Aircraft Structure - d24cdstip7q8pz.cloudfront.net · Load at Mid-spain Simply Supported ... While...

A

I

R

C

R

A

F

T

S

T

R

U

C

T

U

R

E

A

E

2 Career Avenues GATE Coaching by IITians

Aircraft Structure

Table of Contents

1 Aircraft

Materials &

Elasticity

Basic Aircraft Structural Elements

Axial Members

Shear Members

Bending Members

Torsion Members

Aircraft Materials and its Properties

Metals Alloys

Composites

Stresses and associated strains

Rigid Body Motion

Stress

Equilibrium Equation for non uniform Stress Variation

For 2D- Plain stress equilibrium equation

Determination of Stress in Inclined Plane under equilibrium

Tractions or total forces on any surfaces

Stress-Strain characteristics curve

Stress-Strain Relations

Strains induced by Tangential Stress

Stress Strain Relation in three dimension

Orthotropic Materials

Isotropic Materials

Poisson’s Ratio

Elastic Strain Energy

Plane Stress-strain problems

Plane Strain Problem

Plane Stress Problem

Relation between Bulk Modulus (K) and Young’s Modulus (E)

Governing Equations for elasticity problem

Equilibrium Equations

Boundary Conditions


Compatibility Equations for Plane Strain

Compatibility Equations for Plane Stress

Airy Stress Function

Mohr Circle

Equation of Mohr Circle for Stain

2 Torsion

Saint-Venant’s Principle

Torsion of arbitrary cross-section

Rods having circular cross-section

Torsion in thin sheets

Displacement along z – direction in this sheets

Torsional Constant for arbitrary thin-sections

T-section

Open curved thin sections

Torsion in thin closed arbitrary shells

Shear flow

Breth Bratho theorem

Shear Centre

Shear Centre in Open Symmetrical Section

Circular shape thin wall open section

Shear Centre of Channel-section

Multicell thin sheet closed section

3 Shear Forces and Bending

Moments

Simply supported Beam carrying a Concentrated

Load at Mid-spain

Simply Supported Beam with a Concentrated Load at arbitrary point

Simply Supported Beam carrying various Concentrated Load

Simply Supported Beam Carrying a Uniformly Distributed Load

Simply Supported Beam Subjected to a moment

Cantilever Beam having concentrated Load at its

Free End

Cantilever Beam Carrying Uniformly Distributed Load Over the Span

4 Boundary Conditions

Euler’s Theory of Buckling of Columns

Column Hinged at both Ends

Column Fixed at One End and Hinged from Other


Buckling of Vertical

Members

End

Column Fixed at Both Ends

Column Fixed at One End and Free at the Other

Limitations of Euler’s Formula

Rankin Formula

5 Beam Theory

Centroid

Moment of Inertia

Parallel Axis Theorem

Neutral Axis

Simple Bending

Pure Bending

Theory of Simple Bending

Bending Stresses

Deflection of Beam under bending

Section Modulus

Shear Forces and Shear Stress distribution

Rectangular cross-section

Rectangular section titled at 45

Circular Cross-section

I-section

For H-section

For Hollow Circular Section

For Plus Section

For Inverted Channel Section

For Triangle Section

6 Failure Theory

Maximum Principal Stress (Rankine Theory)

The Maximum Principal Strain Theory (St. Venant’s

theory)

Maximum Shear Stress theory or Guest – Columb’s Theory or Tresca’s Theory

Maximum Strain Energy Theory or Beltrami – Haigh’s Theory

Distortion Energy Theory or Von Mises Theory

7

Degree of Freedom

Natural Frequency

Classification of Vibration

Free Vibration

Forced Vibration


Theory of

Vibrations Undamped and Damped Oscillations

Linear and Non-Linear Vibrations

Deterministic and Non-Deterministic

Vibration

Spring Element

Combination of Spring

Spring in Parallel

Springs in Series

Harmonic Motion

Spring-mass system in single degree of freedom

Vibration of Un-damped System

Vibration with Viscous Damping

Critical Damping Constant and Damping Ratio

Logarithmic Decrement

Energy Stored in damper

Two degree of Freedom System

Equations of motion for Forced Vibration

Free Vibration analysis of an Undamped System

Continuous Vibration

Free vibration of Uniform string

String whose both ends are fixed

For contilever

Longitudinal Vibrations in a Bar

Common Boundary Conditions for rod in Longitudinal motion

Torsional Vibration in Beam

Response of a Damped System under Harmonic Force


CHAPTER 1

AIRCRAFT MATERIAL AND ELASTICITY

Like every industry across the world, aircraft industry also have some objective

and goals: such as to provide most economic air transport without

compromising with safety. For that, aircraft structural members needs to meet

objectives.

