Week12 Fuselage Structure

Fuselage Structure

• Similar to aircraft wings, fuselages consist of

thin sheets of material stiffened by large

numbers of longitudinal stringers together with

transverse frames.

• Generally, they carry bending moments, shear forces and torsional loads which induce axial stresses in the stringers and skin together with shear stresses in the skin.

4/6/2013 Aircraft Structural Design 2

Fuselage Structure

• The distance between adjacent stringers is usually small which, combined with the fact that fuselage skin panels are generally thin, means the variation in shear flow in the connecting panel will be small.

• It is therefore 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 stringers/booms carry all the direct stresses while the skin is effective only in shear.


Fuselage Structure

• The direct stress carrying capacity of the skin may be allowed for by increasing the stringer/boom areas as described in the last chapter.

• The analysis of fuselages therefore involves the calculation of direct stresses in the stringers and the shear stress distributions in the skin.


Fuselage Structure

Example # 1

• The fuselage of a light passenger aircraft has the circular cross-section shown in the Figure below. The stringers are equally spaced with cross-sectional areas of 100 mm2 each. If the fuselage is subjected to a downward bending moment of 200 kN.m. applied in the vertical plane of symmetry, at this section, calculate the direct stress distribution.


Fuselage Structure


Solution:

• The section is first idealized using the method described in earlier coverage. As an approximation we shall assume that the skin between adjacent stringers is flat so that we may use either of the two given equations to determine the boom areas. From symmetry B1 =B9, B2 =B8 =B10 =B16, B3 =B7 =B11 =B15, B4 =B6 =B12 =B14 and B5 =B13. Recall the relevant equation:

Fuselage Structure


Fuselage Structure


Fuselage Structure

• Similarly B2 = 216.6 mm2, B3 = 216.6 mm2, B4 =216.7 mm2. We note that stringers 5 and 13 lie on the neutral axis of the section and are therefore unstressed; the calculation of boom areas B5 and B13 is not required. For this particular section Ixy = 0 since Cx (and Cy) is an axis of symmetry. Further, My = 0 so that bending Equation reduces to:


Fuselage Structure


Fuselage Structure Shear

• For a fuselage having an idealized cross-section of the type shown in the previous Figure, the shear flow distribution is given by the following equation in which the direct stress carrying capacity of the skin is assumed to be zero, i.e. tD =0, thus :


Fuselage Structure

• As we know by now, the above Equation is applicable to loading cases in which the shear loads are not applied through the section shear centre so that the effects of shear and torsion are included simultaneously.

• Alternatively, if the position of the shear centre is known (such as in the case of a fully symmetric section), the loading system may be replaced by shear loads acting through the shear centre together with a pure torque, and the corresponding shear flow distributions may be calculated separately and then superimposed to obtain the final distribution.


Fuselage Structure

Example # 2

• Assume the fuselage in the previous Example is subjected to a vertical shear load of 100 kN applied at a distance of 150 mm from the vertical axis of symmetry as shown, for the idealized section, in the Figure below. Calculate the distribution of shear flow in the section.


Fuselage Structure


Fuselage Structure Solution • As in the first Example, Ixy = 0 and, since Sx =0, the shear flow

Equation reduces to:

• And using the Ixx value obtained earlier we get:


Fuselage Structure

• The first term is the ‘open section’ shear flow qb. We therefore ‘cut’ one of the skin panels, say 12, and calculate qb. The results are presented in the Table below.


Fuselage Structure


*Negative sign means flow is actually in the opposite direction

*

Fuselage Structure • The column headed Boom indicates the boom that

is crossed when the analysis moves from one panel to the next.

• qb shear flow distribution is symmetrical about the Cx axis.

• Determine qs,0 by taking moments about a convenient moment centre, say C. Therefore:

Where A=π×381.02 =4.56×105 mm2.


Fuselage Structure

Integral in above Equation represents moments about point C produced by individual shear flows in each wall between two adjacent booms. Since those shear flows are constant their resulting moments are more easily determined by recalling that T = 2Aq (positive ccw). Hence:


Fuselage Structure

100 × 103 × 150= −2𝐴1−2𝑞𝑏,1−2 − 2𝐴2−3𝑞𝑏,2−3− 2𝐴3−4𝑞𝑏,3−4 −⋯2𝐴8−9𝑞𝑏,8−9+ 2𝐴1−16𝑞𝑏,1−16 + 2𝐴16−15𝑞𝑏,16−15+ 2𝐴15−14𝑞𝑏,15−14 +⋯2𝐴10−9𝑞𝑏,10−9+ 2𝐴𝑞𝑠,0

A1-2, A2-3, etc. are areas enclosed between the moment centre (point C in our case) and the two adjacent booms (C-1-2, C-2-3, etc.). Clearly A1-2 =A2-3= · · · =A16- l =4.56×105/16 = 28 500 mm2. Accordingly:


