Unit 5 : LEFM Approach : LEFM concepts, Crack tip plastic zone , Fracture

toughness, Fatigue crack growth, Mean stress effects, Crack growth life estimation.

Questions:
1. Define fracture. Illustrate and name the three modes of fracture.
(Dec. 2011) (04 Marks)
2. Describe the three basic crack displacement modes, with sketches.
(Dec. 06/ Jan. 07) (Dec. 08/ Jan. 09) (May/June 2010) (05 Marks)
3. Under what conditions can LEFM be applied ?(May/June 2008) (04 Marks)
4. Explain the following:
i. Fracture toughness (KC) iv. Energy release rate
ii. Stress intensity factor (K) v. Crack tip plasticity
iii. Modes of crack displacement vi. Threshold stress intensity range
∆௧௛ 
vii. Limitations of LEFM (May/June 2006) (Dec. 06/ Jan. 07) (May/June
2007) (June 2007) (May/June 2010) (June /July 2013) (10 Mark)
5. Explain the following:
i. Plane strain fracture toughness of a material.
ii. Small fatigue cracks
iii. Effect of specimen thickness on fracture toughness.
iv. Mean stress effects on fatigue crack growth behaviour.
v. Monotonic plastic zone and cyclic plastic zone at the crack tip
ௗ௔
vi. − ∆ Sigmoidal curve (Dec. 07/ Jan. 08) (Dec. 08/ Jan. 09)
ௗே
(Dec. 2010) (June /July 2011)
ௗ௔
6. Explain the effect of mean stress on − ∆ Sigmoidal curves.
ௗே
(Dec. 2010) (06 Marks)
7. With a neat sketch, showing the coordinates, write the elastic stress filed
equations around a crack tip. (Dec. 2011) (06 Marks)
8. Define Mode – I stress intensity factor. Distinguish between SCF and SIF.
(Dec. 2011) (05 Marks)
9. Present SIF solutions to five cracked plate geometries.
(Dec. 2011) (05 Marks)
10. Describe the monotonic elastic zone and cyclic plastic zone. Discuss
with a neat sketch, the Dugdale plastic strip modes. (Dec. 2010) (10 Marks)
11. Explain the crack tip plastic zone models. (June 2007) (08 Marks)
12. Discuss the effect of mean stress on fatigue crack growth rates.
13. Derive the equation for the plastic – zone shape near the crack tip using
the plane stress equations. (June /July 2011) (08 Marks)
14. Establish the expression for the stress intensity factor for a thorough
thickness crack in a wide plate under tensile load. `
(Dec. 07/ Jan. 08) (06 Marks)
MMD,Mech Dept., AIT 9964676521 1
Numerical problems
1. A flat plate with a through thickness crack is subjected to a 100 MPa tensile
stress and has a fracture toughness

  of 50 √ . Determine the
critical crack length for this plate assuming the material is linear elastic.
2. Determine the crack length for 4340 tempered steel having following
properties 
 = 60  √ . Design stress, S = 475 MPa, for through
thickness centre crack in a wide plate.
3. A wide thin plate of SAE 1020 cold-rolled steel is subjected to constant
amplitude Uniaxial cyclic loads that produce nominal stresses varying from
 = 200  to  = −50 . The monotonic material properties
are = 630 , = 670 , E = 207 GPa ,  = 104  √ .
What fatigue life would be attained if an initial crack length is 1 mm? State

assumption made. take the Paris Law with A = 6.9 × 10 and n =

3 for this steel. (Dec. 07/Jan.08) (June/July 2013) (Marks 10)
4. A very wide 25mm thick plate of 7075 - T6 Aluminium contains a single
edge crack of length a = 1mm. The plate is subjected to an alternating stress
with  = 175  to  = 0 . Assuming that LEFM applies,
determine
i) Whether plane stress or plane strain condition exists.
ii) Critical crack length at failure.
iii) Number of cycles needed to cause fracture.

Take A = 2.7 × 10 and n = 3.7 , = 560  and 
 =

30  √ . (May/June 2010) (June/July 2011) (Marks 12)
5. A mild steel plate is subject to constant amplitude Uniaxial fatigue load, with
 = 180  to  = −40 . Given yield stress of 500 MPa, UTS
of 600 MPa, Modulus of elasticity 207 GPa and fracture toughness of 
 =
200  √ . the plate contains an initial through thickness crack of 1 mm.
How many cycles of fatigue load can be applied before breakage ?
(May/June 2008)(Marks 10)
6. A supporting plate for a large spring is 95 mm wide, 20 mm thick and 400
mm long. At its ends. A tensile load varies between 855 and 427.5 kN, with a
complete cycle every 12 seconds. The material is ferrite – pearlitic steel, with
a yield strength of 1350 MPa and a fracture toughness of 
 = 110 √ .
A central through crack is discovered and its length measures 15 mm.
Estimate the remaining life of the plate. Take

A = 6.9 × 10 and n = 3. (Dec. 2010)(Marks 14)


MMD,Mech Dept., AIT 9964676521 2

 7. A component in the shape of a larger sheet is to be fabricated from 4340 steel,
which has a fracture toughness of

 = 98.9 √ and a tensile yield
strength are  = 860  . the inspection device used for inspecting the
sheets can not defect flows smaller than 3 mm the part is too heavy as
designed. An engineer has suggested that the thickness be reduced and the
material be heat treated to increase its tensile yield strength to 1515 MPa.
which would result in decreasing fracture toughness to 60.4  √.
Assuming that the stress level does not exceed one half of the yield strength is
the suggested feasible? Explain. (May/June 2010)(Marks 05)
8. A component in the shape of a larger sheet is to be fabricated from 7075- T6
Aluminium, which has a fracture toughness of

