aircraft structures
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Aircraft Structures. Chapter 22- Fuselages. Fuselage Structures. Stresses in an Aircraft Fuselage. Aircraft fuselages consist of thin sheets of material stiffened by large numbers of longitudinal stringers together with transverse frames. - PowerPoint PPT PresentationTRANSCRIPT
Aircraft Structures
Chapter 22- Fuselages
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Fuselage Structures
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Stresses in an Aircraft Fuselage1. Aircraft fuselages consist of thin sheets of material stiffened by large
numbers of longitudinal stringers together with transverse frames.
2. The distance between stringers is usually small, so that the variation in
shear flow in the connection panel is small. Therefore, it is reasonable to
assume that the shear flow is constant between adjacent stringers, so that
the analysis simplifies to the analysis of an idealized section in which the
stingers/booms carry all the direct stresses while the skin is effective only in
shear.
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Bending Moments
Shear Forces
Torsion
Stresses in an Aircraft Fuselage
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22.1 Bending
σ z=M y I xx−M x I xyI xx I yy− I xy
2 x+M x I yy−M y I xyI xx I yy− I xy
2 yRemember
this from chapter 16!
We will use this equation in this chapter to calculate the direct stress in each boom.
B1=tD b
6(2+
σ 2
σ 1)
Remember this from
chapter 20!
To find the broom area:
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Example 22.1The fuselage of a light passenger carrying aircraft has the circular cross-section shown below. The cross-sectional area of each stringer is 100 mm2and the vertical distance given in the figure are to the mid-line of the sectional wall at the corresponding stringer position. If the fuselage is subjected to a bending moment of 200 kN.m applied in the vertical plane of symmetry, at this section, calculate the direct stress distribution.
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Example 22.1 (Cont.)
Boom Areas
Moment of Inertia (I)
Calculate the stresses
How to solve?
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Example 22.1 (Cont.)How to solve?
From Symmetry: B1 =B9 B2 =B8 =B10 =B16 B3 =B7 =B11 =B15 B4 =B6 =B12 =B14 B5 =B13
Boom Areas
B1=tD b
6(2+
σ 2
σ 1)
The stringers 5 and 13 lie on the neutral axis of the section and are therefore unstressed, the calculation of boom areas B5 and B13 does not then rise.
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Example 22.1 (Cont.)
Ixx =2×216.6×381.02+4×216.6×352.02+4×216.6×269.52+4×216.7×145.82
Moment of Inertia (I)
=2.52×108mm4
How to solve?
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Example 22.1 (Cont.)
Calculate the stresses
How to solve?
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22.2 Shear
Remember this from
chapter 20!
We will use this equation to find the shear flow distribution.
Where =0, hence:
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Example 22.2The fuselage of Example 21.1 is subjected to a vertical shear load of 100 kN applied at a distance of 150mm from the vertical axis of symmetry as shown, for the idealized section, in Fig.22.2. Calculate the distribution of shear flow in the section.
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Example 22.2 (Cont.)
Shear flow distribution
Taking moments about some center
Finding the shear stress distribution
How to solve?
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Shear flow distribution
Example 22.2 (Cont.)How to solve?
Open section shear flow qb
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Taking moments about some center
Example 22.2 (Cont.)How to solve?
Remember this from
chapter 17!
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Taking moments about some center
Example 22.2 (Cont.)How to solve?
A= Π x 381.02= 4.56 x105 mm2
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Finding the shear stress distribution
Example 22.2 (Cont.)How to solve?
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22.3 Torsion
Remember this from
chapter 18!
We will use this equation to find the shear stress distribution produce by a pure torque.
Lets apply it to example 22.2:
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22.3 Torsion
From symmetry we get:
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22.3 Torsion
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22.4 Cut-Outs in FuselagesLoads are redistributed in the vicinity of the cut-off.
Reinforcement
Increased Weight Rigid Frames
22.4 Cut-Outs in Fuselages1. In practice , it is necessary to provide openings in these closed stiffened
shells for, for example, doors, cockpits, bomb bays and windows in
passenger cabins.
2. These openings or “cut-outs” produce discontinuities in the otherwise
continuous shell structure, so that loads are redistributed in the vicinity of
the cut-out, thereby affecting loads in the skin, stringers and frames.
22.4 Cut-Outs in Fuselages3. Frequently, these regions must be heavily reinforced, resulting in
unavoidable weight increases. In some cases, door openings in passenger
aircraft, it is not possible to provide rigid fuselage frames on each side of the
opening, because the cabin space must not be restricted. In such situations, a
rigid frame is placed around the opening to resist shear loads and to transmit
loads from one side of the opening to the other.
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22.4 Cut-Outs in FuselagesTo find the effect of cut-outs for windows:
The Trip is Over!
Hope you have enjoyed it
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