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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