IJEIT1412201301_07
-
Upload
baljinder-singh -
Category
Documents
-
view
12 -
download
1
description
Transcript of IJEIT1412201301_07
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
48
Abstract— catastrophic structural failures in many
engineering fields like aircraft, automobile and ships are
primarily due to fatigue. Where any structure experiences
fluctuating loading during service its load carrying capacity
decreases due to a process known as fatigue. Fatigue damage
accumulates during every cycle of loading the structure
experiences during its operation. When this accumulated damage
reaches a critical value, a fatigue crack appears on the structure
under service loading. A structure will have a finite fatigue life
during which fatigue cracks initiate and propagate to critical sizes
leading to catastrophic failure of the structure. Therefore fatigue
life consists of two parts: the first part is the life to the initiation of
fatigue crack and the second part is the fatigue crack propagation
to final fracture. On the other hand fatigue crack growth is the
dominant phase for more ductile structures or material. Large
structure like an aircraft has a large number of components
which are mechanically fastened together to form the total
airframe. Service experience indicates that designing airframes
against fatigue failure is better served if it is assumed that the
airframe has crack-like flaws right from day-one when it enters
the service. Airframe will experience the variable loading during
the service. If damage is present in the structure in the form of a
crack then one needs to calculate the fatigue crack growth life.
This is essential to properly schedule the inspection intervals to
ensure the safety of the structure during its service.
Index Terms— Aircraft, Pressurization, Fuselage Structure,
Fatigue, Fatigue Life, Fatigue Crack Growth, Load Spectrum,
Overload Effect, Finite Element Analysis.
I. INTRODUCTION
Aircraft are members and transverse frames to enable it to
resist bending, compressive and vehicles which are able to fly
by being supported by the air, or in general, the atmosphere of
a planet. An aircraft counters the force of gravity by using
either static lift or by using the dynamic lift of an airfoil, or in
a few cases the downward thrust from jet engines. An aircraft
is a complex structure, but a very efficient man-made flying
machine. Aircrafts are generally built-up from the basic
components of wings, fuselage, tail units and control surfaces.
Each component has one or more specific functions and must
be designed to ensure that it can carry out these functions
safely. Any small failure of any of these components may lead
to a catastrophic disaster causing huge destruction of lives and
property. When designing an aircraft, it’s all about ending the
optimal proportion of the weight of the vehicle and payload. It
needs to be strong and stiff enough to withstand the
exceptional circumstances in which it has to operate.
Durability is an important factor. Also, if a part fails, it
doesn’t necessarily result in failure of the whole aircraft. It is
still possible for the aircraft to glide over to a safe landing
place only if the aerodynamic shape is retained-structural
integrity is achieved. The basic functions of an aircraft’s
structure are to transmit and resist the applied loads; to
provide an aerodynamic shape and to protect passengers,
payload systems, etc., from the environmental conditions
encountered in flight. These requirements, in most aircraft,
result in thin shell structures where the outer surface or skin of
the shell is usually supported by longitudinal stiffening
torsion loads without buckling. Such structures are known as
semi-monocoque, while thin shells which rely entirely on
their skins for their capacity to resist loads are referred to as
monocoque. The load-bearing members of these main
sections, those subjected to major forces, are called the
airframe. The airframe is what remains if all equipment and
systems are stripped away. In most modern aircrafts, the skin
plays an important role in carrying loads. Sheet metals can
usually only support tension. But if the sheet is folded, it
suddenly does have the ability to carry compressive loads.
Stiffeners are used for that. A section of skin, combined with
stiffeners, called stringers, is termed a thin- walled structure,
the airframe of an aircraft is its mechanical structure, which is
typically considered to exclude the propulsion system.
Airframe design is a field of engineering that combines
aerodynamics, materials technology, and manufacturing
methods to achieve balances of performance, reliability and
cost. The fuselage will experience a wide range of loads from
a number of sources. The weight of the fuselage structure and
payload will cause the fuselage to bend downwards from its
support at the wing, putting the top in tension and the bottom
in compression. In maneuvering flight, the loads on the
fuselage will usually be greater than for steady flight. During
negative G-maneuvers, some of the loads are reversed. Also
landing loads may be significant. The structure must be
designed to withstand all loads cases. The bending loads are
higher when the weight is distributed towards the nose and
tail. Therefore, aircraft are loaded close to the canter of
gravity. The larger part of passenger and freighter aircraft is
usually pressurized. The cabin altitude is usually changed
quite slowly, beginning pressurizing long before 2500m,
which is the normal cabin pressure altitude during cruise, is
reached. Combat aircrafts have no need for pressurization of
large areas of the fuselage. Particular problems occur in areas
where the fuselage is required to be non-cylindrical. Internal
Effect of Overload on Fatigue Crack Growth
Behavior of Airframe Structure through FEM
Approach Vinayakumar B. Melmari, Ravindra Naik, Adarsh Adeppa
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
49
pressure will generate large bending loads in fuselage frames.