It should be light weighted so that payload can be high.

The material should possess high stiffness and strength so that it can

withstand high pressure loads and bending loads.

The material should be corrosion free and durable so that the aircraft

can be in service for longer time.

Some of the alloys which are being used in aircraft manufacturing are:-

Steel alloys For its ability to bear stress and it is durable.

Aluminum alloys For its light weight and high strength to weight

ratio.

Titanium alloys For its ability to withstand high stresses.

Composites For its property to withstand high temperature and

applied stresses but the drawback comes in with difficult to machine.

Composites which we normally use in Aircraft industry are carbon fiber,

glasses etc.

Nickel alloys for its ability to withstand high thermal stresses and

therefore, used for turbine blades.

While using structure aircraft, the designer know how to use material properly

for the optimization of strength to weight ratio.

Some of the requirements leads designers to use monocoque structure for

fuselage, in earlier era of aircraft development due to its promising ability of

withstanding high stresses. But time advanced, monocoque structure only are

not able to tolerate the huge fuselage design and we need the help of advanced

design and material to overcome this problem.


Basic Aircraft Structural Elements

In Aircraft Structures, major components are the assemblages of a number of

basic structural elements, which is designed to take a specific type of load such

as, axial, bending or torsional load.

Axial Members

Axial members predominantly experience tensile or compressive loads which

always acts in the axial direction. The stress so produced is uniaxial i.e,

E ... (1.1)

Where E and are young’s modulus and normal strain respectively The total

axial force:

F A EA ... (1.2)

Where A is the cross – sectional area of the member.

And ‘ EA ’ is the axial stiffness of the member which depends upon modules of

the material and cross-sectional area of the number.

Axial stiffness of axial members cannot be varied by changing the shape of

cross-section i.e. a circular rod and a channel having same cross-section which

can carry the same axial load have same stiffness.

When subjected to compressive stresses, due to high length to width ratio,

axial members are more prone to bucking failure. However it can overcome by:-

Increasing bending stiffness

By shortening the length of the buckle mode.

The higher bending stiffness is much better as it has higher bending stiffness

then the circular section. The bending stiffness of these materials is very less

due to slenderness of most axial members. The buckling strength is enhanced

by providing the lateral support along the length with more rigid ribs (in wing)

and frames (in fuselage).


Shear Member

It is thin sheet of material used to carry in–plane shear load. The total shear

force in x direction in fig (a) as shown:-

xV ta G ta

... (1.3)

Where G is the Shear modulus, and is the shear strain.

For a curved panel, the resulting shear force is sub divided into horizontal and

vertical component under the constant shear stress condition which is as

depicted in figure (b):-

xV ta ... (1.4)

yV tb ... (1.5)

Thus the component of the resultant force is:-

x

y

V a

V b .... (1.6)

Bending Member

One of the simple bending member is beam which is used for carrying

longitudinal tensile and compressive stresses. According to simple beam

theory, bending moment M is related to beam deflection is:-

2

2

wM EI

x

... (1.7)

Where EI is the bending stiffness of the beam. The area moment of inertia I

depends on the geometry of the cross – section.

A beam is designed to experience a variety of loads including bending moments

and transverse shear force. Bending stresses are more severe them the

transverse shear force for beams.


In the given figure shown, the fixed end of a beam supported at one end or

cantilever beam is always experience high bending stress and bending moment.

The maximum bending stress is given as:-

3 2

62 2max

12

max

h hM VLVL

bhI bh

… (1.8)

Along the depth the, transverse shear stress varies parabolically and its

maximum value at the symmetry line is:-

3max

2

V

bh … (1.9)

By dividing equation (1.8) by (1.9)

max 4

max

L

h

…(1.10)

For larger span–to–depth ratio, bending stress plays a more dominant role than

the transverse shear stress. Bending stress distribution varies linearly over the

depth with maximum value at the further position from the neutral axis. A

rectangular section bean does not have effective bending member. For a wide

flange beam, the bending force distribution is concentrated for wide flange

because wb t . Also the Transverse shear stress distribution for wide flange

beam and the vertical web is carrying essentially all the constant as shown is

figure below.


