Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

612 CHAPTER 22 Wings Fig.22.25 Correction shear flows in bay③of the wing box of Example 22.7. Forequilibriuminthechordwisedirect ...

Problems 613 Fig.22.26 Final shear flows in bay③(and bay①) of the wing box of Example 22.7. Fig.22.27 Shear flows (N/mm) applied ...

614 CHAPTER 22 Wings Ans. σ 1 =−190.7N/mm^2 σ 2 =−181.7N/mm^2 σ 3 =−172.8N/mm^2 σ 4 =−163.8N/mm^2 σ 5 =140N/mm^2 σ 6 =164.8N/mm^ ...

Problems 615 Fig. P.22.3 Fig. P.22.4 P.22.4 Determine the shear flow distribution for a torque of 56500Nm for the three-cell sec ...

616 CHAPTER 22 Wings Fig. P.22.5 P.22.6 FigureP.22.6showsasinglysymmetric,two-cellwingsectioninwhichalldirectstressesarecarriedb ...

Problems 617 Wall Length (mm) Thickness (mm) Boom Area (mm^2 ) Cell Area (mm^2 ) 12, 56 510 0.559 1, 6 645 I 93000 23, 45 765 0. ...

618 CHAPTER 22 Wings Ans. P 1 =−P 6 =1200N P 2 =−P 5 =2424N P 3 =−P 4 =2462N q 12 =q 56 =3.74N/mm q 23 =q 45 =3.11N/mm q 34 o=0. ...

CHAPTER 23 Fuselage Frames and Wing Ribs....................................................... Aircraftareconstructedprimarilyf ...

620 CHAPTER 23 Fuselage Frames and Wing Ribs Fig.23.1 Cantilever beam of Example 23.1. Fig.23.2 Free body diagrams of stiffeners ...

23.1 Principles of Stiffener/Web Construction 621 Fig.23.3 Equilibrium of stiffener CJG in the beam of Example 23.1. Fromtheequi ...

622 CHAPTER 23 Fuselage Frames and Wing Ribs fromwhich q 4 =23.3N/mm Alternatively,fromtheequilibriumofstiffenerBF, 300 q 4 − 30 ...

23.1 Principles of Stiffener/Web Construction 623 Fig.23.4 Load distributions in flanges of the beam of Example 23.1. Fig.23.5 L ...

624 CHAPTER 23 Fuselage Frames and Wing Ribs Sofarwehavebeenconcernedwithweb/stiffenerarrangementsinwhichtheloadshavebeenapplied ...

23.2 Fuselage Frames 625 23.2 FuselageFrames.................................................................................... ...

626 CHAPTER 23 Fuselage Frames and Wing Ribs Themethodofdeterminingtheshearflowdistributionappliedtotheperipheryofafuselageframe ...

23.3 Wing Ribs 627 Resolvingvertically 300 q 31 − 300 q 23 =15000N (ii) Takingmomentsaboutflange3, 2 ( 50000 + 95000 )q 23 + 2 × ...

628 CHAPTER 23 Fuselage Frames and Wing Ribs Fig.23.11 Equilibrium of rib forward of intermediate stiffener 56. Thus,theshearfor ...

23.3 Wing Ribs 629 Fig.23.12 Equilibrium of stiffener 56. Fig.23.13 Equilibrium of the rib forward of stiffener 31. Finally,wesh ...

630 CHAPTER 23 Fuselage Frames and Wing Ribs Hence, P 1 =P 3 = √ 13533.3^2 +3626.2^2 =14010.7N Thetotalshearforceatthissectionis ...

Problems 631 Fig. P.23.2 P.23.3 Athree-flangewingsectionisstiffenedbythewingribshowninFig.P.23.3.Iftheribflangesandstiffeners ca ...