The structure in these areas must be reinforced to withstand
these loads. Because fuselages are pressurized for safety, the
designer must consider what will happen if the pressurization
is lost. The damage due to depressurization depends on the
rate of pressure loss. For very high rates, far higher loads
would occur than during normal operation. Doors and hatches
are a major challenge when designing an aircraft. Windows,
being small, do not create a severe problem. Depending on
their design, doors will or will not carry some of the load of
the fuselage structure. On the floor of the fuselage also very
high localized loads can occur, especially from small-heeled
shoes. Therefore floors need a strong upper surface to
withstand high local stresses.
II. AIRCRAFT CABIN PRESSURIZATION
Aircraft are flown at high altitudes for two reasons. First, an
aircraft flown at high altitude consumes less fuel for a given
airspeed than it does for the same speed at a lower altitude
because the aircraft is more efficient at a high altitude.
Second, bad weather and turbulence may be avoided by flying
in relatively smooth air above the storms.
Fig .1 High Performance Airplane Pressurization System
Many modern aircraft are being designed to operate at high
altitudes, taking advantage of that environment. In order to fly
at higher altitudes, the aircraft must be pressurized. It is
important for pilots who fly these aircraft to be familiar with
the basic operating principles. In a typical pressurization
system, the cabin, flight compartment, and baggage
compartments are incorporated into a sealed unit capable of
containing air under a pressure higher than outside
atmospheric pressure. On aircraft powered by turbine
engines, bleed air from the engine compressor section is used
to pressurize the cabin. Superchargers may be used on older
model turbine-powered aircraft to pump air into the sealed
fuselage. Piston-powered aircraft may use air supplied from
each engine turbocharger through a sonic venture (flow
limiter). Air is released from the fuselage by a device called
an outflow valve. By regulating the air exit, the outflow valve
allows for a constant inflow of air to the pressurized area. Fig
2.1 a cabin pressurization system typically maintains a cabin
pressure altitude of approximately 8,000 feet at the maximum
designed cruising altitude of an aircraft. This prevents rapid
changes of cabin altitude that may be uncomfortable or cause
injury to passengers and crew. In addition, the pressurization
system permits a reasonably fast exchange of air from the
inside to the outside of the cabin. This is necessary to
eliminate doors and to remove stale air. Fig 1
Fig 2 Standard Atmospheric Pressure Chart [11]
Pressurization of the aircraft cabin is an accepted method
of protecting occupants against the effects of hypoxia. Within
a pressurized cabin, occupants can be transported
comfortably and safely for long periods of time, particularly if
the cabin altitude is maintained at 8,000 feet or below, where
the use of oxygen equipment is not required. The flight crew
in this type of aircraft must be aware of the danger of
accidental loss of cabin pressure and be prepared to deal with
such an emergency whenever it occurs.
III. GEOMETRIC CONFIGURATION OF THE
FUSELAGE
A segment of the fuselage is considered in the current
study. The structural components of the fuselage are skin,
bulkhead and Longerons. Geometric modeling is carried out
by using SOLIDWORKS 2012 software. The total length of
the structure is 1500mm and diameter is 2200mm. It contains
4nos Z section (Bulkhead) and 40nos L section (Longerons).
Fig 3 CAD Model of the Fuselage Part
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
50
IV. FINITE ELEMENT MODEL OF THE FUSELAGE
Fig 4 Meshing of the fuselage model
Elements used in the Fuselage structure:
Number of grid points =148320
Number of CBEAM elements =3200
Number of CQUAD4 elements =136640
Number of CTRAI3 elements =1280
Loads and boundary conditions:
Fig 5 Loads and Boundary Conditions
Edge nodes of fuselage are constrained and applying the
inside pressure.