Torsion member

Application of shear Stress in the plane of cross – section in the production of

torque force in that plane. The torque so produced varies linearly along the

radial direction. It is related to twist angle per unit length as given by:

T Gj ... (1.11)

Where j is the torsional constant.

Aircraft Structure

Torsional Constant, J is equal to polar moment of inertia of the circular and

hollow circular cross-section of the member, therefore,

4 4 2 2 21 1

2 2pJ I b a b a b a b a ... (1.12)

where ‘GJ’ is torsional stiffness.

If the wall thickness is ‘t=b-a’ is very small as compared with the inner radius,

then j is:-

32J tr ... (1.13)

where 3 / 2r a b .

For thin walled cylinder, the torsional stiffness is proportional to the 3/2 power

of the area 2r .

The material near the inner – cavity in a thick – walled cylinder is

underutilized. It is obvious that a thin – walled tube will be more proficient for

torques than a solid cylinder and a tube. It can be found that the torsional

stiffness of the tube is almost 50 times that of the solid cylinder.


Aircraft materials and its properties

Metal Alloys

Most commonly used materials in aircraft structure are aluminum, titanium

and steel alloys. Some new jet plane, such as Boeing 787 have gained its

momentum in the field of composites that comprises of utmost 50% of their

structural weight.

Selection of aircraft material depends on many considerations which

categorized by cost or by structural performance based on structural

performance, we can categorize as,

Density (weight)

Stiffness (Young’s modulus)

Strength (Ultimate and yield Strength)

Durability (Fatigue)

Damage tolerance (fracture toughness and crack growth)

Corrosion

Combination of multiple materials as well, able to deliver all the desired

properties rather than any single material. Basic mechanical properties of some

metallic aircraft structural materials are shown in table 1.

Table 1:

Material

Property

E u Y

GPa(msi) MPa (ksi) MPa (ksi) 3 3/ /g cm lb in

Aluminium

2024-T3

7075-T6

72(10.5)

71(10.3)

0.33

0.33

449(65)

538(78)

324(47)

490(71)

2.78(0.10)

2.78(0.10

Titanium

Ti – 6Al –

4V

110(16)

0.31

925(134)

869(126)

4.46(0.16)

Steel

AlSl4340

300 M

200(29)

200(29)

0.32

0.32

1790(260)

1860(270)

1483(212)

1520(220)

7.8(0.28)

7.8(2.28)


Here u and

y are tensile ultimate stress and tensile yield stress respectively.

v is the passion’s ratio i.e. the ratio of lateral strain to longitudinal strain.

From table 1, we can se the ultimate and yield stress for titanium alloys is

almost double than that of aluminum 7075 – T6 its corrosion resistance is

superior to both aluminum and steel alloys. But aluminum is required for high

temperature application. Similar to it, Titanium is used for blades, as due to

rotation, high stress exerts on the blade, so we want high blade strength

materials and it should also be corrosion free.

Composites

For structural application we cannot use fibers in isolation. That is why to

improve its property, we can introduce fibers in same type of material matrix,

in which it can orient properly. Mechanical properties of fibers are shown in

table 2.

Table 2:

Material

Property

E u

( )GPa msi ( )GPa ksi 3/g cm

E – glass 77.0(11) 2.50(350) 2.54

S- glass 85.0(12) 3.50(500) 2.48

Silcom Carbide(Nicalom) 190.0(27) 2.80(400) 2.55

Carbon (Hercules HS4) 240.0(35) 3.60(510) 1.80

Carbon (Hercules HMS) 360(51) 2.20(310) 1.80

Carbon (Toray T300) 240.0(35) 3.50(500) 1.80

Boron 385.0(55) 3.50(500) 2.65

Kevlar – 49 (Aramid) 130.0(18) 2.80(400) 1.45

Kevlar – 29 65.0(9.5) 2.80(400) 1.45


Longitudinal mechanical properties of other composites are shown in table 3.

