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




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

http://www.flightlearnings.com/pressurized-aircraft-part-one/798/














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




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.




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.




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




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




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




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


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


Kopening= 12.344659 MPa


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


Kopening=8.22976 MPa


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




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




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




59

Engineering, Bheemanna Khandre institute of technology, Belgaum,

Karnataka, India. One international journal published and one national

conference given.

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IJEIT1412201301_07