Aircraft Design Day3
Aircraft Design Day3
Aircraft Design Day3
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
Laboratory
Flight Airplane Design Development
Test Data Test Data
L=W L+L=W+F
W
L F Az
g
Az
L L W 1
g A
n 1 1
g
L
W
DESIGN LOAD FACTOR
(6 to 8)
(2 to 3)
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
Vs12 S
L W C Lmax ………….(1)
296
From (1)
296 W
Vs1 S ………….(2)
C Lmax
Vsn2 S
nW C Lmax ………….(3)
296
From (1) & (3) we get
Vs2
n C Lmax ………….(5)
296 W / S
V-n DIAGRAM
MANEUVERING V-n DIAGRAM
GUST V-n DIAGRAM
LIMITING LOAD FACTOR VALUES
S.No Weight of the Limiting load factor
airplane
1 < 4118 3.8
C) TORQUE
STRUCTURAL INDEX
• STRUCTURAL INDEX OFFERS THE DESIGNER
A GUIDE TO DESIGN THE OPTIMUM TYPE OF
STRUCTURE
• PRELIMINARY SIZING
• PRODUCTION STRESS ANALYSIS
• FORMAL STRESS ANALYSIS FOR
CERTIFICATION
PRELIMINARY SIZING
Step1 : Recognize the function and configuration of the
component
F 0 Y
FREE BODY DIAGRAM
M 0 L P
C T
A B
1200 lb
40000 lb-in
STRUCTURE W=8000 lb
M B 0
F 0
F
Y
x 0 1200 * 10 P * 150 8000 * 6 40000
T 1200 lb 15P 7600
L P 8000
P 506 L 7494
TRUSS
TRUSSES ARE CLASSIFIED AS
- STATICALLY DETERMINATE
- STATICALLY INDETERMINATE
m= Number of members
m 2 j 3 j= Number of joints
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 P
C P P P
A B A B C D
RAH RBH
RAV RBV
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
- METHOD OF SECTIONS FREE BODY DIAGRAM
4000 lb
STRUCTURE
2000 lb A B C
A 1000 lb
R1
M D 0
D E G H
1000 * 30 4000 * 10 2000 * 10 R3 * 20
R2 R3
R3 4500 lb
F X 0 F Y 0
R 2 R3 5000
R1 2000
R 2 500 lb
TRUSS ANALYSIS (Contd …)
JOINT D FAD
F X 0 FDE 2000 2000
F Y 0 FAD 500 D FDE
500
JOINT A
Y
F 0 FAE Sin 45
FAD
A FAB
FAE 707lb 2000 lb
X
F 0 FAE Cos 45
FAB 2000
FAE
FAB 2500 lb FAD=500
FEG
FBE 500lb FDE=2000
FEG 2500 lb
TRUSS
JOINT B
ANALYSIS (Contd …)
4000 lb
F Y 0 FBG Sin45 FBE 4000
FAB=2500
FBC
FBG 4950lb
B
XF 0 FBG Cos 45
FAB FBC FBE=500
FBC 1000 lb FCG
FBG=4950
JOINT G FGH
FEG=2500
X
F 0 FBG Cos 45
FEG FGH
G
FGH 1000 lb
4500
F Y 0 FBG Sin45 FCG 4500
FCG 1000lb
1000 lb
FCH
JOINT H
Y
F 0 FCH Sin 45
1000 FGH=1000
FCH 1414lb H
TRUSS ANALYSIS
(METHOD OF SECTIONS)
4000 lb
2000 lb A B C
1000 lb
R1
G
D E G H
R2 R3=4500
TRUSS ANALYSIS
(METHOD OF SECTIONS)
4000 lb
M G 0
2000 *10 4000 *10 500 * 20 FBC * 10
2000 lb A B FBC
FBC 1000 lb
FBG F Y 0
R2 =500 lb F X 0
FEG FBG Cos 45 2000 2000 1000
FEG 2500 lb
TRUSS ANALYSIS
(METHOD OF SECTIONS)
M B 0
4500 *10 1000 * 20 FEG *10 FBC C
B
FEG 2500 lb FBG 1000 lb
F Y 0
FBG Sin45 4500 1000
FEG G
FBG 4950 lb H
R3=4500
F X 0
FBC FBG Cos 45 FEG
FBC 1000 lb
TRUSS WITH MEMBERS IN BENDING
200 lb
RAX RBX1 E
RAY=100 lb RBY1=100 lb D
RBX1 200 lb
RCX
RBY2=100 lb RCY=100 lb
TRUSS WITH MEMBERS IN BENDING
F Y 0 200 lb F Y 0
FCE Sin30 100 B F FBE 200 lb
FAB BC
100 lb FCE 200 lb
F X 0
FBC C FAB FBC 173.2 lb
F X 0 FBE
FBC FCE Cos30 FBE=200 lb
FCE
FBC 173.2 lb FAE=200 lb FCE=200 lb
FED E
F Y 0
FAE
F X 0
FD FED 400 lb
FAX FAB FAE Cos 30 FD D
FAX 346.4 lb
FINITE ELEMENT METHOD
• A GENRALIZED MATHEMATICAL
PROCEDURE FOR CONTINUUM PROBLEMS
POSED BY MATHEMATICALLY DEFINED
STATEMENTS 10
9 11
9 10
9 7 8 11 12
4 5 6 7 8
1 2 5 6
3 4
5 6
1 2 3
TRUSS ELEMENT
Y x2
x1
U1 U2
X
r=-1 r=1
dr 2 2 AE
1
1 L
K 2 1 1 dr
U 2 U1 L 1
1 2
L AE 1 - 1
K
1
B 1 1 L - 1 1
L
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