DAY 3

AIRCRAFT AS SEEN BY DIFFERENT GROUPS

Aerodynamic group

Stress group

Production group

FLOW CHART OF AN A/C DESIGN

Design Specification

Design Criteria

Basic Loads

Laboratory Development

Test Data

Flight Test Data

Airplane Design

Approved Type Certificate

Certification Test Program

DESIGN CYCLE OF AN A/C

LOAD FACTOR

L=W L+L=W+F

LOAD FACTOR :

A factor which defines load in terms of weight

W

L

g

An

11

zAg

WF L

g

AW z1LL

DESIGN LOAD FACTOR

(2 to 3)

(6 to 8)

WING LOAD CASES• CASE A

– Large angles of attack corresponding to CL max while reaching a load factor of nmax (steep climb)

• CASE A’– Reaching a load factor of nmax at

maximum airspeed

• CASE B– Reaching a load factor of 0.5nmax

at maximum airspeed with Ailerons deflected

• CASE C– Ailerons deflected at maximum

airspeed (dive)

• CASE D & D’– Acrobatic maneuvers with a

negative load factor

WING LOAD REGIONS

Region I : n=nmax

Region II : 0n<nmax

Region III : 0nnmin

Region IV : n=nmin

CONSTRUCTION OF V-n DIAGRAM

From (1)

From (1) & (3) we get

296

VL

2s1SCW

maxL

maxLCS

W296Vs1

………….(1)

296

V2snSCnW

maxL

………….(2)

S/WCn

maxL 296

V2s

………….(3)

………….(4)

………….(5)

ns1sn VV

V-n DIAGRAM

MANEUVERING V-n DIAGRAM

GUST V-n DIAGRAM

LIMITING LOAD FACTOR VALUES

lim

24,0002.1

10,000itto

nW

S.No Weight of the airplane

Limiting load factor

1 < 4118 3.8

2 4118 - 50000

3 > 50000 2.1

AIRCRAFT STRUCTURAL DESIGN

PHASES OF AIRCRAFT STRUCTURAL DESIGN

• Specification of function and design criteria

• Determination of basic external loads

• Calculation of internal element loads

• Determination of allowable element strengths and margins of safety

• Experimental demonstration or subtantiation test program

CLASSIFICATION• CRITICAL

– ALL PRIMARY STRUCTURESFUSELAGEWING

EMPENNAGEAILERON / FLAPS/STABILIZERS

• NEAR CRITICAL PARTS– TRAILING EDGE PANELS OF WINGS &

EMPENNAGE – RADOME

• NON-CRITICAL PARTS– FAIRINGS / INTERIOR PANELS

CRITERIA FOR MATERIAL SELECTION

• Light weight

• Stiffness

• Toughness

• Resistance to corrosion

• Fatigue

• Environmental heat

• Availability

• Easy to fabricate

MATERIALS

• Aluminum alloy

• Titanium

• Steel

• Composites

SIGN CONVENTIONS

C) BENDING MOMENTB) SHEAR FORCEA) AXIAL FORCE

F) ANGLE & ROTATIOND) TORQUE E) SHEAR FLOW

SIGN CONVENTIONS (REACTION LOADS)

A) AXIAL & SHEAR

C) TORQUE

B) BENDING MOMENT

• STRUCTURAL INDEX OFFERS THE DESIGNER A GUIDE TO DESIGN THE OPTIMUM TYPE OF STRUCTURE

• STRUCTURAL INDEX IS USEFUL IN DESIGN WORK BECAUSE IT CONTAINS– THE INTENSITY OF THE LOAD– DIMENSIONS, WHICH LIMIT THE SIZE OF THE

STRUCTURE

STRUCTURAL INDEX

STRUCTURAL INDEX

A) SHEAR

G) WING BOXF) TUBEE) PANEL

D) WIDE COLUMNC) COLUMNB) TORSION

SEQUENCE OF STRESS WORK

• PRELIMINARY SIZING

• PRODUCTION STRESS ANALYSIS

• FORMAL STRESS ANALYSIS FOR CERTIFICATION

PRELIMINARY SIZINGStep1 : Recognize the function and configuration of the

component

Step2 : Basic loads (static / fatigue / fail safe / crash)

Step3 : Material selection (static / fatigue / fracture toughness)

Step4 : Fastener and repairability

Step5 : Efficient structure (Fabrication / Configuration / assembly / installation / stiffness)

Step6 : Cost ( Manufacturing / assembly/ performance / market)

EQUILIBRIUM OF FORCES• THE FORCE EQUILIBRIUM EQUATIONS ARE

0

0

0

M

F

F

Y

X

STRUCTURE

FREE BODY DIAGRAM

PL

1200 lb40000 lb-in

W=8000 lb

A BC T

lb 1200

0

T

Fx

506P

0067 15P

400006*8000150*10*1200

0

P

M B

9447 L

8000

0

PL

FY

TRUSSTRUSSES ARE CLASSIFIED AS

- STATICALLY DETERMINATE- STATICALLY INDETERMINATE

32 jm m= Number of membersj= Number of joints

m 2j-3 Structure is unstablem 2j-3 Structure is statically indeterminate

TRUSSES (Contd…)• Identify whether the structure is statically

determinate / indeterminate

C

A B

RAH

RAV

RBH

RBV

P PPPPP

A B C D

ASSUMPTIONS IN TRUSS ANALYSIS

• The members of the truss are straight, weightless and lie in one plane

• The members of the truss meeting at a point are considered as joined together by a frictionless pin