 = 24.2  √ and a
tensile yield strength are  = 495  . determine the larger edge crack
that could be tolerated in the sheet if the nominal stress does not exceed one
half of the yield strength. (May/June 2010)(Marks 05)
9. A very wide 60 mm thick plate of mill annealing Ti-6Al-4V Titanium alloy
with through thickness central crack of length 2 = 10 , is subjected to a
nominal stress of 600 MPa normal to the crack plane. Determine
i. The applied stress intensity factor level – K
ii. The effective stress – intensity factor level
 based on the
plasticity correction.
iii. If the load is cyclically varying between  = 600  &
 = 0 , determine the total number of cycles to cause the
failure.
iv. Comment on the applicability of LEFM.
(June/July 2009)(Marks 12)
10. A Uniaxial loaded with plate of low carbon steel with yield
strength.  = 400 ,  = 550 ,

 = 110  √ , ∆
 =
5  √ is subjected to a fluctuating load with  = 250  &
 = 50  the initial through thickness central crack length 2 =

5.0 . The constants for the Paris equation are A = 6.9 × 10 and

n = 3 and for the Walker’s modified equations are = 0.5  = 0.3.
determine
i. K at the beginning and compare it with ∆
 
 . what
conclusion you can make ?
ii. The plane stress plastic zone size 2  . Check the applicability of
LEFM approach.
iii. Critical crack length  .
iv. Total number of life cycles before fracture.
(June 2007) (Dec. 09/Jan.10) (Marks 12)
MMD,Mech Dept., AIT 9964676521 3
 Fracture mechanics approach:-
It is based on the assumptions that there exists crack in a structure or
component. The cracks may be man- made, i.e. Holes, notches, slots or cut out,
etc. The crack may be due to manufacturing defects like slag, inclusion, voids or
the presence of foreign particles.
A crack may also be nucleated and grown during the service of component.
Therefore fracture mechanics deals with crack which is lightly to grow at certain
load condition. Use of fracture mechanics yields to lower F.O.S for a part. This
reducing the cost and weight of a part and increase the reliability.
LEFM : Linear Elastic Fracture Mechanics :
A brittle material is the one in which material remains elastic even at the
crack tip where stress are high (But most of engineering material do not fall in the
category). Eg. Diamond known to be brittle even in the vicinity of crack tip.
Analysis of brittle material is for simpler than the analysis of material having a
plastic zone at crack tip.
Presence of plastic zone incorporate two types of
− ∈ approach behaviour. i.e
plastic behaviour inside plastic zone and elastic behaviour outside it.
Therefore one has to keep crack of probably arising in the interface between two
zones.

Define Fracture. Illustrate and name the three modes of fracture.

Describe the three basic crack displacement modes with sketches.
Solution : Fracture in Engg: - Rupture of material too weak to sustain the force
on it. The fracture of a work piece during forming can result from flaws in the
metal. The structure or parts subjected to vibrations or other cyclic loading must be
designed to avoid the fatigue fracture.
Loading modes of fracture failure:
Mode –I In this opening mode, the dominant displacement is normal
to the crack surface it is most common displacement.
Crack surface
y y
y Crack tip

x
x x

Mode II (Sliding Mode) z

z Mode III (Tearing Mode)
Mode I (Opening Mode) z
Modes of crack extension

MMD,Mech Dept., AIT 9964676521 4

Mode II or Sliding Mode or in Plane shear mode. In this the displacement mode
is in the plane of plate.
Mode III or tearing mode or out plane mode displacement is parallel to the crack
point.
Mixed Mode fracture:

ߚ
2a
y

Modes of crack extension

Mixed Mode I and II loading due to a crack on an inclined plane.

A mixed mode crack extension i.e more than one mode present in the type crack
tends to grow on plane of maximum shear. Fig. shows an example of mode I and
mode II crack extension. This involves axial loading of a crack inclined with
respect to X - axis when
= 90଴ . ூ is maximum and ூூ is zero. Even for
conditions
is more but less than 90଴ mode I contribution tends to dominant the
crack tip stress at field. Mode III associated with pure shear condition, E.g. A
round notched bar loaded in tension.

Parameter Basic of Symbol Name

Development
G ( V/mm) Energy based In honour of Griffith
Crack density
force

K ( MPa √ ) Stress based In honour of Keis

Stress intensity
factor
J ( N-mm) Energy based J.R. Rose (for EPFM)
Integral
Crack Tip Open Displacement
Displacement

MMD,Mech Dept., AIT 9964676521 5

G – Energy release rate:
Energy released per unit increases in area during crack growth. Here the rate
is defined with respect to change in crack area.
R – Crack resistance
Energy required by the crack per unit increase in area or extension.
R is replaced with surface energy because, during crack growth in elastic
deformation also occurs up to certain depth of the crack surfaces. R in fact sum of
the energies required to
i) To form a new surface, i.e. above and below crack.
ii) Cause an elastic deformation.
∵ Energy release rate of crack must be greater than crack resistance to have change
of crack to grow.
∵ G ≥ R ( for crack to grow)
If G exceeds R, the crack acquires KE and crack grow at faster speed.
Therefore, to avoid fracture failure, plastic zone at crack tip is best suited. i.e. a
larger amount of energy is required to advance crack tip, became energy is
continuously dissipated by advancing crack tip to elastic deformation.
Consequently crack remains sleeping or dormant. By confining our attention only
to elastic material (Brittle) we would be able to obtain closed forms to many
problems and also able to learn the singularity ( infinite stress at the crack tip)
௄మ
Energy release rate G = Plane stress condition
ா