Stress analysis of the fuselage structure:
Fig 6 Stress Analysis of the Fuselage Structure
Displacement of the fuselage structure:
Fig 7 Displacement of the Fuselage Structure
Table 1 Maximum Displacement, Maximum Stress and
Hoop Stress for Different Pressure in the Fuselage
Structure
Sl
no
Applied
Pressure
Max
stress
in
Mpa
Ma
displ
in mm
Hoop
Stress
Theo
Hoop
FEM
%
error
1 6psi 38.45 0.41 22.66 25.51 11.2%
2 7psi 44.83 0.474 25.49 29.14 9.1%
3 8psi 51.2 0.541 30.21 33.45 9.7%
4 9psi 57.68 0.609 33.99 35.62 4.6%
5 10psi 90.93 0.659 37.77 39.25 3.8%
6 11psi 100.1 0.725 41.55 44.24 6.08%
7 12psi 108.8 0.791 45.32 48.16 5.89%
8 13psi 118.7 0.857 49.63 51.99 4.53%
9 14psi 127.5 0.923 52.87 54.41 2.83%
10 15psi 136.3 0.989 56.65 58.62 3.36%
12 16psi 145.2 1.01 60.42 62.10 2.70%
13 17psi 153.1 1.11 64.21 66.71 3.75%
14 18psi 161.9 1.19 67.98 69.92 2.77%
15 19psi 170.6 1.25 71.76 73.97 2.98%
16 20psi 181.5 1.32 75.54 76.12 0.76%
Theoretical calculation for Hoop stress
σhoop th= p*r/t (eq 1)
Where
Cabin pressure (p) = 6psi=0.0042kg/mm 2
=0.041202Mpa
Radius of curvature of the fuselage (r) =1100mm
Thickness of the skin (t) = 2mm
Substitution of the values in eqn 1 we get
σhoop th=2.31kg/mm 2=
22.66Mpa
V. LOCAL ANALYSIS OF THE STIFFENED PANEL
From the Global finite element analysis carried out on the
fuselage segment at the maximum stress location, Based on
the maximum stress location a local analysis is carried out by
considering a stiffened panel.
Geometric configuration for local analysis
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
51
Fig 8 Geometric Configuration for Local Analysis
Finite element mesh of the stiffened panel
Fig 9 Finite Element Mesh Of the Stiffened Panel
Elements used in the stiffened panel
Number of grid points =22091
Number of CBAR elements =84
Number of CQUAD4 elements =21174
Number of CTRIA3 elements = 52
Loads on the stiffened panel
For skin:
For 6psi pressure the hoop stress is,
σhoop th=2.31kg/mm 2=
22.66Mpa
UDL is applied on the axial x direction,
Ps =Load on the skin
σhoop th=2.31kg/mm 2
A= Cross sectional area of the skin in mm 2
A= w*t= 375*2= 750 mm 2
Ps = σhoop th *A= 1732.5kg/mm 2 UDL=1732.5/375
= 4.62kg/mm
For Bulkhead:
Pbh =Load on the bulkhead
σhoop th=2.31kg/mm 2
A= Cross sectional area of the skin in mm 2
A= (L1+L2+L3)*tbh= (17+94+27)*3= 414 mm 2
Ps = σhoop th *A= 956.34kg/mm 2 Ps =956.34/138
Ps =6.93kg/mm
Load applied on the stiffened panel
Fig 10 Loads Applied On the Stiffened Panel
Applying the UDL on skin is 4.62kg/mm and on bulkhead is
6.93kg/mm
stress in the stiffened panel:
Fig 11 Stresses in the Stiffened Panel
Displacement in the stiffened panel:
Fig 12 Displacement in the Stiffened Panel
VI. INTRODUCTION TO FATIGUE CRACK
GROWTH
Fatigue may be defined as a mechanism of failure based on
the formation and growth of cracks under the action of
repeated stresses. Normally, small cracks will not cause
failure, but if the design is insufficient in relation to fatigue,
the cracks may propagate to such an extent that failure of the
considered detail occurs. Fatigue is associated with stresses
that vary with time often in a repeated manner. There many
possible sources of time varying stress e.g.
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
52
Fluctuating live loads
Acceleration forces in moving structures
Pressure changes
Temperature fluctuations
Mechanical vibrations
Environmental loading (wind, waves, current)
VII. STUDY OF OVERLOAD EFFECT
When one single high stress is interspersed in a constant
amplitude history, the crack growth immediately after the
“overload” is much slower than before the overload. After a
period of very slow growth immediately following the
overload, gradually the original growth rates are resumed.
This phenomenon is known as “retardation”. A negative load
following the overload reduces retardation but does not
eliminate it. Crack growth analysis for variable amplitude
loading is not very well possible without an account of
retardation effects. Before such an account can be made,
retardation must be explained .Consider a crack subjected to
constant amplitude loading at R=0; during the first cycle the
load varies from A to C through B. Before the loading starts,
imagine a little dashed circle (figure 6.1) at the crack tip.
Indicating the material that will undergo plastic deformation
in the future plastic zone The plate is then loaded to B.
Imagine that one could now remove the plastic zone and put it
aside (figure 6.1). After unloading to C the situation of figure
8.1.d is reached: all material is elastic the plastic material
having been cut out so that all strains and displacements are
zero after unloading. Hence, after unloading the hole at the
crack tip in figure 8.1d is equal in size to the dashed circle in
figure 8.1a the plastic zone has become permanently
deformed: it is larger than before the loading started. Thus it
will not fit in the hole of figure 6.1 in order to make it fit it
must be squeezed back to its original size for which a stress at
least equal to the yield strength is needed. The plastic zone in
fatigue loading is very small. Most of the fatigue crack growth
takes place at low values of k the rates are so high that little
crack growth life is left; most of the life is covered at low k.
