582 CHAPTER 21 Fuselages
Fig.21.3
Shear flow (N/mm) distribution in fuselage section of Example 21.2.
Itisimmaterialwhetherornotthesectionhasbeenidealized,since,inbothcases,theboomsareassumed
nottocarryshearstresses.
Equation(21.2)providesanalternativeapproachtothatillustratedinExample21.2forthesolution
of shear loaded sections in which the position of the shear center is known. In Fig. 21.1, the shear
centercoincideswiththecenterofsymmetrysothattheloadingsystemmaybereplacedbytheshear
loadof100kNactingthroughtheshearcentertogetherwithapuretorqueequalto100× 103 × 150 =
15 × 106 NmmasshowninFig.21.4.Theshearflowdistributionduetotheshearloadmaybefound
using the method of Example 21.2 but with the left-hand side of the moment equation (iii) equal to
zeroformomentsaboutthecenterofsymmetry.Alternatively,usemaybemadeofthesymmetryof
thesectionandthefactthattheshearflowisconstantbetweenadjacentbooms.Supposethattheshear
flowinthepanel21isq 21 .Then,fromsymmetryandusingtheresultsofTable21.2,
q 98 =q 910 =q 161 =q 21
q 32 =q 87 =q 1011 =q 1516 =30.3+q 21
q 43 =q 76 =q 1112 =q 1415 =53.5+q 21
q 54 =q 65 =q 1213 =q 1314 =66.0+q 21Theresultantoftheseshearflowsisstaticallyequivalenttotheappliedshearloadsothat
4 (29.0q 21 +82.5q 32 +123.7q 43 +145.8q 54 )= 100 × 103