E = Strain energy (not Young’s Modulus)

௄మ
G = 1 – νଶ  Plane strain condition
ா

Fatigue crack growth and damage tolerant design, they fall into are often needed.
1) Stress intensity factor 
2) The fracture toughness௖ 
3) Applicable fatigue crack growth rate expression.
4) Initial crack size௜  ௢ 
5) Final or critical crack size௙  ௖ 

MMD,Mech Dept., AIT 9964676521 6

 LEFM method is used to determine crack growth in material under the
assumption → Material condition is predominantly linear elastic during fatigue
process.
But for crack growth and fracture condition. that violate basic assumption & uses
the Elasto-Plastic Fracture Mechanics (EPFM). Approach to describe fatigue and
fracture process.
Note :-
1) Energy released (G) is energy based & is applicable to brittle or less ductile
material.
2) Stress intensity factor (K) is stress based & is also applied for developed for
brittle or less ductile material.
3) J - integral has been developed to deal with ductile material (it, formulation is
quite general and can be applied to brittle material also).
4) CTOD - Crack Tip Opening Displacement was also developed for ductile
material and the name itself suggest that is displacement based.
Stress Intensity Factor : [K]
Definition : - It is defined as the intensity of load transmitted through crack tip
region induced by the introduction of crack in a flaw free body.
y
The stress field at any point near the
crack tip can be described as

 = 1 + sin + sin
ߪ௬ 
cos
ߪ௫ √   

 = 1 − sin + sin

Crack r ߬௫௬ cos
√   
ߠ
a
 =
x 
cos sin cos
√   

Elastic stresses near the crack tip
⁄ ≪ 1  
= 
= 
= 0 For plane stress


=  ( +  )  For plane strain

= 
= 0

Fig. shows through the thickness of crack in a elastic stress near crack tip( ⁄ << 1)

Consider a through thickness sharp crack in a Linear elastic isotropic body

subjected mode I loading, in a arbitrary stress element in the vicinity (near) of the
crack tip with co-ordinate r and
related to crack tip and crack plane is also shown
in Fig.
MMD,Mech Dept., AIT 9964676521 7
Fig shows elastic normal and elastic shear stress in the vicinity of crack tip is
dependent on ‘
’ ‘’  ‘’ only. The magnitude of these stress at a given point
are entirely dependant on K. for this reason. K is called “stress field parameter” or
“stress intensity factor”.
The value of K depends on loading, crack shape , mode of crack displacement
component, specimen or structure configuration.
From above Fig. it is observed that elastic stress distribution in Y direction for
 = 0 , as radius r approaches to zero. The stress approaches infinitely at crack tip
and then stress singularity exists at
= 0, since infinite stress cannot exists in a
physical body, the elastic solution must be modified to account for same crack tip
plasticity.
If plastic zone radius ‘

’ , at small crack tip is small relative to local geometry,
little or no modification to the stress intensity factor, K, is needed. Thus, an
important restriction to the use of linear elastic fracture mechanics is that the plastic
zone at the crack tip must be small relative to geometrical dimension of specimen
or part. So K is the fundamental parameter of LEFM.
[Note : stress concentration factor   is different from stress intensity factor  ]
K expression for common crack member:
For expression refer data handbook.

Crack Tip Plastic Zone:

Surface

Mid section
Surface Plane stress
Mode I
y Plane strain
Crack z Mode I
tip x
Fig. Plastic zone size at the tip of a through - thickness crack

1 
ଶ
 
2 ௬
r =

The above fig. shows resultant monotonic plastic zone shape for Mode I using von
Misses criteria for plane stress condition, A much larger plastic zone exists
compared to that of plane strain condition.
MMD,Mech Dept., AIT 9964676521 8
This is because

having different values for plane stress than for plane strain then


= 0. This will restrict plastic flow resulting smaller plastic zone compared to
plane stress.
Therefore let us assume that a through thickness crack exists in a thick plate. The
plate free surface have

= 
= 
= 0 . Therefore free surface must be
plane stress condition.
However the interior region of the plane plate near the crack tip is closer to plane
strain condition as a result (Small plastic zone size) of elastic constraints away from
the crack. Thus, the size of plastic zone along the crack tip varies similarly to that
as shown in above figure. Among several model available to describe plastic zone
shape may be circular, strip extending behind the crack tip, or a butterfly shape.
The above figure is just one of several models butterfly shape.
Note: Most plastic zone model were developed using classical yield criterion
model to determine boundaries where the material begins to yield.
Two such models for approximating the plastic zone size are discussing in next
topics

Irwin’s Plastic Zone Model:

By substituting the yield strength  for
 or ahead of the crack tip =0 can
be determined from the equation

= 
= 
= 0 . For plane stress condition


=  ( +  ) 

= 
 = 0
For plane strain


= for  = 0

… … (1)
√

This identifies the plastic zone boundary as a result

r = 2  , the plastic zone size in fig (a) is expressed as

r = … … (2)
 ೤

Because of plastic relaxation, redistribution stress field in the plastic zone, the
actual plastic stress plastic zone size as estimates by Irwin approximate twice the
value of equation (2).
(Note: He argued that plasticity at the crack tip causes the crack to behave as it is
larger than the true physical size).