In the following crack growth analysis for variable amplitude
loading will be illustrated on the basis of the wheeler model.
This model is used here not because it is believed to be better
than any other,but because it is very simple, so that is can be
used easily for illustrations. In this chapter first we used to
calculate the SIF for stiffened panel crack growth is
calculated by the use of vsKeffective from the paper
NASA/TM-2005-213907 for Al 2024-T351 Crack growth
rate curve we get the crack growth rate per cycle. Then Over
load plastic zone size and Retardation factor are calculated,
finally Calculating the crack growth rate within the overload
region is carried out.
Fig. 13. Residual stress at crack tip (a)Load at A; (b) Load at B;
(c) Load at B: plastic zone removed;(d) Load at C: plastic zone
still out; (e) Load at C: plastic zone back in.
VIII. STRESS INTENSITY FACTOR CALCULATIONS
THROUGH STRESS ANALYSIS OF THE UN
STIFFENED PANEL
Virtual crack closure technique for determining stress
intensity factor
There are different methods used in the numerical fracture
mechanics to calculate stress intensity factors (SIF). The
crack opening displacement (COD) method and the force
method were popular in early applications of FE to fracture
analysis. The virtual crack extension (VCE) lead to increased
accuracy of stress intensity factor results The virtual crack
extension method requires only one complete analysis of a
given structure to calculate SIF. The total energy release rate
or J-integral is computed locally, based on a calculation that
involves only elements affected by the virtual crack
extension. Both the COD and VCE methods can be used to
calculate SIF for all three fracture modes. However,
additional complex numerical procedures have to be applied
to get results. The equivalent domain integral method which
can be applied to both linear and nonlinear problems renders
mode separation possible. The VCCT, originally proposed in
1977 by Rybicki and Kanninen, is a very attractive SIF
extraction technique because of its good accuracy, a relatively
easy algorithm of application capability to calculate SIF for
all three fracture modes. The VCCT has a significant
advantage over other methods; it has not yet been
implemented into most of the large commercial
general-purpose finite element codes. This technique can be
applied as a post processing routine in conjunction with
general-purpose finite element codes. The VCCT is based on
the energy balance. In this technique, SIF are obtained for
three fracture modes from the equation.
GI= β --- (Eq7.1)
Where
GI= the energy release rate for mode I,
KI = stress intensity factor for mode I in MPa ,
E =elastic modulus in MPa,
γ = Poisson ratio,
β = 1 for plane stress,
β = 1 – γ2 for plane strain.
The energy released in the process of crack
expansion is equal to work required to close the crack to its
original state as the crack extends by a small amount c.
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
53
W = (Eq7.2)
Where
u = relative displacement,
σ =stress,
r = distance from the crack tip,
Δc = change in virtual crack length.
Therefore, the energy release rate is,
G = (Eq7.3)
Figure 14 shows a FE model in the vicinity of a crack tip
before the virtual crack closure, while Figure 7.2 shows the
same finite elements after the closure. In the finite element
(FE) calculations, changes in the crack length as it progresses
are governed by size of finite element. Moreover, the
interaction between the neighboring FE exists only in shared
nodes. Therefore, the element edges do not transfer forces.
Generally, work needed to close the crack stating from the
state shown in fig.14 to reach the state shown in fig 15 is
equal to
W= (Eq7.4)
Fig 14 2D Finite Element Models in The Vicinity Of a Crack Tip
Before the Virtual Closure
Table 2 Displacement vector for UN stiffened Panel
G= (Eq7.2)
Where
G =Strain energy release rate
F =Forces at the crack tip in kg or N
𝜟c=change in virtual crack length in mm
T = thickness of skin in mm
U = relative displacement,
Then the SIF is calculated by FEM method by substituting
Eq7.2 in below Eq7.3
KI= MPa (Eq7.3)
Where
KI= stress intensity factor (SIF)
E =young’s modulus
=7000kg/mm2=68670 MPa
G=Strain energy release rate
Theoretically SIF value is calculated by
KI = * f ( MPa (Eq7.4)
And
f ( = (Eq7.5)
Where
= Crack length in mm
f ( =Correction factor
b=Width of the plate (200 mm)
From the stress analysis of the stiffened panel it can be
observed that a crack will get initiated from the maximum
stress location but there are other possibilities of crack
initiation at different locations in the stiffened panel due to
discrete source of damage. It may be due to bird hit, foreign
object hit etc. Fine meshing is carried out near the crack to get
accurate results which is shown in Fig 16 and Fig 17 .Other
than the crack region coarse meshing is carried out. To get the
mesh continuity from fine-mesh to coarse-mesh different
quad and tria elements are used.
