝑛 1 − 𝑛ൗ2
1 16.34(−0.5) − 2.5(−0.5)
∆𝑁 =
5.76 × 10−11 0.649 0.9 100 𝜋 3 0.5
∆𝑁 = 12049 𝑐𝑦𝑐𝑙𝑒𝑠 75
Assignment W5.3
Loading Spectrum (for 1 flights) Data Crack Initiation Data for crack propagation
Smax (MPa) Smin (MPa) n Ultimate stress, Su (MPa) 450 Initial crack size, 𝑎0 5 mm
250 0 10 Fatigue Limit, Se (MPa) 100 𝑚𝑚Τ𝑐𝑦𝑐𝑙𝑒 5.76 E-11
𝐶𝑒𝑙𝑏𝑒𝑟
275 -100 1 𝑀𝑃𝑎 𝑚𝑚0.5 𝑛
300 0 10
290 -50 1 A 0.39
75 0 500 B 0.61
50 -10 1000 n 3
𝐾1𝑐 𝑀𝑃𝑎. 𝑚0.5 37
Thickness, t 2
Yield Stress (MPa) 380
(a) Determine when will a fatigue crack initiate?
Limit stress (MPa) 200
(b) Calculate the time the crack takes to grow from initial crack size R 0.1
to the critical crack size if the Smax=300 MPa

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AIRCRAFT STRUCTURES II

Fatigue Life Damage Estimation

Ts. Dr. Haris Ahmad Bin Israr Ahmad

➢ PART 1: OVERVIEW OF DAMAGE TOLERANCE CONCEPT

➢ PART 2: FATIGUE DAMAGE INITIATION

➢ PART 3: FATIGUE DAMAGE PROPAGATION

Not Fail Safe Fail Safe

Repeated load test at low stress level

To quantify the fatigue endurance of a material, Wohler proposed

𝑅1 < 𝑅2 < 𝑅1 𝑆𝑚1 < 𝑆𝑚2 < 𝑆𝑚3

The higher is the mean stress, the lower is 12

Typical aircraft spectrum:

Complex Spectrum - Miner's rule basis

• The Rainflow method is usually used to analyze a complex spectrum

𝑆𝑒 = 𝐸𝑛𝑑𝑢𝑟𝑎𝑛𝑐𝑒 𝑙𝑖𝑚𝑖𝑡 𝑜𝑟 𝐹𝑎𝑡𝑖𝑔𝑢𝑒 𝑙𝑖𝑚𝑖𝑡

𝐹𝑜𝑟 𝑆𝑜𝑑𝑒𝑟𝑏𝑒𝑟𝑔 𝑅𝑒𝑙𝑎𝑡𝑖𝑜𝑛,

𝐵𝑎𝑠𝑞𝑢𝑖𝑛’𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑆𝑎𝑟 = 𝐴𝑁 𝐵 (1)

log(𝑆𝑒 ) = log 𝐴 + 𝐵 log 1000,000 = log 𝐴 + 6𝐵 (4)

log 𝑆𝑒 − log 0.9𝑆𝑢

𝐺𝑟𝑜𝑢𝑝 𝑆𝑚𝑖𝑛 𝑆𝑚𝑎𝑥 𝑆𝑚 𝑆𝑎 𝑆𝑎𝑟 𝑛𝑖 𝑁𝑖 𝑑 = 𝑛𝑖 /𝑁𝑖

𝐺𝑟𝑜𝑢𝑝 𝑆𝑚𝑖𝑛 𝑆𝑚𝑎𝑥 𝑆𝑚 𝑆𝑎 𝑆𝑎𝑟 𝑛𝑖 𝑁𝑖 𝑑 = 𝑛𝑖 /𝑁𝑖

𝑀𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝐺𝑜𝑜𝑑𝑚𝑎𝑛, 𝐵𝑎𝑠𝑞𝑢𝑖𝑛′ 𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛,

𝐺𝑟𝑜𝑢𝑝 𝑆𝑚𝑖𝑛 𝑆𝑚𝑎𝑥 𝑆𝑚 𝑆𝑎 𝑆𝑎𝑟 𝑛𝑖 𝑁𝑖 𝑑 = 𝑛𝑖 /𝑁𝑖

4 80 220 150 70 101.8 3

𝑇𝑦𝑝𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 𝑓𝑜𝑟 𝑃:

Engineering/Industrial Approach - IQF general law

Engineering/Industrial Approach - IOF general law:

𝑁𝑇𝑜𝑡𝑎𝑙 = 𝑁𝐼𝑛𝑖𝑡𝑖𝑎𝑡𝑖𝑜𝑛 + 𝑁𝑃𝑟𝑜𝑝𝑎𝑔𝑎𝑡𝑖𝑜𝑛

Typical questions answered by Fracture Mechanics

• What is the residual strength of a structure with a given crack length ?

(Opening) (Shearing) (Tearing)

Analysis of Crack Bodies - Stress Intensity Factor (SIF)

𝑆𝑡𝑟𝑒𝑠𝑠 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟

The SIF, is a key parameter of Fracture

The fracture is thus defined by a relationship between the applied load

• Plastic zone sizes at fracture are much larger in thin

• Fracture toughness 𝑲𝒄 of metals is also dependent on:

• Besides that, 𝑲𝒄 can be very sensitive to metallurgical conditions such as:

Let consider the following fracture mechanics data:

𝑲𝟏𝒄 = 𝟏𝟏𝟎𝟎 𝑴𝑷𝒂. 𝒎𝒎𝟎.𝟓

𝑺𝒍𝒊𝒎𝒊𝒕 = 𝟐𝟓𝟎 𝑴𝑷𝒂

(a) Determine the Kc

𝑎𝑐 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑐𝑟𝑎𝑐𝑘 𝑠𝑖𝑧𝑒

𝐶𝑟𝑎𝑐𝑘 𝑝𝑟𝑜𝑝𝑎𝑔𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒

𝐶𝑜𝑚𝑚𝑜𝑛𝑙𝑦 𝑢𝑠𝑒𝑑 𝑙𝑎𝑤: 𝑷𝒂𝒓𝒊𝒔′ 𝒍𝒂𝒘

log(10−3 ) = log 𝐶 + 𝑛 log 800 … … . (1)

log(10−4 ) = log 𝐶 + 𝑛 log 350 … … . (2)

𝐹𝑜𝑟 𝑎 𝑔𝑖𝑣𝑒𝑛 ∆𝐾:

𝐹𝑟𝑎𝑐𝑡𝑢𝑟𝑒 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤ℎ𝑒𝑛 𝐾𝑚𝑎𝑥 𝑟𝑒𝑎𝑐ℎ𝑒𝑠 𝐾𝑐 :

• The plastic zone is constrained by the

• Compression stresses arise in the

𝐾𝑚𝑖𝑛 = 0 ∆𝐾𝑒𝑓𝑓 = 𝐾𝑚𝑎𝑥 − 𝐾𝑜𝑝 = 𝑈 𝑅 . ∆𝐾

𝑪 𝒅𝒆𝒑𝒆𝒏𝒅𝒔 𝒐𝒏 𝑹 → 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒄𝒖𝒓𝒗𝒆𝒔

(a) Calculate the critical crack size, 𝑎𝑐𝑟

𝐹𝑎𝑖𝑙𝑢𝑟𝑒 ≈ 𝐾𝑙𝑖𝑚𝑖𝑡 = 𝑆𝑙𝑖𝑚𝑖𝑡 𝜋𝑎𝑐𝑟 = 𝐾𝐶

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