MMD,Mech Dept., AIT 9964676521 9

Because of this stress distribution could not simply terminate above the yield
strength
௬ . Thus stress distribution for
௬ (refer fig a ) must be shift to right to
accommodate the plastic deformation and satisfy equilibrium equations. Thus under
monotonic loading the plane stress in the plane of crack is

2௬ = 2    =  
ଶ ଶ
ଵ ௞ ଵ ௞
… . . . (3) from plane stress
ଶగ ௌ೤ గ ௌ೤

For plane strain condition equation (4) usually 1/3rd of the plane stress value
  Where ௬ plastic zone radius
ଶ
ଵ ௞
2௬ ≅ … … (4)
ଷగ ௌ೤

Dugdale Plastic zone strip model:

y
Sy Plastic zone Sy

R
a
2c R

S
Fig. Schematic of Dugdale plastic zone strip model

This model is widely used and formulated by Dugdale in the plane stress condition.
He modelled plastic region extending behind the physical crack length as narrow
strip having a distance R shown in Fig. According to this model on interval stress
௬ acting over the distance R recloses. The crack to a length of 2c this interval
stress equals the material yield strength
௬ . Dugdale showed the plastic zone

 
ଶ
గమ ௖ ௌ
distance R as R ≅ … … (5)
଼ ௌ೤
By rearranging equation (6) in terms of k and
௬ gives

 
ଶ
గ ௄
R ≅ … … (6 )
଼ ௌ೤
K = S√
(Note: For plane stress plastic zone size in the crack plane for equation 3 and that
from equation 6 is within 20%)
From above fig the crack edge R in front of physical crack carry the
௬ tends to
close the crack.
MMD,Mech Dept., AIT 9964676521 10
The size of R chosen such that stress singularity is appears this means that stress
intensity


Due to uniform stress distance S has to be compensated by stress intensity due to
wedge form  
∴ 
= − 

→ S√ +
 → 2
 
 cos
 

∴ S√ +
 
= 2  cos
 
మ మ 
Neglecting the higher order terms, = మ
೤

  ∴  = S√
 మ 
= మ =
೤ ೤

Limitations of LEFM:
If the plastic zone area  at the plane zone tip is small relative to the local
geometry little or no modification of  is needed. Under monotonic loading
without significant violation of LEFM principle use of K is that  ≤


 ≤

For cyclic loading condition, approximately (Cyclic plastic zone size is

smaller than the monotonic plastic zone)

Also  ≤
 
and where t = thickness, w-a is the uncracked ligament along
the plane of crack.
Therefore limitation of LEFM is that the plastic zone size at crack tip must be small
relative to the crack length as well as the geometrical dimensions of the specimen
or fact.
Net nominal stress in the crack plane must be less than the yield strength i.e, in
actual usage the net nominal stress should be less than 80% of the yield strength.
(For the case plasticity correction of nearly 20% is required for , the time EPFM
must be used).
Fracture toughness:
 for linear stress condition
Definition: Critical stress intensity factor  or fracture toughness is one when
critical value of  refers to the condition in which a crack extend in rapid manner
 =  
೎
without an increase in load. f


MMD,Mech Dept., AIT 9964676521 11

Where
௖ = Applied nominal stress at crack instability
௖ = Crack length at instability
௖ = Fracture toughness. (It depends on material strain rate, environment
,temperature, their crack length).
௖ gives qualitative design parameters to prevent brittle type fracture, also ௖
represents critical value of K for given load crack length and geometry required to
cause fracture. It also represents stress at last cycle of the fatigue fracture.
Mixed
Plane Plane
mode
stress Strain

ࢉ , √

100
Thin
80 B

B Medium Thick
60
B
Fracture toughness

40
K Ic
20

0
0 10 20 30

Thickness (B), mm
Fig. Effect of specimen thickness on fracture toughness

Fig shows relationship between ௖ and thickness of material. It also shows the
appearance of fracture surface for different thickness of single edge notch
specimens.
The fracture toughness values would be higher for dull or notched type, crack front.
Thin parts have a high ௖ value accompanied by appreciable “shear lips” or slant
fracture. (This type is a high energy fracture) As the thickness is increased, the
percentage of shear lips or slant fracture decreases, as does ௖ ( i.e., ௖ also
decrease). This type of fracture is called “mixed mode”, implying slant and flat
fracture.
For thick parts, entire fracture surface is flat (Fig 3) and ௖ approaches as minimum
value. Any further increase in thickness does not alter the fracture surface
appearance. The minimum value of the ௖ is called plain strain fracture toughness
(ூ௖ ). For thin section where appreciable shear lip occurs, the crack tip region most
closely experiences a plane stress situation. Then plastic zone sizes of the fracture
is much larger in thin parts than in thick parts.
Plane strain fracture toughness ூ௖ is considered a true material property because it
is independent of thickness. Plane strain fracture toughness value to be considered
ଶ
a and t ≥ 2.5   .
௄಺೎
valid fracture toughness it is required that,
ௌ೤
MMD,Mech Dept., AIT 9964676521 12
Thickness value determination the basis of s calculated yield strength to young’s
ௌ೤
modulus ( if stress intensity factor
>
௖ the crack grows Kc is found by
ா
experimental method for specimen. Stress intensity is a parameter to measure the
severity of stress at crack tip but
௖ is limit on K.
MPa
580 1120 1680 2240
240
Fracture toughness
ࡵࢉ , √

240

200 Vim + Var 200

√
160 Var 160

ࡵࢉ ,
120 Air 120

80 80
Aluminium
alloys
Titanium
40 alloys Steels 40

0 0
0 40 80 120 160 200 240 280 320 360

Yield strength (Sy) , ksi

Fig. Locus of plane strain fracture toughness versus yield strength

Vacuum Induction Melting[VIM ]or Vacuum arc Melting [VAR] of the base
material produces high
ூ஼ value for yield strength than just single vacuum arc or
air melting of steel(AIR). Thus, low impurity provides better fracture toughness.
Fracture toughness
ூ஼ 
ூ is very sensitive to metallurgical process.
Fracture toughness
ࡵࢉ , √

200

Fracture ሺࡷ ሻ
ࡵࢉ
Yield strength (S ) , MPa

toughness 1000
150
y

Yield stress, (S y ) 750

100

500

50
250

0
-150 -100 -50 0 50
Temperature, 0 C
Fig. Variation of with ࡷࡵࢉ temperature for low alloy nuclear pressure vessel steel A533B

FIG: Variation of
ூ஼ with temperature result for a low - alloy nuclear vessel steel.
As temperature decreases
஼ decreases and yield structure increases, even for
uncracked or unnotched tensile strength increase with decrease in temperature. The
crack or flaw resistance can be drastically reduced, increased strain rate tends to
cause change in
஼ similar to those resulting from a decrease in temperature.