Fig. 15 Fine Element Mesh for Un Stiffened Panel at The Center
Of Skin Near The Crack
Fig.16 Close up View of Fine Mesh for UN Stiffened Panel at the
Center of Skin near the Crack
Table 3 Comparison of Analytical (FEM) SIF Values with
Theoretical SIF Values for Un-Stiffened Panel for 1mm Element
Size
Width of the plate: 500mm
Length of the plate: 1000mm
Thickness of the plate: 2mm
POI
NT
ID.
TY
PE
T1 T2 T3 R1 R2 R3
5482
0
G -1.941
106E-
03
1.040
564E-
01
0.0 0.0 0.
0
1.0100
68E-03
5486
3
G -1.940
486E-
03
1.075
943E-
01
0.0 0.0 0.
0
-1.0098
16E-03
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
54
Applied load: 1500Kg
Validation of FEM approach for stress intensity
factor (SIF) calculation
Fig 17 Stress in the UN stiffened panel crack tip
MVCCI METHOD
Fig.18.Nodes and Elements near the Crack Tip
Consider crack length, 2a=20 mm
SIF calculation by Theoretical method
KI = * f ( (From Eq7.4)
Where
=P/A=1.5Kg/mm2
=10 mm
f ( =1.000740667which is calculated by using Eq7.8
Substituting above values in Eq7.7 .SIF value will be
KItheoretical = 2.6027MPa
Strain energy relies rate is calculated by Eq7.2 which is
G=
For relative displacement adding the displacement of nodes
54820 and 54863 in T2 direction the displacement is obtained
in the f06 file created by the MSC Nastran (solver) software
shown in Table 3
Table 4 Grid point force balance for un stiffened Panel
For the relative displacement
u= 1.075943E-01 - 1.040564E-01
u= 0.003538
For Forces at the crack tip in kg or N, adding any one side of
elements (Elm 35312, 54120 or Elm306, 54100) forces acting
on the crack tip in T2 direction. The Forces at the crack tip is
obtained in the f06 file created by the MSC Nastran (solver)
software shown in table 7.2
For Forces at the crack tip F= 5.739905+5.562092
F= 11.3020
SR
.N
o
Crac
k
lengt
h
”2a”
in
mm
Kfemin
MPa√
m
Kthwitho
ut
consider
ing the
C.F in
MPa√m
Correcti
on factor
C.F
Kthwith
consideri
ng the
C.F in
MPa√m
%erro
r
1 10 1.828
3
1.843 1.00018 1.8433 0.81%
2 20 2.595
0
2.6008 1.0007 2.6027 0.295
%
3 30 3.192
1
3.1943 1.00168 3.1996 0.234
%
4 40 3.699
2
3.6884 1.0030 3.6996 0.01%
5 50 4.147
4
4.1238 1.00482 4.1436 0.09%
6 60 4.558
7
4.5174 1.00704 4.5492 0.208
%
7 70 4.942
1
4.8794 1.00974 4.92690 0.307
%
8 80 5.304
7
5.2163 1.01290 5.2836 0.397
%
9 90 5.651
6
5.5327 1.01659 5.6244 0.481
%
10 100 5.986
3
5.8320 1.02080 5.9533 0.551
%
POIN
T-ID
ELE-
ID
SOUR
CE
T1 T2 T3
378 306 QUAD
4
2.533442E+0
0
-5.738164E+
00
0.0
378 3531
2
QUAD
4
2.534859E+0
0
5.739905E+0
0
0.0
378 5410
0
QUAD
4
-2.536320E+
00
-5.563833E+
00
0.0
378 5412
0
QUAD
4
-2.531982E+
00
5.562092E+0
0
0.0
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
55
Where
F = 11.3020Kg =110.87N
U = 0.003538 mm
𝜟c= 1 mm
T = 2 mm
Substitute all values in Eq7.5 then
G= 0.09806Mpa
Now Analytical SIF is calculated by Eq7.3 which is
KI fem=
Where
E=7000kg/mm2=68670 MPa
Substituting G and E values in Eq7.6
KI fem=2.5950MPa
The above calculation is carried for different crack length
considering a known load. A stress intensity factor value
calculated by FEM (using VCCT technique) and stress
intensity values calculated by theoretical method for
un-stiffened panel is tabulated. From the table 4 and the fig
19 it is clear that SIF values obtained by using FEM (by
using MVCCI technique) for un-stiffened panel agrees with
the SIF values calculated theoretically by Eq7.7.Therefore
FEM (by using MVCCI method) for finding SIF value is
valid.