100 × 103 × 150

= 2 × 28500−𝑞𝑏1−2 − 𝑞𝑏2−3 −⋯𝑞𝑏8−9

+𝑞𝑏1−16 + 𝑞𝑏16−15 +⋯𝑞𝑏10−9+ 2 × 4.56 × 105𝑞𝑠,0

Substituting the values of qb from the Table we obtain: 100 × 103 × 150

= 2 × 28500 −262.4 + 2 × 4.56 × 105𝑞𝑠,0

Which gives:

𝑞𝑠,0 = 32.8𝑁

𝑚𝑚(𝐶𝐶𝑊)


Fuselage Structure • Complete shear flow distribution follows by adding

qs,0 to qb, giving final distribution shown below.


Solution may be checked by comparing resultant of shear flow distribution parallel to the Cy axis to the applied load (100 kN):

The small discrepancy is attributed to numerical rounding.


Fuselage Structure Buckling of Curved Fuselage Panels

• The same familiar equation is used to predict the critical buckling stress for curved panels, i.e.,

• Where K this time is found using different charts which give those values as a function of parameter Z given by:

𝑍 = 𝑏2

𝑟𝑡1 − 𝜈2 1 2

• And b is always the shorter side of the panel.


Fuselage Structure Pressurization Stresses

Recall that a pressure vessel subjected to an internal pressure p develops hoop and axial stresses that are given by:

𝜎𝐻 =𝑝𝑟

𝑡

And

𝜎𝐴 =𝑝𝑟

2𝑡


Fuselage Structure • Stringers reduce the axial load in the panels such that the

new formula for axial stresses in the skin panels is:

𝜎𝐴 =𝑝𝑟

2𝑡

1 + 2𝜈𝑘

1 + 𝑘

𝑘 =𝑁𝐴𝑠2𝜋𝑟𝑡

N is the total number of stiffeners, As is cross sectional area of each stringer. The stringers axial stress is:

𝜎𝑠 =𝑝𝑟

2𝑡

1 − 2𝜈

1 + 𝑘

Finally, the axial stress in the panels can also be written as:

𝜎𝐴 =𝑝𝑟

2𝑡− 𝑘𝜎𝑠


Fuselage Structure Quiz: Obtain axial stress σA in the skin panels (t = 1 mm) of a fuselage section with r = 510 mm and 16 equally-spaced stiffeners having a cross sectional area equal to 100 mm2 each, if an internal pressure p equal to 65 kPa is applied. Assume all materials to be aluminum with a Poisson’s ratio ν = 0.30.

𝜎𝐴 =𝑝𝑟

2𝑡

1 + 2𝜈𝑘

1 + 𝑘



Fuselage Structure Solution:

𝜎𝐴 =𝑝𝑟

2𝑡

1 + 2𝜈𝑘

1 + 𝑘

Where k is :


𝑘 =16 ∗ 100 𝑚𝑚2

2𝜋(510 𝑚𝑚)(1𝑚𝑚)= 0.499


Fuselage Structure

𝜎𝐴 =𝑝𝑟

2𝑡

1 + 2𝜈𝑘

1 + 𝑘=

0.065𝑁

𝑚𝑚2 ∗ 510 𝑚𝑚

2 ∗ 1 𝑚𝑚

1 + 2 ∗ 0.3 ∗ 0.499

1 + 0.499

= 14.37 𝑀𝑃𝑎


Fuselage Structure

𝜎𝑠 =𝑝𝑟

2𝑡

1 − 2𝜈

1 + 𝑘=

0.065𝑁

𝑚𝑚2 ∗ 510 𝑚𝑚

2 ∗ 1 𝑚𝑚

1 − 2 ∗ 0.3

1 + 0.499= 4.42 𝑀𝑃𝑎

And σH :

𝜎𝐻 =𝑝𝑟

𝑡=0.065

𝑁𝑚𝑚2 ∗ 510 𝑚𝑚

1 𝑚𝑚= 33.15 𝑁

𝑚𝑚2

𝜎𝐻 = 33.15 𝑀𝑃𝑎


Fuselage Structure Buckling under combined stresses:

Combined Longitudinal Compression and Shear:

𝑅𝑠2 + 𝑅𝐿 = 1

where the R values are ratios of applied to critical stresses (𝑅𝐿 =

𝜎𝑎𝑝𝑝

𝜎𝐶𝑅, 𝑅𝑠 =

𝜏𝑎𝑝𝑝

𝜏𝐶𝑅).

𝑀. 𝑆.= 2

𝑅𝐿𝑅𝐿

2 +4𝑅𝑠

2 − 1


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Week12 Fuselage Structure