MMD,Mech Dept., AIT 9964676521 13

Influence of the Fracture toughness on allowable stress and crack size.

ࡵࢉ
0.8 S y =
√ 

Increasing KIc
Stress, S (MPa)

Higher K Ic

Lower K Ic

Crack length , a ( m )
Fig. Influence of fracture toughness on allowable stress or crack size

Fig shows how changes in fracture toughness influences the relationship

between allowable nominal stress and allowable crack size, which is a plot of

಺಴
S = . The allowable stress in the presence of given crack size is directly
√
proportional to the fracture toughness while the allowable crack size for given
stress is proportion to the square of fracture toughness. This increases in  has a
much larger influence on allowable crack size than the allowable stress.
For monotonic loading of components containing existing crack has higher the
fracture toughness results in larger allowable crack size than on allowable stress.
For  value corresponding to the lower  curve, the dashed region below the
lower  curve identify the no fracture region for combined stress and crack
length. But region above this curve identifies the fracture region for monotonic

಺಴
region. From the above equation i.e. S = a large stress is necessary to cause
√
fracture in the presence of smaller.
Note: However remember that limitation of LEFM that allowable stress should not
exceed 80% of the yield strength. Thus 0.8  as identified in Fig.


Fatigue Crack Growth: Sigmoidal Curve  − ∆ 
Explain the procedure for obtaining - ∆ K Plot.
Fig shows crack length v/s applied cycle curves for three identical test specimens
subjected to different repeated stress levels with  >  >  . All specimens
contain the same identical crack length  ,and each test the minimum stress was
zero. Refer S –N Plot. With the higher stress the crack growth rate that are
represented by slope of the curves are higher at a given crack length and fatigue
crack growth life (total number of applied cycles,  ) is shorter.

MMD,Mech Dept., AIT 9964676521 14

The crack lengths at fracture are shorter at the higher stress levels. for the given
initial crack size, the life to fracture depends on the magnitude of applied stress and
fracture resistance of the material.

 ∆
௙ , ௙  =
 ∆
Crack length, a

ଵ >
ଶ >
ଷ
∆

ଷ

ଶ

ଵ

> ∆

ao S

Time
Applied cycles, N
Fig. Fatigue crack length versus applied cycles. fracture is indicated by X

, is simply the slope of the  versus  curve at a


The fatigue crack growth rate,


 in above Fig.

 ∆
given crack length or given number of cycles, as identified by

 ∆

The corresponding applied stress intensity factor ∆ K is calculated knowing the

crack length (a) and the applied stress range (∆ )
Therefore ∆ = ∆  = K  − K 
=   √  -  √ 
= (  −   )√ 
∆ = ∆ √ 
Because stress intensity factor K = S√  is undefined in compression   is
taken as zero if   is compression.
The correlation of constant amplitude loading is usually log-log plot of fatigue
crack growth rate,


, 

  , v/s the opening mode stress intensity factor
range ∆  ∆ in MPa√ . As ∆  primarily depends on stress intensity
range ∆, Crack length  and geometry (for example, )

= f(∆ ,  ,


Many models of the form ) = f (∆)


MMD,Mech Dept., AIT 9964676521 15



Sigmoidal − ∆ curve



ࡷࢉ or
in stability

݀ܰ
݀ܽ
Crack growth rate,
n


= A (∆)௡


log scale

for linear
portion

∆ࡷ࢚ࢎ

Region I Region II Region III

Stress intensity factor range, ∆

log scale

Schematic sigmoidal behaviour of fatigue crack

∆
KI or mode I predominant mode of macroscopic fatigue crack growth mode II and
III have second order effects on both crack direction and crack growth rate. The

typical log-log plot of versus ∆ shows a sigmoidal shape that can be divided

into 3 major regions. Region I is near the threshold region and indicates a threshold
value ∆ , below which there is no observable crack growth. This threshold
occurs at crack growth rate on the order of 1 × 10 /  .
Microstructure, mean stress, frequency and environment mainly controls region I
crack growth.
Region II :

Fig shows a linear relationship between og and log ∆  which corresponds to


the formula = A (∆) .First suggested by Paris, Gomez and Anderson.

where n is the slope of line and A coefficient found by extending the straight line to
∆ = 1 MPa √ . Region II (Paris region) fatigue crack growth corresponding to
stable microscopic crack growth that is typically controlled by environment. In the
region II less influence of microstructure and means stress level as affected in
region I.
Region III:
The fatigue crack growth rate are very high and they approach instability and little
fracture growth life (  ) is involved. this region is controlled primarily by fracture
MMD,Mech Dept., AIT 9964676521 16
toughness



, which is in turn depends on the microstructure, mean stress
and environment as sown in fig. The fatigue crack growth time spent in region III is
small compared to that in region I and II.
Therefore the stress intensity factor range is principle controlling factor in fatigue
crack growth.

