Aircraft Design Day3

DAY 3

AIRCRAFT AS SEEN BY DIFFERENT GROUPS Stress group

Aerodynamic group

Production group

FLOW CHART OF AN A/C DESIGN Design Specification Design Criteria Basic Loads Flight Test Data

Airplane Design

Certification Test Program Approved Type Certificate

Laboratory Development Test Data

DESIGN CYCLE OF AN A/C

LOAD FACTOR LOAD FACTOR : A factor which defines load in terms of weight

L+∆ L=W+F

L=W

W ∆L = F = Az g  Az  ∆L + L = W 1 +  g  

∴

A ∆L n =1+ =1+ g W

DESIGN LOAD FACTOR (6 to 8)

(2 to 3)

WING LOAD CASES • CASE A – Large angles of attack corresponding to CLmax 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 : 0≤ n
CONSTRUCTION OF V-n DIAGRAM L = W = C Lmax From (1)

Vs1 =

Vs12 S 296

296 W

S

………….(2)

Vsn2 S 296

………….(3)

C Lmax

nW = C Lmax

………….(1)

From (1) & (3) we get

Vsn = Vs1 n n = C Lmax

Vs2 296 W / S

………….(4)

………….(5)

V-n DIAGRAM

MANEUVERING V-n DIAGRAM

GUST V-n DIAGRAM

LIMITING LOAD FACTOR VALUES S.No

Weight of the airplane

Limiting load factor

1

< 4118

3.8

2

4118 - 50000

3

> 50000

nlim it

24, 000 = 2.1 + ( Wto + 10, 000 )

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 STRUCTURES FUSELAGE WING EMPENNAGE AILERON / 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

A) AXIAL FORCE

D) TORQUE

B) SHEAR FORCE

E) SHEAR FLOW

C) BENDING MOMENT

F) ANGLE & ROTATION

SIGN CONVENTIONS (REACTION LOADS)

A) AXIAL & SHEAR

B) BENDING MOMENT

C) TORQUE

STRUCTURAL INDEX • 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

A) SHEAR

B) TORSION

E) PANEL

F) TUBE

C) COLUMN

D) WIDE COLUMN

G) WING BOX

SEQUENCE OF STRESS WORK • PRELIMINARY SIZING • PRODUCTION STRESS ANALYSIS • FORMAL STRESS ANALYSIS FOR CERTIFICATION

PRELIMINARY SIZING Step1 : 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

∑F = 0 ∑F = 0 ∑M = 0 X

Y

FREE BODY DIAGRAM L

A STRUCTURE

∑F

x

=0⇒

T = 1200 lb

∑M

B

P

C B 1200 lb 40000 lb-in

T

W=8000 lb

=0 ⇒

1200 * 10 + P * 150 −8000 * 6 = 40000 15P = 7600 P = 506

∑F

Y

=0⇒

L + P = 8000 L = 7494

TRUSS TRUSSES ARE CLASSIFIED AS - STATICALLY DETERMINATE - STATICALLY INDETERMINATE m= Number of members j= Number of joints

m = 2 j −3

m < 2j-3 Structure is unstable m > 2j-3 Structure is statically indeterminate

TRUSSES (Contd…) • Identify whether the structure is statically determinate / indeterminate P

P

C

A

A

B RBH

RAH RAV

RBV

P

P

P

B

C

D

P

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 ANALYSIS TRUSSES CAN BE ANALYSED BY - METHOD OF JOINTS FREE BODY DIAGRAM - METHOD OF SECTIONS 4000 lb STRUCTURE C 2000 lb A B A

