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Fuselage Structure
➒Given fuselage structure and determine
the ultimate bending strength.
➒ Given loads and determine the
maximum stresses applied to the
fuselage structure.
➒ Given loads and determine the shear
flow distribution.
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
The figure shows a fuselage cross-
section. The stringers are arranged
symmetrically w.r.t. the fuselage z-axis.
The skin and stringer were made from
aluminum 2024. The stringer Hight is 1
in. while there area is 0.12 𝑖𝑛2.
Calculate the maximum stresses due to a
design bending moment 1.6 E6 Ib.in.?
Determine the fuselage Margin of safety
if the Stringer-skin allowable strength is
32000 psi?
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution strategy
β€’ The given problem is a trail and error problem, because the
neutral axis position depends on the applied compressive
stresses (unknown), and the applied stresses depends on the
fuselage effective area (unknown) which is also depends on
the stringer-skin stresses.
β€’ Initially, an effective width should be assumed either as a
factor of the skin thickness π‘Šπ‘’π‘“π‘“ = 30 βˆ— π‘‘π‘ π‘˜π‘–π‘› or by
assuming linear stress distribution 𝑀𝑒𝑓𝑓 = 1.9𝑑
𝐸
πœŽπ‘™π‘–π‘›π‘’π‘Žπ‘Ÿ
β€’ Due to the fuselage symmetry about the z-axis only one-half
of the fuselage will be considered in the present analyses.
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure – Trial-1
β€’ Calculate total areas of stringer and
effective skin. All sheets in tension
side are assumed to be effective.
𝐴 = 𝐴𝑠𝑑 + 30 βˆ— π‘‘π‘ π‘˜π‘–π‘› βˆ— π‘‘π‘ π‘˜π‘–π‘›
Area under compression:
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure – Trial-1
β€’ Stringer and skin initial position
Skin centroid position
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure – Trial-1 Buckled skin contribution
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure – Trial-1
Buckling stress
Buckled skin contribution
Z
Buckling
coefficient
r/t 𝐾𝑐
343.75 55.86 19.6
343.75 44.23 16
750 20.27 9
1187.5 12.80 6
1187.5 28.62 12
1187.5 33.35 14
𝒓 =
π’“π’”π’•π’‚π’•π’Šπ’π’1 + π’“π’”π’•π’‚π’•π’Šπ’π’2
2
, 𝒁 =
𝒃2
𝒓 βˆ— 𝒕
1 βˆ’ 𝒗2 ,
πˆπ’„π’“ =
π’Œπ’„π…2
𝑬
ሻ
12(1 βˆ’ 𝒗2
𝒕
𝒃
2
Z can be obtained
from Figure C9.1
This formula
can be used
for Poisson’s
ration 0.3
ACE - Dr Mohamed Elfarran
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure –Trial-1
Effective area
Buckled skin contribution
β€’ Calculate the effective factor
based on assumed stresses.
β€’ Then calculate the effective area
for the buckled skin.
In our analysis, we will not assume any
stresses, instead in the initial trail we will
assume all the skins as in tension.
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure – Trial-1 All together
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure –
Trial-1
Skin without buckling
Buckled skin
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure –
Comments
β€’ The importance of trial – 1 is to determine the skin in
tension and compression which depends on the neutral
axis position which also depends on the skin effective
width.
β€’ In the initial trail we will assume all the stringers and
skins are effective.
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure –
Trial-2 Skin without buckling
Buckled skin
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure –Trial-2 All together
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure –Comments
β€’ The results of trail – 1, give the neutral axis position 3.38” below the fuselage
center line, and a second moment of area 1470 𝑖𝑛4
.
β€’ The results of trail – 2, give the neutral axis position 0.135” above the location
from the first trail, and a second moment of area 1489 𝑖𝑛4
.
β€’ The error between trail – 1 and trail – 2 can be calculated to be
β€’ Error based on the centroid position πœ–% =
𝑍2 βˆ’ 𝑍1
𝑍2
π‘₯100 =
27.6βˆ’24.2
24.2
π‘₯100 = 14%
ACE - Dr Mohamed Elfarran
Ultimate bending strength - Example
Solution Procedure –Trial-2 All together
ACE - Dr Mohamed Elfarran

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Ace 402 lec 8 max stress

  • 1.
