578 CHAPTER 21 Fuselages
Fig.21.1
(a) Actual fuselage section; (b) idealized fuselage section.
B 11 =B 15 ,B 4 =B 6 =B 12 =B 14 ,andB 5 =B 13 .FromEq.(19.1),
B 1 = 100 +
0.8×149.6
6
(
2 +
σ 2
σ 1
)
+
0.8×149.6
6
(
2 +
σ 16
σ 1
)
thatis,
B 1 = 100 +
0.8×149.6
6
(
2 +
352.0
381.0
)
× 2 =216.6mm^2
Similarly,B 2 =216.6mm^2 ,B 3 =216.6mm^2 ,B 4 =216.7mm^2 .Wenotethatstringers5and13lieon
theneutralaxisofthesectionandarethereforeunstressed;thecalculationofboomareasB 5 andB 13
doesnotthenarise.Forthisparticularsection,Ixy=0,sinceCx(andCy)isanaxisofsymmetry.Further,
My=0sothatEq.(15.18)reducesto
σz=
Mxy
Ixx
inwhich
Ixx= 2 ×216.6×381.0^2 + 4 ×216.6×352.0^2 + 4 ×216.6× 26952
+ 4 ×216.7×145.8^2 =2.52× 108 mm^4
ThesolutioniscompletedinTable21.1.
21.2 Shear...................................................................................................
ForafuselagehavingacrosssectionofthetypeshowninFig.21.1(a),thedeterminationoftheshear
flow distribution in the skin produced by shear is basically the analysis of an idealized single cell