Aircraft Structures

Chapter 22- Fuselages

2

Fuselage Structures

3

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.

4

Bending Moments

Shear Forces

Torsion

Stresses in an Aircraft Fuselage

5

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 y

Remember

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:

6

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.

7

Example 22.1 (Cont.)

Boom Areas

Moment of Inertia (I)

Calculate the stresses

How to solve?

8

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.

9

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?

10

Example 22.1 (Cont.)

Calculate the stresses

How to solve?

11

22.2 Shear

Remember

this from

chapter 20!

We will use this equation to find the shear flow distribution.

Where =0, hence:

12

Example 22.2

The 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.

13

Example 22.2 (Cont.)

Taking moments about some center

Finding the shear stress distribution

How to solve?

14

Shear flow distribution

Example 22.2 (Cont.)How to solve?

Open section shear flow qb

15

Taking moments about some center

Example 22.2 (Cont.)How to solve?

Remember

this from

chapter 17!

16

Taking moments about some center

Example 22.2 (Cont.)How to solve?

A= Π x 381.02= 4.56 x105 mm2

17

Finding the shear stress distribution

Example 22.2 (Cont.)How to solve?

18

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:

19

22.3 Torsion

From symmetry we get:

20

22.3 Torsion

21

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.

24

22.4 Cut-Outs in FuselagesTo find the effect of cut-outs for windows:

The Trip is Over!

Hope you have enjoyed

it 25