Fuselage Design

2. 3
Aircraft loads | 32
Fuselage structure
• Revisit stresses in pressure vessel
• And the Hooke’s law for plane
stress or biaxial stress condition
Pressurization (case I)
∆p
dϕ
ϕ
R
σlong
∆p
Fy
long
circ
circ
long circ
long
E E
E E
σ
σ
ε ν
σ σ
ε ν
= −
= −
;
2
circ long
pR pR
t t
σ σ
= =
ACE - Dr Mohamed Elfarran

3. 4
Aircraft loads | 32
Fuselage structure
• With σcirc =2σlong this is
• For a metallic pressure vessel with ν ≈ 0.3 this means that
εcirc =4.25 εlong
Pressurization (case I)
1
2
1
2
circ
circ
circ
long
E
E
σ ν
ε
σ
ε ν
 
= −
 
 
 
= −
 
 
ACE - Dr Mohamed Elfarran

4. 5
Aircraft loads | 32
Fuselage structure
• For a sphere under pressure
• The strain in a sphere is equal in all directions
Pressurization (case I)
1 2 2
pR
t
σ σ
= =
( )
1
1
2
sphere
pR
t E
ε ν
= −
ACE - Dr Mohamed Elfarran

5. 6
Aircraft loads | 32
Fuselage structure
• What ratio tsphere/tcylinder is needed to connect cylinder to sphere in a
pressure vessel?
• Avoid discontinuities in εcirc to avoid these deformation mismatches
• Thus εcirc = εsphere !
Pressurization (case I)
ACE - Dr Mohamed Elfarran

6. 7
Aircraft loads | 32
Fuselage structure
• Derivation
• Thus
• With Rcylinder = Rsphere this means tsphere ≈ 0.4 tcylinder
Pressurization (case I)
cylinder sphere
ε ε
=
( )
( )
2 1
2 1
1
sphere sphere
cylinder cylinder
R t
R t
ν
ν
−
= =
−
( )
1 1
1 1
2 2
cylinder sphere
cylinder sphere
pR pR
t E t E
ν
ν
 
− = −
 
 
ACE - Dr Mohamed Elfarran

7. 8
Aircraft loads | 32
Fuselage structure
• Aircraft pressure cabin
• Role frames & stringers (5 - 10%)
• Other disturbances (doublers around cut-outs)
• Fracture of fuselage could be disastrous: ‘exploding balloon’
• No limit load (formal definition: once in the lifetime of the aircraft),
“limit” load occurs every flight during pressurization of the cabin
⇒ safety factor is 2
Pressurization (case I)
ACE - Dr Mohamed Elfarran

8. 9
Aircraft loads | 32
Fuselage structure
• Aircraft pressure cabin
• Design pressure differential pd = pcabin - ph
• Cabin altitude: pcabin related to altitude of 2400 - 3000 m
• Safety factor j=2: pult = 2 pd = 2(pcabin - ph)
• Example
• Pressure differential is 45 kPa
• Aircraft altitude of h =9050 m ⇒ cabin altitude of 2440 m
• Aircraft altitude of h =10350 m ⇒ cabin altitude of 3050 m
Pressurization (case I)
ACE - Dr Mohamed Elfarran

9. 10
Aircraft loads | 32
Fuselage structure
• The axial stresses in a tube under torsion
Torsional loading (case II)
ACE - Dr Mohamed Elfarran

10. 11
Aircraft loads | 32
Fuselage structure
• Function of cylinder against torsional loading
• Rectangular shape
• Deforms like
• Equilibrium if F1d1 = F2d2 = Fndn
⇒ M1 = M2 = MT
Torsional loading (case II) F1
F2
d1
d2
M1
M2
ACE - Dr Mohamed Elfarran

11. 12
Aircraft loads | 32
Fuselage structure
• Torsion causes shear stress & strain
• Equilibrium: τoutside = τcross section
Torsional loading (case II)
τ
G
τ
γ =
MT
MT
γ
ACE - Dr Mohamed Elfarran