Table 3:

Material

Property

E u

( )GPa msi ( )GPa ksi 3/g cm

Carbon/epoxy

T300/5208

IM6/3501 – 6

AS4/3501 – 6

140.0(20)

177.0(25.7)

140.0(20)

1.50(210)

2.86(414)

2.10(300)

1.55

1.55

1.35

Born/aluminum

B/Al 2024

210.0(30)

1.50(210)

2.65

Glass/epoxy

S2 glass/epoxy

43.0(6.2)

1.70(245)

1.80

Aramid/epoxy

Kev 49/epoxy

70.0(10)

1.4(200)

1.40

Stresses and associated strains

In the figure given above structural axial member having length 0L which is

subjected to stresses and gets elongated. The strain in the member can be

calculated as:-

o

L

L

... (1.14)

Where

1 0L u u ... (1.15)


Where 1 and

0 are the axial displacement component.

Strain in 1 – D case (or in axial member):-

, ,xx yy zz

u v w

x y z

... (1.16)

A 3 – D body under a complex stress system not only experience axial stresses

but also experiences shear stresses, therefore, normal strain components are

not sufficient to describe as a general state of deformation in a 3 – D body

completely. Moreover, shear strain components are also needed to describe the

distortional deformation.

Let use consider a 2 – D case

From figure above displacement may be v and u which may be positive or

negative.

Here 1

v

x

and 2

u

y

1 2xy

v u

x y

... (1.17)

similarly

yz

v w

z y

... (1.18)

and

xz

w u

x z

... (1.19)


Rigid body motion

For a rigid body motion, a body must experience a displacement without

introducing strains in the body. If the body is moved from one position to

another without getting deformed then the motion of the body cane be termed

as rigid body motion.

0

0

0

u u constant

u v constant

u w constant

... (1.20)

Represent a rigid translation motion and do not yield any strains.

Another example of a rigid body moving around in a circle. The following

displacements represent a rotation of rigid body in the x – y plane.

u y

v x

0w ... (1.21)

0, 0,xx yy

u v

x x

... (1.22)

Questions: 1.1 Consider 2a D body (a unit square’ PQRS’) in the x –y plane as

shown in figure. After deformation, the four corner points move to

A, B, C and D respectively. Assume the displacement field is

given by:

0.03u x

0.015v y

What are the total strains and the new coordinates of P, Q, R

and S.

Solution: Here

0.03xx

u

x

0.015yy

v

y


0xy

v u

x y

New position of ‘P’ after deformation

' 0.03 0 0x X

' 0.015 0 0y X

New coordinates of P which is A: (0,0)

Similarly new position of Q R and S

New coordinates of Q which is C : (1.03,1.015)

New coordinates of S Which is D: (0,1.015)

Stress

When a body is subjected to a force field, than a restoring force develops in the

body which opposes the acting force or helps the body to retain its shape. This

restoring force per unit area is termed as stress i,e:-

P

A ... (1.23)

Where A is cross – section area. If A = 1 unit area. Then P

For a 3 – D body having an area A with a unit normal vector, the stress is

defined as

0limA

Ft

A

... (1.24)

Thus, t can be considered as the force per unit area acting on the given plane

surface.

Tensile stress always acts normal to the surface whereas, shear stresses

always acts parallel to the surface.


Equilibrium Equation for non uniform stress Variation

Consider an infinitesimal element having dimensions dx Xdy X dz subjected to a

non – uniform stress field. The various stress components acting on the

different faces of the elements are as shown in figure below.

For equilibrium, 0F

. . . . . . . . . . . . 0yxxx zx

xx xx yx yx zx zxx y z z y x z z z x y x yx y z

... (1.25)

Dividing the whole equation by . .x y z we get

0xyxx xz

x y z

... (1.26)


0yy yx yz

y y z

... (1.27)

0zyzxzz

z x y

... (1.28)

Above three equations are equilibrium equations about any point in a body. If a

body is said to be in equilibrium, the stress field must satisfy these equation

everywhere in the body.

Question 1.2: A horizontal rectangular plate ABCD is hinged at points A, B

and D. AC reactions RA, RB, and RC at points A, B and D,

respectively, are

[GATE 2013]

(a) Indeterminate

(b) p,-p,p

(c) 0,p,0

(d) p/3,p/3,p/3

Solution: (b)

From force equilibrium equation, we get:-

A B CR R R P

From moment equilibrium equation, we get:-

AB: 0CP BC R AD

CR P

AD: 0BP CD R AB

AR P

From force equation:

BP R P P P

, , , ,A B CR R R P P P

Aircraft Structure - d24cdstip7q8pz.cloudfront.net · Load at Mid-spain Simply Supported ... While...

Documents

Transcript of Aircraft Structure - d24cdstip7q8pz.cloudfront.net · Load at Mid-spain Simply Supported ... While...