∑M

D

R1

=0⇒

1000 * 30 + 4000 *10 + 2000 *10 = R3 * 20 R3 = 4500 lb

∑F

X

=0⇒

R1 = 2000

1000 lb

∑F

Y

=0⇒

R 2 + R3 = 5000 R 2 = 500 lb

D R2

E

G R3

H

TRUSS ANALYSIS (Contd …) JOINT D

∑F ∑F

X

= 0 ⇒ FDE = 2000

Y

= 0 ⇒ FAD = 500

2000

D

 F = 0 ⇒ F Sin 45 = FAD ∑ Y AE

FAE = 707lb

 F = 0 ⇒ − F Cos 45 + FAB = 2000 ∑ X AE

FAB = 2500 lb

∑F

Y

2000 lb

A



FBE = 500lb  F = 0 ⇒ F Cos 45 + FDE = FEG ∑ X AE

FAB FAE

FAD =500

= 0 ⇒ FAE Sin45 = FBE

FEG = 2500 lb

FDE

500

JOINT A

JOINT E

FAD

FAE =707 FBE FEG

FDE =2000 E

TRUSS ANALYSIS (Contd …) JOINT B ∑F

= 0 ⇒ FBG Sin45 + FBE = 4000 

Y

FBG = 4950lb  F = 0 ⇒ F Cos 45 − FAB = FBC ∑ X BG FBC = 1000 lb

FAB =2500

 F = 0 ⇒ F Cos 45 − FEG = FGH ∑ X BG

Y

G 4500

= 0 ⇒ FBG Sin 45 + FCG = 4500 

FCG = 1000lb

JOINT H

 F = 0 ⇒ F Sin 45 = 1000 ∑ Y CH

FCH = 1414lb

FCG FGH

FEG =2500

FGH = 1000 lb

∑F

FBC

B FBE =500 FBG =4950

JOINT G

4000 lb

FCH

1000 lb

FGH =1000 H

TRUSS ANALYSIS (METHOD OF SECTIONS) 4000 lb 2000 lb A

C

B

1000 lb R1 D

E R2

G

G R3=4500

H

TRUSS ANALYSIS (METHOD OF SECTIONS) ∑M

4000 lb 2000 lb A

B

FBC

G

=0⇒

− 2000 * 10 + 4000 * 10 − 500 * 20 = FBC * 10 FBC = 1000 lb

∑F

Y

FBG R1=2000 lb D R2 =500 lb

E

=0⇒

FBG Sin 45 = 4000 − 500 FBG = 4950 lb

FEG G

∑F

X

=0⇒

FEG = FBGCos 45 − 2000 + 2000 − 1000 FEG = 2500 lb

TRUSS ANALYSIS (METHOD OF SECTIONS) ∑M

B

=0⇒

4500 * 10 − 1000 * 20 = FEG *10 FEG = 2500 lb

∑F

Y

FBG = 4950 lb X

=0⇒

FBC = FBGCos 45 − FEG FBC = 1000 lb

C 1000 lb

=0⇒

FBG Sin 45 = 4500 − 1000

∑F

FBC B FBG

FEG

G R3=4500

H

TRUSS WITH MEMBERS IN BENDING 100 lb A

RAX

200 lb

RBX1

RBY1 =100 lb RAY =100 lb 200 lb RBX1 RCX RBY2 =100 lb

RCY =100 lb

200 lb

100 lb

B

C

E D

TRUSS WITH MEMBERS IN BENDING ∑F

Y

=0⇒

200 lb B F BC FAB

FCE Sin30 = 100

100 lb FCE = 200 lb FBC C

∑F

X

FBC = FCE Cos30

FCE

Y

100 lb FAX

A FAB FAE

∑F

=0⇒

FBE = 200 lb X

FAB = FBC = 173.2 lb

FBE =200 lb FCE =200 lb FAE =200 lb

FBC = 173.2 lb

∑F

=0⇒

Y

FBE

=0⇒

∑F

E

FED

=0⇒

FAE Sin30 = 100

FED = F CE + FAE Cos 60  + FBE Cos 60

FAE = 200 lb

FED = 400 lb

∑F

X

FED

=0⇒

FAX = FAB + FAE Cos 30 FAX = 346.4 lb



FD

D

FD = FED = 400 lb

FINITE ELEMENT METHOD • A GENRALIZED MATHEMATICAL PROCEDURE FOR CONTINUUM PROBLEMS POSED BY MATHEMATICALLY DEFINED STATEMENTS 10 9

11

9 9

7

8

1

5 2

4

5

6

10 11 6 5

3 1

12 7 6

4 2

3

8

TRUSS ELEMENT Y

x1

x2 U1

U2 r=1

r=-1

Natural coordinate system r Global coordinate system X 2

2

X = ∑ hi X i

U = ∑ hiU i

1 h1 = (1 − r ) 2 1 h2 = (1 + r ) 2

1 h1 = (1 − r ) 2 1 h2 = (1 + r ) 2

1

1

X

TRUSS ELEMENT (Contd…) dU dU dr ε= = dX dr dX dU 1 = dr 2 dX 1 L = ( X 2 − X1 ) = dr 2 2 ( U 2 − U1 ) ε= L 1 B = [ − 1 1] L

1

K = ∫ BDBT dX −1

1 1 − 1 K = ∫ [ − 1 1] AE  L L 1 −1 1

 Jdr 

1 − 1  L AE K = 2 ∫ [ − 1 1]  dr  L −1  1  2 - 1 AE 1 K= 1 L - 1

STATIC ANALYSIS KU = F K-STIFFNESS MATRIX U–DISPLACEMENT VECTOR F- FORCE VECTOR SOLUTION METHODS A) SPARSE DECOMPOSITION B) CHOLESKEY FACTORIZATION C) PCG METHOD D) FRONTAL SOLVER

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