12. 13
Aircraft loads | 32
Fuselage structure
• Torsion causes shear stress
• Equilibrium: τoutside = τcross section
• Torsional moment
• Define q = τ t ⇒ MT = 2qA
Torsional loading (case II)
τ
G
τ
γ =
2
2 2
T
M r t r r t
τ π τπ
= =
MT
MT
γ
circumference
circle
area
circle
ACE - Dr Mohamed Elfarran

13. 14
Aircraft loads | 32
Fuselage & wing structure
• Thin walled box, stringers not loaded, torsion moment MT
• Moment around an arbitrary point:
• with
• All ‘fractions’ together:
T
q ds a M
= ∆
2
a ds A
= ∆
2
T
M qA
=
2
T
M
q
A
=
Torsional loading (case II)
ACE - Dr Mohamed Elfarran

14. 15
Aircraft loads | 32
Fuselage & wing structure
• For a thin walled closed box (enclosed area A) the shear flow is
• All cross sections
have the same
shear-flow q
• A torsion box does
not need to have a
circular cross section
2
T
M
q
A
=
enclosed area ≡ A
enclosed area ≡ A
enclosed area ≡ A
enclosed area ≡ A
Torsional loading (case II)
ACE - Dr Mohamed Elfarran

15. 16
Aircraft loads | 32
Fuselage & wing structure
• The torsion box will not “function” at a cut-out
• A stiff frame with rigid corners around the cut-out is needed
Torsional loading (case II)
ACE - Dr Mohamed Elfarran

16. 17
Aircraft loads | 32
Wing structure
• Consider elementary spar
• Diamond shaped deformation is resisted
by sheets
• Frame exercises a shear stress on the sheets
• As reaction: sheets exercise shear stress on the frame
• Assumption: bars of frame very stiff,
no deformation
• Without sheets: Frame is not functioning
Bending of wing spar
ACE - Dr Mohamed Elfarran

17. 18
Aircraft loads | 32
Wing structure
• If frame functions, sheets in equilibrium
• High forces: pleat formation
• However: Structure still functions !
• Resistance to shear is the resistance to tension and compression
under 45 degrees. Relationship: E, ν, G
Bending of wing spar
ACE - Dr Mohamed Elfarran

18. 19
Aircraft loads | 32
• Break down the elements ⇒ book keeping
• Global
• Horizontal equilibrium: F - F1x - F2x = 0
• Vertical equilibrium: F1y - F2y = 0
• Moment equilibrium: F h + F1y w = 0
• Local
• Equilibrium in elements …
Wing structure
Bending of wing spar
F1y
F1x
F2y
F2x
ACE - Dr Mohamed Elfarran

19. 20
Aircraft loads | 32
• Break down the elements ⇒ book keeping
• For shear there is physically no difference but we need a sign
convention
Wing structure
Bending of wing spar
_ +
ACE - Dr Mohamed Elfarran

20. 21
Aircraft loads | 32
Wing structure
• In bending, the function of spar webs (shear) is essential! The
webs have to be supported on the upper and lower side by caps in
order to realize equilibrium
• webs transfer (external) transverse forces into shear-flows
• caps transfer shear-flows in normal forces
Bending of wing spar
ACE - Dr Mohamed Elfarran

21. 22
Aircraft loads | 32
Wing structure
• Global equilibrium (always check!)
• Horizontal: √
√
√
√
• Vertical: √
√
√
√
• Moment: √
√
√
√
Bending of wing spar
4 4
0
U L
N N
− =
1 2 3 3
0
F F F q h
+ + − =
( ) ( )
4
1 1 2 3 2 2 3 3 3
0
U
N h F l l l F l l F l
− + + − + − =
ACE - Dr Mohamed Elfarran

