22219477 Aircraft Design Day4

8/3/2019 22219477 Aircraft Design Day4

1/20

DAY 4


2/20

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


3/20

V-n DIAGRAM


4/20

SIGN CONVENTIONS

C) BENDING MOMENTB) SHEAR FORCEA) AXIAL FORCE

F) ANGLE & ROTATIOND) TORQUE E) SHEAR FLOW


5/20

SIGN CONVENTIONS

(REACTION LOADS)

A) AXIAL & SHEAR

C) TORQUE

B) BENDING MOMENT


6/20

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

lb1200

0

=

=T

Fx

506P

006715P

400006*8000150*10*1200

0

=

=

=+

=P

MB

9447L

8000

0

=

=+

=PL

FY


7/20

TRUSSTRUSSES ARE CLASSIFIED AS

- STATICALLY DETERMINATE- STATICALLY INDETERMINATE

32 = jm m= Number of membersj= Number of jointsm < 2j-3 Structure is unstablem > 2j-3 Structure is statically indeterminate


8/20

TRUSSES (Contd)

Identify whether the structure is statically

determinate / indeterminate

C

A B

RAH

RAV

RBH

RBV

P PPPPP

A B C D


9/20

TRUSS ANALYSIS

TRUSSES CAN BE ANALYSED BY- METHOD OF JOINTS

- METHOD OF SECTIONS


10/20

SPACE TRUSS The six equilibrium eqns are

0

00

=

==

Z

Y

X

F

FF

0

0

0

=

=

=

Z

Y

X

M

M

M

RCosF

RCosF

RCosF

Z

Y

X

=

=

=

222

zyX FFFR ++=Resultant force

Component forces

Z

F

Y

F

X

F

L

R ZYX ===Geometric relation


11/20

LANDING GEAR

Resolving the forces in V direction

Resolving the forces in D direction

lbcos*F

F

AV

V

193161520000

0

==

=

lbsin*F

F

AD

D

517615200000

==

=


12/20

LANDING GEAR (FREEBODY DIAGRAM)

C.

.

.

.

... .H

A A

BB

C

FBH

FCG

19318 lb19318 lb

5176 lb

I

I

J

J

FJV

FIV

FJD

FID

FGV

FEDFES

FES

TE

TEFED

FEV

FEV

FIS

FGS

E

E

G

G

5176 lb

8"

3"

3"

20"

23"

44"

45

V

D

S


13/20

LANDING GEAR (FORCE ESTIMATION)

A

B

C

FBH

FCG

19318 lb

FEDFES

TE

FEV

in-lb41408

08*5176

0

=

=

=

E

E

EV

T

T

M

V

D

S

lb14003

03*)45(*20*)45(*44*5176

0

=

=

=

BH

BHBH

ES

F

SinFCosF

M

lb11370

020*)7.33(*3*)7.33(*8*19318

0

=

=

=

CG

CGCG

ED

F

SinFCosF

M

lb19760

45Cos7.3319318

0

=+=

=

EV

BHCGEV

V

F

FCosFF

F

lb4730

45Cos5176

0

= +=

=

ED

BHED

D

F

FF

F

5176 lb

56.3

45

lb6309

33.7Sin

0

==

=

ES

CGES

S

FFF

F


14/20

LANDING GEAR (FORCE CALCULATIONS)

I

J

FJV

FIV

FJD

FID

FCG =11370 lb

FES =6309 lb

TEFED =4730 lb

FEV =19760 lb

FIS

E

G

00 == ISS FF

lb1361-

203*4730

0

=

==

JD

JDE

IV

F

FT

M

lb6091

13614730

0

=+=

=

ID

ID

D

F

F

F

lb12109

*202*)7.33(*18*)7.33(*3*19760

0

=

=++

=

JV

JVCGCG

ID

F

FSinFCosF

M

lb17110

7.3319760

0

=

+=

=

EV

JVCGIV

V

F

FCosFF

F


15/20

WING STRUCTURES

WING STRUCTURES


16/20

WING STRUCTURES

(FREEBODY DIAGRAM)

FAZ

FAY

FGZ

A

C

G

60

100

180

16 lb/in

lb2592

90*180*16*100

0

=

=

=

GZ

GZ

A

F

F

M

FRONT SPAR

Using Geometric relation

lbFGY 4320100*60

2592 ==

lb288

2592180*16

0

==

=

AZ

AZ

Z

F

F

F

FGY

Z

F

Y

F

X

F

L

R ZYX ====

WIN TRU TURE


17/20

WIN TRU TURE(FREEBODY DIAGRAM)

FBZ

FBY

FEZ

B E

100

180

4 lb/inREAR SPAR

Using Geometric relation

lb648

90*180*4*100

0

=

=

=

EZ

EZ

B

F

F

M

lb1080

lb324

60648

10030

=

=

==

EY

EX

EYEX

F

F

FF

lb72

648180*4

0

==

=

BZ

BZ

Z

F

F

F

FEX

FEY

WING STRUCTURES (TRUSS)


18/20


G

EB

30 30 40 40 40A

L

KI

JP

O

N

MFGY =4320

FBY

FAY

17515075 200200

FEY =1080

FEX =324 100

X

Y

00 == PNY FF

1000 == POX FFJOINT O

167

10013.53

0

=

=

=

ON

ON

X

F

CosF

FJOINT P JOINT E

JOINT N

JOINT M300

02001.53

0

=

=++

=

NM

ONNM

X

F

CosFF

F

133

13.53Sin

0

=

=

=

NE

ONNE

Y

F

FF

F

50013.53

0

=

=

=

ME

MENM

X

FCosFF

F

lbF

SinFF

F

OM

ONOM

Y

133

13.53

0

==

=

lbF

FSinFF

F

MG

OMMNMG

Y

533

13.53

0

= +=

=

824

52413.53

0

=

+=

=

GE

MEGE

X

F

CosFF

F

lbF

FSinFF

F

MG

NEMEEL

Y

547

108013.53

0

=+= =



19/20


lbF

SinFF

F

LJ

GJLJ

Y

547-

551

13.53

0

=

=

=

G

EB

30 30 40 40 40A

L

KI

JP

O

N

MFGY =4320

FBY

FAY

17515075 200200

FEY =1080

FEX =324 100

X

Y

lb1373

13.53

0

=

=

=

GL

GEGL

X

F

FCosF

F

lbF

SinFF

F

GK

GLGK

Y

533

5951

432013.53

0

==

=

JOINT G

lb998

017513.53

0

=

=++

=

LK

LKGL

X

F

FCosF

F

JOINT L

JOINT K

lb1412

045

0

==

=

KJ

LKKJ

X

F

FCosF

F

lbF

SinFF

F

KI

KJKI

Y

05950

6948

45

0

==++

=



20/20


G

EB

30 30 40 40 40A

L

KI

JP

O

N

MFGY =4320

FBY17515075 200200

FEY =1080

FEX =324 100

X

Y

JOINT J

JOINT I

lb1148

015045

0

=

=++

=

IJ

IJKJ

X

F

FCosF

F

lbF

SinFF

F

JB

KJJB

Y

551

1550

45

0

=+=

=

lb1624

45

0

=

=

=

IB

IJIB

X

F

FCosF

F

lbF

SinFF

F

IA

KJIA

Y

8096

6

=

=

=94845

0

lbFF

F

IAAy

Y

8096==

= 0JOINT A

JOINT B

lb2698

45

0

=+=

=

BY

JBIBBY

X

F

FCosFF

F

FAY

FAX

lbF

F

BA

Y

1225-=

= 0

lbF

F

AX

Y

1225=

= 0

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22219477 Aircraft Design Day4