580 CHAPTER 21 Fuselages
Fig.21.2
Idealized fuselage section of Example 21.2.
Thefirsttermontheright-handsideofEq.(ii)isthe“opensection”shearflowqb.Wetherefore“cut”
oneoftheskinpanels,say12,andcalculateqb.TheresultsarepresentedinTable21.2.
Note that in Table 21.2, the column headed Boom indicates the boom that is crossed when the
analysis moves from one panel to the next. Note also that, as would be expected, theqbshear flow
distributionissymmetricalabouttheCxaxis.Theshearflowqs,0inthepanel12isnowfoundbytaking
momentsaboutaconvenientmomentcenter,sayC.Therefore,fromEq.(16.17),
100 × 103 × 150 =
∮
qbpds+ 2 Aqs,0 (iii)inwhichA=π×381.0^2 =4.56× 105 mm^2 .Sincetheqbshearflowsareconstantbetweenthebooms,
Eq.(iii)mayberewrittenintheform(seeEq.(19.10))
100 × 103 × 150 =− 2 A 12 qb,12− 2 A 23 qb,23−···− 2 A 161 qb,16l+ 2 Aqs,0 (iv)inwhichA 12 ,A 23 ,...,A 161 aretheareassubtendedbytheskinpanels12,23,...,16latthecenterC
ofthecircularcrosssectionandcounterclockwisemomentsaretakenaspositive.ClearlyA 12 =A 23 =
··· =A 161 =4.56× 105 / 16 =28500mm^2 .Equation(iv)thenbecomes
100 × 103 × 150 = 2 × 28500 (−qb 12 −qb 23 −···−qb16l)+ 2 ×4.56× 105 qs,0 (v)SubstitutingthevaluesofqbfromTable21.2inEq.(v),weobtain
100 × 103 × 150 = 2 × 28500 (−262.4)+ 2 ×4.56× 105 qs,0