22. 23
Aircraft loads | 32
Wing structure
• Web plate 1 and upper and lower cap 1
• Equilibrium of forces in cap 1
• q1 is shear flow in web plate 1
Bending of wing spar
2
2
1 1
1
1 1 1
1
1 1 1
0
U
L
F q h
F
N q l l
h
F
N q l l
h
− =
= =
= =
ACE - Dr Mohamed Elfarran

23. 24
Aircraft loads | 32
Wing structure
• Web plate 2 and upper and lower cap 2
• Equilibrium of forces in cap 2
Bending of wing spar
3 2
3
2 2 1 2 1
1 1 2
2 2 1 2
1 1 2
1 2
...
U U
L
F q h q h q h F
F F F
N N q l l l
h h
F F F
N l l
h h
= − = −
+
= + = +
+
= = +
ACE - Dr Mohamed Elfarran

24. 25
Aircraft loads | 32
Wing structure
• Web plate 3 and upper and lower cap 3
• Equilibrium of forces in cap 3
Bending of wing spar
4 3
4
3 3 2 3 1 2
3 3
1 2 3
1 1 2
1 2 3
1 2 3
1 1 2
1 2 3
...
U U
L
F q h q h q h F F
N N q l
F F F
F F F
l l l
h h h
F F F
F F F
N l l l
h h h
= − = − −
= +
+ +
+
= + +
+ +
+
= = + +
ACE - Dr Mohamed Elfarran

25. 26
Aircraft loads | 32
Wing structure
• Shear flow in the webs:
• Transverse force building up:
Bending of wing spar
n
n
D
q
h
=
1 2
1
...
n
n n n
D F F F F
= + + + = ∑
ACE - Dr Mohamed Elfarran

26. 27
Aircraft loads | 32
Wing structure
• Shear flow in the webs:
• Transverse force building up:
• Normal force in the caps at the location of stringers:
Bending of wing spar
n
n
D
q
h
=
1 2
1
...
n
n n n
D F F F F
= + + + = ∑
1
1
1 m
m n n
n
N D l
h
+
=
= ∑
Next
slide
ACE - Dr Mohamed Elfarran

27. 28
Aircraft loads | 32
Wing structure
• Equilibrium in A:
• Equilibrium in B:
• Normal force increases linearly from outboard to inboard !
Bending of wing spar
3
2
0
x
U U
N N q x
− − = ( )
2 1
2 1
x
U
D D
N l x l
h h
= − +
( )
2 1
2 1
x
U
D D
N l x l
h h
= − +
( )
2
2 2
0
x
U U
N N q l x
− − − =
ACE - Dr Mohamed Elfarran

28. 29
Aircraft loads | 32
Wing structure
• The bending moment at location x in the spar is:
• We see that that for l2
• This is valid on every location x on the spar:
Bending of wing spar
( )
2 2 1 1
x
U x
N h M D l x D l
≡ = − +
2
x
dM
D
dx
= −
x
x
dM
D
dx
= −
ACE - Dr Mohamed Elfarran

29. 30
Aircraft loads | 32
Wing structure
• External forces Fn
• Transverse shear forces known at
every location
• Normal forces in caps known at
every location
Bending of wing spar
1
n
n n
D F
= ∑
M
N
h
=
n
n
D
q
h
=
1
m
n n n
n
M D l
+
= ∑
ACE - Dr Mohamed Elfarran

30. 31
Aircraft loads | 32
• Shear stress in webs
• Normal stress in spar caps
• In a spar hAx is called the moment of resistance (W)
Wing structure
Lift, engine and fuselage weight
x x
x
x x
q D
t ht
τ = =
x x
x
x x
N M
A hA
σ = ± = ±
ACE - Dr Mohamed Elfarran

31. 32
Aircraft loads | 32
• Three cases studied
• Fuselage shell due to pressurization
• Fuselage shells due to applied torsional load
• Wing spars due to applied bending loads
Summary
Loads to stresses
ACE - Dr Mohamed Elfarran