Fig.19 Comparison of Theoretical SIF Value With Analytical
SIF Value
Methodology of finding SIF values for un-stiffened panel
using FEM was extended to get SIF values for stiffened panel.
IX. STRESS INTENSITY FACTOR CALCULATIONS
THROUGH STRESS ANALYSIS OF THE STIFFENED
PANEL
Fig 20 Stress in the Stiffened Panel Crack Tip
Table 5 Calculation of crack growth rate for 250 mm crack
in the stiffened panel.
The crack growth rate is calculated or obtained through
vs𝜟k curve from the respective material. Therefore to
obtain the (crack growth rate) one should first calculate
the 𝜟keffective.
𝜟keffective is calculated by using the Eq 8.1 and Eq 8.2 (Ref:
“The practical use of fracture mechanics” by David Broek)
[17].
𝜟Keffective= 𝜟Kmax-𝜟Kopening MPa (Eq8.1)
𝜟Kopening=𝜟Kmax(0.5+0.4R) MPa (Eq8.2)
R→0
SR.No Crack length
”2a” in mm
Kfemin MPa√m
1 10 2.7073
2 20 3.8507
3 30 5.3582
4 40 5.6338
5 50 6.0779
6 60 6.6572
7 70 6.8367
8 80 7.6994
9 90 8.1174
10 100 8.5105
11 110 9.1091
12 120 9.5588
12 130 9.8287
13 150 10.421
14 180 11.773
15 200 13.430
16 220 14.216
17 230 15.120
18 240 15.882
19 250 16.666
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
56
Where
𝜟Kmax= maximum stress intensity factor
𝜟Kman= minimum stress intensity factor
R = = 0 because 𝝈minimum→0
We know that
Kmax =16.666MPa .
Kmin=0.0
Substituting the values in Eq 8.2 we get
Kopening=8.33 MPa
Substituting the values in Eq 8.1 we get
Keffective= 8.33MPa
From the paper Ref [12] NASA/TM-2005-213907 for Al
2024-T351 Crack growth rate curve we get the crack growth
rate per cycle
For Keffective=8.33MPa ,
We get 7.39×10-5
mm/cycle
To growth a crack of 1mm it requires 13532 cycles.
After 13532 cycles the crack size will be 251mm which is
considered for the next analysis and overload cycle is applied.
B. Calculation of stress intensity factor (SIF) for 101 mm
crack in the stiffened panel with overload is applied.
Considering a crack length 0f 251 mm on the skin and the
overload corresponding to 9 PSI which is L=2598.75kg ≈
6.93kg/mm uniformly distributed load is applied at the remote
edge of the panel. All the loads and boundary conditions are
same.
SIF calculation by Analytical method (FEM) for
251mm crack length Table 6 Displacement vector
POINT ID. TYPE T1
39396 G 2.583312*E-01
40911 G 2.341051*E-01
For the relative displacement
(2.583312*E-01-2.341051*E-01) = 0.0242261 mm
For Forces at the crack tip in kg or N, adding any one side of
elements (Elm 35037, 35039 or Elm36045, 36047) forces
acting on the crack tip in T1 direction. The Forces at the crack
tip are obtained in the f06 file created by the MSC Nastran
(solver) software. These values are tabulated and shown in
table 7.
Table 7 Grid point Force balance
POINT-ID ELEMENT-ID SOURCE T1
39399 35037 QUAD4 37.4300
39399 35039 QUAD4 37.27135
39399 36045 QUAD4 -36.4621
39399 36047 QUAD4 -38.1642
For Forces at the crack tip
F=(37.4300+37.27135)=74.5866Kg.
Where
F = 74.5866 Kg
U = 0.0242261 mm
𝜟c= 1 mm T = 2 mm
Substitute all values in Eq7.2 then
G= 0.904861 Kg/mm
Now Analytical SIF is calculated by Eq7.3 which is
KI fem=
Where
E=7000kg/mm2
Substituting G and E values in Eq7.6
KI fem=24.689318 MPa
C. Calculation of crack growth rate for 251 mm crack
in the stiffened panel for 9PSI load
We know that
Kmax =24.689318 MPa .
Kmin=0.0
Substituting the values in Eq 8.2 we get
Kopening= 12.344659 MPa
Substituting the values in Eq 8.1 we get
Keffective= 12.344659 MPa
From the paper NASA/TM-2005-213907 for Al 2024-T351
Crack growth rate curve we get the crack growth rate per
cycle
For Keffective=12.344659 MPa ,
We get 2.85×10-4
mm/cycle
With 1 cycle of 9PSI load the crack growth increment is
0.000285mm.
After the application of one overload cycle with crack
increment of 0.000285 mm the total crack size will be
251.000285mm.