This allows the fatigue crack growth v/s ∆
date obtained under constant

amplitude conditions with simple specimen configurations to be used in design
situations.
If one knows the stress intensity factor expression, K, for the given component and
loading the fatigue crack growth life of component can be obtained by integrating
the sigmoidal curve between the limits of initial crack size and final crack size. The
greatest usage of this method has been in damage tolerant design for many
aerospace application, as well as nuclear energy same along with fractographic
failure analysis. However, as with all constant amplitude material fatigue
properties, these data alone do not provide sufficient information on fatigue crack
growth interaction or load sequence effects.

Mean stress Effect (Paris law):

Paris law (also known as the Paris erdogon law) relates the stress intensity factor
range to sub criteria crack growth under a fatigue stress region as such it is the most
popular crack growth model used in fracture mechanics the basics formula.

= A (∆
) where a = crack length , N=No. of load cycles.


A and n are material constants.


= crack growth rate


∆
range of the stress intensity factor

General influence of mean stress effect on fracture crack growth behaviour is
shown schematically in above fig. the stress ratio

೘೔೙ ೘೔೙
R =

೘ೌೣ
=
೘ೌೣ
is used as principle parameters consider positive stress ratio  ≥ 0.

The increasing R ratio ( means increasing mean stress) has tendency to increase the
crack growth rate in all portions of the sigmoidal curve.
The region III where


 controls the substantial difference in crack growth
rate occurs for different R ratios.

MMD,Mech Dept., AIT 9964676521 17

 Stress intensity factor range,
10 100
1

R = 0.
R = 0. 2
R = 0.3
R = 0.4
Crack growth rate, R = 0.5

Crack growth rate,

R = 0.
R = 0. 2
R = 0.3
R = 0.4
R = 0.5

1
10 100
Stress intensity factor range,

Fig. Schematic mean stress influence on fatigue crack growth rates

The upper transition region on the curves are shifted to lower ∆
values as R,
hence

 increases. the effect of R ratio on fatigue crack growth behaviour is
strongly material dependant. A common used equation depending mean stress
effects in region II and region III in the following equation.



 
(∆) 
(∆)
= 
= 
… … ()
  ∆  

Where ′ & N  are empirical material fatigue crack growth rate constants &

applicable fracture toughness for the material and the thickness. The above

equation is modification of the Paris equation = A (∆
) that incorporates

mean stress and region III fatigue crack growth behaviour.
It should be recognized that ′ & N  are different from the coefficient and slope
rates A and n in the Paris equation.

A shown
 the denominators approaches to zero; thus, the crack growth rate

becomes very large. This describes in region III type of crack growth behaviour. if
′ & N  are known, the various stress ratio curves generally collapse into a single
curve.
Another equation used by Walker for mean stress effects with R ≥ 0.


 
(∆)
=  ( ) = ′′ (∆
) … … (2)


MMD,Mech Dept., AIT 9964676521 18

Where A and n are the Paris coefficient and slope for R=0 and
is a material


constant. ( Paris equation = A (∆) and equation (2) are similar). Where


′′ =

 ೙(భషഊ)

In equation (1) ‘R’ does not affect the slope ‘n’. But in equation (2) effect f R on
fatigue crack growth is material dependant. It is necessary to determine material
constant (
). The value of
for various metals range from 0.3 to nearly 1. ( if data
not available a value of
= 0.5 is often used)

The effect of mean stress on ∆

The R increasing from 0 to about 0.8 the threshold ∆  decreases by factor
of about 1.5 to 2.5. This has the effect of shifting the curve towards the abscissa
which reduces ∆  for a given crack length by the same factors the effect of R
ratio on the threshold gives ∆   = ∆  1 −  () where is an
empirical constant fitted to that for positive R values other than zero.

Cyclic plastic zone size


 
Equations 2  = describe the monotonic plastic zone size developed at
 ೤
the crack tip. this is the plastic zone size developed due to the maximum load
applied in the loading cycle(point A in Fig) produces the monotonic plastic zone
size and stress distribution shown in Fig. However, maximum load is reduced
during the loading cycle to the minimum load(point B in Fig). The local stress near
the crack tip is reduced to a value less than that observed for the monotonic plastic
zone size. This unloading causes the development of the cyclic plastic zone size,
2  ′ , and the corresponding stress distribution, as shown in. Fig.3. thus Fig illustrate
monotonic (Fig.2) and cyclic plastic zone (Fig.3) sizes at the maximum and
minimum loads of a single load cycle.
For the assumed elastic perfectly plastic behaviour , the maximum stress developed
in (Fig.2) is  . The maximum stress change developed during unloading can be as
large as 2 . Thus, (Fig.3) shows the sum of the inelastic loading stress
distribution plus the elastic unloading stress distribution. note that the stress within
the cyclic plastic zone , 2  ′ , are compressive stress decreases and then becomes
tensile.
Thus, if a region yields in tension during loading, as shown in (Fig.2) after
unloading a portion of that region is in compression, as shown in (Fig.3).
MMD,Mech Dept., AIT 9964676521 19
the size of the cyclic plastic zone where yielding occurs can be approximated by
using 2



& ∆  
Cyclic Plastic zone size For plane stress condition. It gives,
 
2
≈   =  
′  ∆  ∆
This is R = 0.
  ೤  ೤

The cyclic plastic zone for plane strain condition using analysis similar to those
yields a value 1 3 as large as that of corresponding cyclic zone for plastic zone for
plane stress.
Cyclic Plastic zone size For plane strain condition. It gives,
 
  =  
′  ∆  ∆
2
≈ This is R = 0.
  ೤   ೤

MMD,Mech Dept., AIT 9964676521 20

Numerical problems:
1. A flat plate with a through thickness crack is subjected to a 100 MPa tensile
stress and has a fracture toughness
ூ௖  of 50 √ . Determine the
critical crack length for this plate assuming the material is linear elastic.
Solution:
At fracture, K ୍ୡ = K ୍ = S ௖