  • 2. Fuselage Structure ➒Given fuselage structure and determine the ultimate bending strength. ➒ Given loads and determine the maximum stresses applied to the fuselage structure. ➒ Given loads and determine the shear flow distribution. ACE - Dr Mohamed Elfarran
  • 3. Ultimate bending strength - Example The figure shows a fuselage cross- section. The stringers are arranged symmetrically w.r.t. the fuselage z-axis. The skin and stringer were made from aluminum 2024. The stringer Hight is 1 in. while there area is 0.12 𝑖𝑛2. Calculate the maximum stresses due to a design bending moment 1.6 E6 Ib.in.? Determine the fuselage Margin of safety if the Stringer-skin allowable strength is 32000 psi? ACE - Dr Mohamed Elfarran
  • 4. Ultimate bending strength - Example Solution strategy β€’ The given problem is a trail and error problem, because the neutral axis position depends on the applied compressive stresses (unknown), and the applied stresses depends on the fuselage effective area (unknown) which is also depends on the stringer-skin stresses. β€’ Initially, an effective width should be assumed either as a factor of the skin thickness π‘Šπ‘’π‘“π‘“ = 30 βˆ— π‘‘π‘ π‘˜π‘–π‘› or by assuming linear stress distribution 𝑀𝑒𝑓𝑓 = 1.9𝑑 𝐸 πœŽπ‘™π‘–π‘›π‘’π‘Žπ‘Ÿ β€’ Due to the fuselage symmetry about the z-axis only one-half of the fuselage will be considered in the present analyses. ACE - Dr Mohamed Elfarran
  • 5. Ultimate bending strength - Example Solution Procedure – Trial-1 β€’ Calculate total areas of stringer and effective skin. All sheets in tension side are assumed to be effective. 𝐴 = 𝐴𝑠𝑑 + 30 βˆ— π‘‘π‘ π‘˜π‘–π‘› βˆ— π‘‘π‘ π‘˜π‘–π‘› Area under compression: ACE - Dr Mohamed Elfarran
  • 6. Ultimate bending strength - Example Solution Procedure – Trial-1 β€’ Stringer and skin initial position Skin centroid position ACE - Dr Mohamed Elfarran
  • 7. Ultimate bending strength - Example Solution Procedure – Trial-1 Buckled skin contribution ACE - Dr Mohamed Elfarran
  • 8. Ultimate bending strength - Example Solution Procedure – Trial-1 Buckling stress Buckled skin contribution Z Buckling coefficient r/t 𝐾𝑐 343.75 55.86 19.6 343.75 44.23 16 750 20.27 9 1187.5 12.80 6 1187.5 28.62 12 1187.5 33.35 14 𝒓 = π’“π’”π’•π’‚π’•π’Šπ’π’1 + π’“π’”π’•π’‚π’•π’Šπ’π’2 2 , 𝒁 = 𝒃2 𝒓 βˆ— 𝒕 1 βˆ’ 𝒗2 , πˆπ’„π’“ = π’Œπ’„π…2 𝑬 ሻ 12(1 βˆ’ 𝒗2 𝒕 𝒃 2 Z can be obtained from Figure C9.1 This formula can be used for Poisson’s ration 0.3 ACE - Dr Mohamed Elfarran
  • 9. ACE - Dr Mohamed Elfarran
  • 10. Ultimate bending strength - Example Solution Procedure –Trial-1 Effective area Buckled skin contribution β€’ Calculate the effective factor based on assumed stresses. β€’ Then calculate the effective area for the buckled skin. In our analysis, we will not assume any stresses, instead in the initial trail we will assume all the skins as in tension. ACE - Dr Mohamed Elfarran
  • 11. Ultimate bending strength - Example Solution Procedure – Trial-1 All together ACE - Dr Mohamed Elfarran
  • 12. Ultimate bending strength - Example Solution Procedure – Trial-1 Skin without buckling Buckled skin ACE - Dr Mohamed Elfarran
  • 13. Ultimate bending strength - Example Solution Procedure – Comments β€’ The importance of trial – 1 is to determine the skin in tension and compression which depends on the neutral axis position which also depends on the skin effective width. β€’ In the initial trail we will assume all the stringers and skins are effective. ACE - Dr Mohamed Elfarran
  • 14. Ultimate bending strength - Example Solution Procedure – Trial-2 Skin without buckling Buckled skin ACE - Dr Mohamed Elfarran
  • 15. Ultimate bending strength - Example Solution Procedure –Trial-2 All together ACE - Dr Mohamed Elfarran
  • 16. Ultimate bending strength - Example Solution Procedure –Comments β€’ The results of trail – 1, give the neutral axis position 3.38” below the fuselage center line, and a second moment of area 1470 𝑖𝑛4 . β€’ The results of trail – 2, give the neutral axis position 0.135” above the location from the first trail, and a second moment of area 1489 𝑖𝑛4 . β€’ The error between trail – 1 and trail – 2 can be calculated to be β€’ Error based on the centroid position πœ–% = 𝑍2 βˆ’ 𝑍1 𝑍2 π‘₯100 = 27.6βˆ’24.2 24.2 π‘₯100 = 14% ACE - Dr Mohamed Elfarran
  • 17. Ultimate bending strength - Example Solution Procedure –Trial-2 All together ACE - Dr Mohamed Elfarran