With the crack length of 251mm other iteration is carried out
with the load of 6psi.
Kmax =16.45953MPa .
Kmin=0.0
Substituting the values in Eq 8.2 we get
Kopening=8.22976 MPa
Substituting the values in Eq 8.1 we get
Keffective= 8.23MPa
From the paper Ref [12] NASA/TM-2005-213907 for Al
2024-T351 Crack growth rate curve we get the crack growth
rate per cycle
For Keffective=8.23MPa ,
We get 7.39×10-5
mm/cycle
Fig. 21 A Typical Load Spectrum With An Over Load
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
57
The overload plastic zone size is given by the following
Equation
Rpc =Rpo=
(Eq8.3)
Where
Rpo= over load plastic zone size
Kmax= maximum SIF value
= yield strength
Considering Fty=345 Mpa
And Kmax=16.66585154Mpa
Substituting the values in Eq8.3
Rpo =8.150×10-4
mm.
Followed by the 251.000285mm crack and 6psi load
Kmax=16.45953Mpa
Substituting in Eq8.3
Rpc==
Rpc= 3.622570124×10-4
mm.
Where
Rpc= current plastic zone size.
D. Calculations for the Retardation factor
As explained in the section 8.2 about the crack growth
retardation because of the overload plastic zone size the
retardation factor is calculated using the following equation
ØR= here Eq(8.4)
Where
ØR= Retardation factor
ai=current crack length
ao= after the overload crack length
Wheeler parametric value =1.4
ØR= 0.2111902.
E. Calculating the crack growth rate within the overload
region
As the crack grows within the overload plastic size the effect
of over load on the crack growth rate also varies. Therefore
the crack growth rate is calculated by dividing the overload
region by four parts each one having region as
2.0375×10-4
mm
Fig. 22 Four Divisions Considered Within the Overload Plastic
Zone Size
A) First Region.
To calculate crack growth rate in the First region following
data are considered
Load of 6psi with crack size now become
ai= 251+(2.0375×10-4
×1)=251.0002038mm
And ao, Rpc ,Rpo will be as follows
ao= 251.000285+(2.0375×10-4
×1)=251.0004888mm
Rpc=3.622570124×10-4
mm
Rpo=(2.0375×10-4
×3)=6.1125×10-4
mm
Retardation factor for first region from Eq 8.4
ØR1= here
ØR1 =0.404191924
Therefore total overload plastic zone size is 7.39×10-5
by
multiplying the retardation factor we get the da/dN crack
growth size for the First region is calculated.
da/dN= 7.39×10-5
× ØR1
da/dN=8.150×10-4
×0.404191924
da/dN=3.294164181×10-4
mm/cycle
To grow the first region crack it required 3036 cycles.
B) Second Region.
To calculate crack growth rate in the Second region
considering the following data
Load of 6psi with crack size now become
ai= 251+(2.0375×10-4
×2)=251.0004075mm
And ao, Rpc ,Rpo will be
ao =251.000285+(2.0375×10-4
×2)=251.0006925mm
Rpc=3.622570124×10-4
mm,
Rpo=(2.0375×10-4
×2)=4.075×10-4
mm,
Retardation factor for second region from Eq 8.4
ØR2= here
ØR2==0.40367898
Therefore total overload plastic zone size is 5×10-5
by
multiplying the retardation factor we get the da/dN crack
growth size for Second region is calculated.
da/dN = 7.39×10-5
×ØR2
da/dN = 8.150×10-4
×0.403678698
da/dN =3.289981×10-4
mm/cycle
To grow the 1mm crack it required 3039 cycles.
C) Third Region.
To calculate crack growth rate in the Third region we taken
fallowing data
Load 6psi with crack size now become
ai= 251+(2.0375×10-4
×3)=251.0006113mm,
And ao, Rpc ,Rpo become
ao =251.000285+(2.0375×10-4
×3)=251.0008963
Rpc=3.622570124×10-4
mm,
Rpo=(2.0375×10-4
×1)= 2.0375×10-4
,
The overload plastic zone size after the second region is
2.0375×10-4
is less than the current plastic zone size.
Therefore the assumed Third regions for the calculation of
retardation factor need not to be considered. Shown in table 8.
Crack Growth Calculations
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
58
SIF
Number of
cycles
required
for a crack
growth
increment
of 1mm.