• All the members axis intersect at the centre of the pin

• All the external loads are only applied to the joints and in the plane of truss

TRUSS ANALYSISTRUSSES CAN BE ANALYSED BY

- METHOD OF JOINTS- METHOD OF SECTIONS

STRUCTURE

FREE BODY DIAGRAM

B CA

D GE

R1

H

1000 lb

2000 lb

4000 lb

A

20001

0

R

FX

lb 4500 3

20*310*200010*400030*1000

0

R

R

MD

lb 500 2

500032

0

R

RR

FY

R3R2

TRUSS ANALYSIS (Contd …)

A

D

E

2000

500

2000 lb

5000 ADY FF

20000 DEX FFJOINT D

JOINT E

JOINT A

lbF

FSinFF

AE

ADAEY

707

450

lb 2500

2000450

AB

ABAEX

F

FCosFF

lb 2500

450

EG

EGDEAEX

F

FFCosFF lbF

FSinFF

BE

BEAEY

500

450

FDE

FAD

FEG

FBE

FAB

FAEFAD=500

FDE=2000

FAE=707

TRUSS ANALYSIS (Contd …)

B

G

4500

H

1000 lb

4000 lbJOINT B

JOINT G

lbF

FSinFF

BG

BEBGY

4950

4000450

lb 1000

450

BC

BCABBGX

F

FFCosFF

lb 1000

450

GH

GHEGBGX

F

FFCosFF

lbF

FSinFF

CG

CGBGY

1000

4500450

lbF

SinFF

CH

CHY

1414

1000450

JOINT H

FAB=2500FBC

FBG=4950

FBE=500

FCH

FGH=1000

FGHFEG=2500

FCG

TRUSS ANALYSIS (METHOD OF SECTIONS)

B CA

D E

R1

R2 R3=4500

H

1000 lb

2000 lb

4000 lb

GG

TRUSS ANALYSIS (METHOD OF SECTIONS)

lb 1000

10*20*50010*400010*2000

0

BC

BC

G

F

F

M

BA

D E

R1=2000 lb

2000 lb

4000 lb

G lbF

SinF

F

BG

BG

Y

4950

500400045

0

R2 =500 lb

FBC

FBG

FEG

lb 2500

10002000200045

0

EG

BGEG

X

F

CosFF

F

TRUSS ANALYSIS (METHOD OF SECTIONS)

R3=4500

H

1000 lb

G

FBC

FBG

FEG

CB

lb 2500

10*20*100010*4500

0

EG

EG

B

F

F

M

lb 1000

45

0

BC

EGBGBC

X

F

FCosFF

F

lbF

SinF

F

BG

BG

Y

4950

1000450045

0

TRUSS WITH MEMBERS IN BENDING

RAX

RBY1=100 lbRAY=100 lb

RBX1

RCXRBX1

RBY2=100 lb

200 lb

200 lb

A

E

D

CB

RCY=100 lb

100 lb 200 lb 100 lb

TRUSS WITH MEMBERS IN BENDING

A

E

D

C

B

100 lb

200 lb

100 lb

FCE

FBC

lbF

SinF

F

CE

CE

Y

200

10030

0

lb 2.173

30

0

BC

CEBC

X

F

CosFF

F

FABFBC

lb 2.173

0

BCAB

X

FF

F

lb 200

0

BE

Y

F

F

FAE

FABFAX

lbF

SinF

F

AE

AE

Y

200

10030

0

lb 4.346

30 Cos

0

AX

AEABAX

X

F

FFF

F

FBE

FAE=200 lb

FBE=200 lbFCE=200 lb

FED

lbFF EDD 400

lbF

CosFCosFFF

ED

BEAECEED

400

6060

FED

FD

FINITE ELEMENT METHOD• A GENRALIZED MATHEMATICAL

PROCEDURE FOR CONTINUUM PROBLEMS POSED BY MATHEMATICALLY DEFINED STATEMENTS

4

321

1110

875

5

6

6

9

9

1 2

3 4

7 8

9

65

10

11 12

TRUSS ELEMENT

r=-1 r=1

Y

X

x2

x1 U2U1

Natural coordinate system rGlobal coordinate system X

rh

rh

XhX ii

12

1

12

1

2

1

2

1

rh

rh

UhU ii

12

1

12

1

2

1

2

1

TRUSS ELEMENT (Contd…)

1 1L

1B

L

UU

LXX

dr

dXdr

dUdX

dr

dr

dU

dX

dU

12

12 22

12

1

1 1-

1- 1K

1

11 1K

1

1

L

11 1

L

1K

K

L

AE

drLL

AE

JdrAE

dXBDBT

2

1

12

1

1

1

1

STATIC ANALYSIS

FKU

SOLUTION METHODSA) SPARSE DECOMPOSITIONB) CHOLESKEY FACTORIZATIONC) PCG METHODD) FRONTAL SOLVER

K-STIFFNESS MATRIXU–DISPLACEMENT VECTORF- FORCE VECTOR

PRACTICEWORK

Prob. 1 : Solve the given truss using method of joints, method of sections and compare

PRACTICEWORK

2. Solve the given aircraft structure using method of joints, method of sections and compare