50 = 100 × × ௖
௖ = 0.0796 = 79.6
  ℎ = 2௖ = 159

2. Determine the crack length for 4340 tempered steel having following
properties ூ௖ = 60  √ . Design stress S = 475 MPa, for through
thickness centre crack in a wide plate.
Solution: At fracture, K ୍ୡ = K ୍ = S ௖

60 = 475 × × ௖
௖ = 0.00507 = 5.07
  ℎ = 2௖ = 10.14
For the above problem , if allowable design stress reduces to 320 MPa. i.e., S = 320
MPa , 2௖ = ?
K ୍ୡ = K ୍ = S ௖

60 = 320 × × ௖
௖ = 0.01119 = 11.19
  ℎ = 2௖ = 22.38
Then to attain 22.38 mm crack dimension with , S = 475 MPa, K ୍ = ?
= K୍ = S ௖

K ୍ୡ = 475 × × 11.19 × 10ିଷ

K ୍ୡ = 89.06 MPa√m
Thus material exhibiting higher fracture toughness has higher residual strength for
a given crack.
MMD,Mech Dept., AIT 9964676521 21
 3. A component in the shape of a larger sheet is to be fabricated from 7075- T6
Aluminium, which has a fracture toughness of

 = 24.2  √ and a
tensile yield strength are  = 495  . determine the larger edge crack
that could be tolerated in the sheet if the nominal stress does not exceed one
half of the yield strength. (May/June 2010)(Marks 05)
Solution: Material 7075 Aluminium,

 = 24.2  √ &  = 495 
1 1
S =  = × 495 = 247.5 
2 2
Final crack length, K  = K  = S 
1
 1 24.2
 = = = 0.002426 
 247.5 × 1.12
 = 2.426 
4. A component in the shape of a larger sheet is to be fabricated from 4340 steel,
which has a fracture toughness of

 = 98.9 √ and a tensile yield
strength are  = 860  . the inspection device used for inspecting the
sheets can not defect flows smaller than 3 mm the part is too heavy as
designed. An engineer has suggested that the thickness be reduced and the
material be heat treated to increase its tensile yield strength to 1515 MPa.
which would result in decreasing fracture toughness to 60.4  √.
Assuming that the stress level does not exceed one half of the yield strength is
the suggested feasible? Explain. (May/June 2010)(Marks 05)
Solution:
i. Material 4340 steel,

 = 98.9  √ and  = 860 
1 1
 =  ;  = × 860 = 430 
2 2
1
 1 98.9
 = = = 0.01342 
 430 × 1.12
 = 13.42 
ii.

 = 60.4  √ and  = 1515 
1 1
 =  ;  = × 1515 = 757.5 
2 2
1
 1 60.4
 = = = 1.6133 × 10 
 757.5 × 1.12
 = 1.6133 
The suggestion is feasible, because the critical crack length is decrease as the
appreciable changes in fracture toughness & yield strength altered as given in
problem.
MMD,Mech Dept., AIT 9964676521 22
 5. A wide thin plate of SAE 1020 cold-rolled steel is subjected to constant
amplitude Uniaxial cyclic loads that produce nominal stresses varying from


 = 200  to

 = −50 . The monotonic material properties
are
 = 630 ,
 = 670 , E = 207 GPa ,  = 104  √.
What fatigue life would be attained if an initial crack length is 1 mm? State
assumption made. take the Paris Law with A = 6.9 × 10 and n =

3 for this steel. (Dec. 07/Jan.08) (June/July 2013) (Marks 10)
Solution: Since the problem states that the plate is thin
Thin plate ↔ Plane stress condition
Assume corrosion environment & room temperature effect is not involved.
The initial value of 
 is calculated with

 = 200  &  = 1 
For a through thickness edge crack

K  = 1.12 S √  = 1.12 × 200 × × 0.001

K  = 12.6  √
1 



2r =

  = 0.000127  = 0.127 
1 12.6
2r =
630
2 r = 0.127 
r = 0.0635 
Initial crack length,  = 1 
Plastic zone radius, r = 0.0635  r ≤ a8

≤ 18 ∴=1

0.0635 ≤ 0.125 mm, ∴ Application of LEFM is applicable.

The Paris crack growth rate equation is often a reasonable expression for region II

& even region III Paris equation, = A (∆)


This equation is developed for R = 0, the small compressive stress -50 MPa, will
not have much effect on crack growth behaviour and can be neglected.
Thus ∆
=

 −

 = 200 − 0 = 200 
∆ = 
 − 
 = 12.6 − 0 = 12.6 √
From table no (6.19) ,  = 6.9 × 10 ; = 3
MMD,Mech Dept., AIT 9964676521 23


= A (∆)
= 6.9 × 10 × 12.6 = 1.38 × 10 


This initial fatigue crack growth rate falls within the limit of the Paris region of
crack growth rate the final crack length     can be obtained from setting
 at fracture equal to 
∆ = ∆ S √ 

= A (∆ S √ )
=  ∆ S
 
 =  

  


   



 = 
 
1
 ∆ S    






 =  
 =   
  
Integration,  
 ∆        





1
 ∆ S
 
=





 

= "
 
 ! ≠ 2,  
  

   
 

  −  

   
   
 
=
− #!2$ + 1

  −  

   
   
∴  =
 

#− !2 + 1$  ∆ S
 




' *
∴  =
2 & 1
− 
 ))
1
&
! − 2 ∆ S
  
& )


 