Constant amplitude
loading (6PSI) for 250
mm crack
SIF= 16.6658Mpa
7.39𝖷10-5cycles 13532
Overload (without
considering the
overload plastic zone
size applied)(9PSI)
SIF=24.3446 Mpa
at 251mm crack length
2.85𝖷10-4cycles 3509
After the overload plastic zone size effect
With considering the
retardation factor
0.2111902
0.2111902𝖷2.85𝖷10-5 =
6.0189207𝖷10-5
cycles. 16615
Table 8 Crack growth calculations
IX. CONCLUSION
Damage tolerance design philosophy is generally used in
the aircraft structural design to reduce the weight of the
structure. Stiffened panel is a generic structural element of the
fuselage structure. Therefore it is considered for the current
study. A FEM approach is followed for the stress analysis of
the stiffened panel. The internal pressure is one of the main
loads that the fuselage needs to hold. Stress analysis is carried
out to identify the maximum tensile stress location in the
stiffened panel. A local analysis is carried out at the maximum
stress location with the rivet hole representation. The crack is
initiated from the location of maximum tensile stress. MVCCI
method is used for calculation of stress intensity factor. A
crack in the skin is initiated with the local model to capture the
stress intensity factor. Stress intensity factor calculations are
carried out for various incremental cracks. When the SIF at
the crack tip reaches a value equivalent to the fracture
toughness of the material, then the crack will propagate
rapidly leading to catastrophic failure of the structure. A load
spectrum consisting of constant load cycles with a over load
in-between is considered to study the over load effect on the
crack growth rate. The crack growth rate for constant
amplitude load cycles is carried out by considering the crack
growth rate data curve (da/dN vs.𝜟K). Plastic zone size due
to constant amplitude load cycles and over load cycle is
calculated. The calculations have indicated that the over load
will reduce the rate of crack growth due to large plastic zone
size near the crack tip. The effect of large plastic zone due to
over load is estimated by calculating the crack growth
retardation factor.
REFERENCES
[1] Full-scale Testing of the Fuselage Panels By Dr. John G.
Bakuckas, Dr. Catherine A. Bigelow,Dr. Paul W. Tan, Prof.
Jonathan Awerbuch, Prof. Alan C. Law, Prof. T. M. Tan.
[2] Experimental/numerical techniques for aircraft fuselage
structures containing damage By Padraic E. O’Donoghue,
Jinsan Ju .
[3] New techniques for detecting early fatigue damage
accumulation in aircraft structure by Curtis A. Ride out and
Scott J. Ritchie IEEEAC paper #1244, Version 2, November
29, 2006.
[4] Elangovan.R“A analytical determination of residual strength
and linkup strength for curved panels with multiple site
damage”.
[5] Garcia a N, “Simplifying MSD modeling by using continuing
damage assumption and parametric study: The role of rivet
squeeze force.
[6] A. Murphya, M. Pricea, C. Lynchb, A. Gibsona “The
computational post-buckling analysis of fuselage stiffened
panels loaded in shear”.
[7] A. Cornec , W. Schönfeld, K.-H. Schwalbe, I. Scheider
“Application of the cohesive model for predicting the residual
strength of a large scale fuselage structure with a two-bay
crack”.
[8] Andrzej Leski, 2006, “Implementation of the virtual crack
closure technique in engineering FE calculations”. Finite
element analysis and design 43, 2003, Pages 261-268.
[9] An accurate and fast analysis for strongly interacting multiple
crack configuration including kinked (v) branched (y) cracks
by A.K. Yavuz, S.L. Phoenix, S.C. TerMaath
[10] Investigation of mixed mode stress intensity factors non
homogeneous materials using an interaction integral method by
Hongjun Yu, Linzhi Wu, Licheng Guo, Shanyi Du, Qilin.
[11] http://www.flightlearnings.com/2010/04/14/pressurized-aircra
ft-part-one/.
[12] NASA/TM-2005-213907 for 7075-T6 and 2024-T351
Aluminum Alloy Fatigue Crack Growth Rate Data by Scott C.
Forth and Christopher W. Wright Langley Research Center,
Hampton, Virginia William M. Johnston, Jr. Lockheed Martin
Corporation, Hampton, Virgini.
AUTHOR BIOGRAPHY
Mr. Vinayakumar B. Melmari is currently
working as assistant professor in Dept. of
Mechanical Engineering, Shaikh college of
Engineering and technology, Belgaum, Karnataka,
India. Two Year industrial experience and one
international journal published and two national
conferences given.
Mr. Ravindra Naik is currently working as
assistant professor in Dept. of Mechanical
Engineering, AGMR Engineering and technology,
Varur, Hubli, and Karnataka, India. Ten Year
industrial experience and one international journal
published and one national conference given.
Mr. Adarsh Adeppa is currently working as
assistant professor in Dept. of Mechanical
ISSN: 2277-3754 ISO 9001:2008 Certified
International Journal of Engineering and Innovative Technology (IJEIT)
Volume 2, Issue 7, January 2013
59
Engineering, Bheemanna Khandre institute of technology, Belgaum,
Karnataka, India. One international journal published and one national
conference given.