%   (
Use equation to obtain the critical length at fracture
  
 = + - = + - = 0.068  = 68 
1 1 104
 ,   200 × 1.12

=  

య ሺయషమሻൗ − ሺయషమሻൗ
×. × షభమ  × య ×  ൗమ × . య . మ .  మ

= 4631.397 ×  
1 1
−
0.0316 0.2608
 = 128804
MMD,Mech Dept., AIT 9964676521 24
 6. A very wide 25mm thick plate of 7075 - T6 Aluminium contains a single
edge crack of length a= 1mm. The plate is subjected to an alternating stress
with

 = 175  to

 = 0 . Assuming that LEFM applies,
determine
i) Whether plane stress or plane strain condition exists.
ii) Critical crack length at failure.
iii) Number of cycles needed to cause fracture.
Take A = 2.7 × 10 and n = 3.7 ,
 = 560  and  =
30  √. (May/June 2010) (June/July 2011) (Marks 12)
Solution :
i. Since the problem states that the plate is thick plane strain condition
ii. Critical crack length to cause failure
K  = 1.12 S √ 
1  
1 30 
 = = = 7.457 × 10 


 175 × 1.12
 = 7.457 
iii. Number of cycles needed to cause fracture

∆ S = 175 


 = 1.12 S √  = 1.12 × 175 × √ × 0.001


 = 10.98 MPa√ 
 = 0 MPa√

∆ K = 
 − 
 = 10.98 − 0 = 10.98 MPa√

 
2  1 1 
∴ =   −  
 − 2 ∆ S  
 
 
 

  

 
  
 = య.ళ ሺయ.ళషమሻൗ − ሺయ.ళషమሻൗ
.×. × షభభ  ×య.ళ × ൗమ ×.య.ళ . మ . మ

= 17.34 × 290.5

 = 5026.812 

MMD,Mech Dept., AIT 9964676521 25

 11. A Uniaxial loaded with plate of low carbon steel with yield
strength. 
= 400 ,  = 550 ,  = 110  √ , ∆ =
5  √ is subjected to a fluctuating load with   = 250  &
  = 50  the initial through thickness central crack length 2 =
5.0 . The constants for the Paris equation are A = 6.9 × 10

and

n = 3 and for the Walker’s modified equations are = 0.5  = 0.3.
determine
i. K at the beginning and compare it with ∆   . what
conclusion you can make ?
ii. The plane stress plastic zone size 2
. Check the applicability of
LEFM approach.
iii. Critical crack length  .
iv. Total number of life cycles before fracture.
(June 2007) (Dec. 09/Jan.10) (Marks 12)
1. A wide thin plate of SAE 1020 cold-rolled steel is subjected to constant
amplitude Uniaxial cyclic loads that produce nominal stresses varying from
  = 300  to   = 100 . The monotonic material properties
are 
= 630 ,  = 670 , E = 207 GPa ,  = 104  √.
What fatigue life would be attained if an initial through thickness crack of 1
mm length existed in the plate? State assumption made. Take the Paris Law

with A = 6.9 × 10 and n = 3 for this steel. Take = 0.5 in the

walker’s relationship. (Dec. 07/Jan.08) (June/July 2013) (June/July
2014) (Marks 10)
Solution: Since the problem states that the plate is thin
Thin plate ↔ Plane stress condition
Assume corrosion environment & room temperature effect is not involved.
The initial value of   is calculated with   = 300  &  = 1 
For a through thickness edge crack

K  = 1.12 S √ = 1.12 × 300 ×  × 0.001

K  = 18.82  √
1 

2r =  
 

K  = 1.12 S √ = 1.12 × 100 ×  × 0.001
K  = 6.27  √
MMD,Mech Dept., AIT 9964676521 26
 ∆
=

 −

 = 18.82 − 6.27 = 12.55 √

walker’s relationship.
 6.9 × 10
 =
′′
( )
= = 1.27 × 10
1− 1 − 0.33 ( . )

 (∆
)
ᇲ

= = ′′ (∆
)
1 − ( )


= ′′ (∆
) = 1.27 × 10 × (12.55) = 2.51 × 10
m
cycle
For final crack length
Use equation to obtain the critical length at fracture


 =   =   = 0.0305  = 30.5 
1 1 104
 
   300 × 1.12

1 
 
 
2r =

  = 2.93 × 10 
1 18.82
 630
2r =

2 r = 2.93 × 10 
r = 1.46 × 10 
Initial crack length,  = 1  Plastic zone radius,r = 1.46 × 10 
r ≤ a8 r ≤ 18 ∴=1

1.46 × 10 ≤ 0.125 mm, ∴ Application of LEFM is applicable.

The Paris crack growth rate equation is often a reasonable expression for region

II & even region III Paris equation, = A (∆
)


This equation is developed for R = 0, the small compressive stress -50 MPa, will
not have much effect on crack growth behaviour and can be neglected.
Thus ∆ = 
 − 
 = 300 − 100 = 200 
From table no (6.19) ,  = 6.9 × 10 ; = 3

= A (∆
) = 6.9 × 10 × 12.6 = 1.38 × 10 

MMD,Mech Dept., AIT 9964676521 27

 This initial fatigue crack growth rate falls within the limit of the Paris region of
crack growth rate the final crack length

   can be obtained from
setting  at fracture equal to 

 
2  1 1 
∴ =  − 
 − 2 ′′ ∆ S        

  
 
 


=  

య ሺయషమሻൗ − ሺయషమሻൗ
×1.27 × 10−11  × య ×  ൗమ × . య . మ .  మ

1 1
=× − 
0. 0.

= 65000 

MMD,Mech Dept., AIT 9964676521 28