13.1 Aircraft Inertia Loads 383
LetNandSbetheaxialandshearloadsatthesectionAA,asshowninFig.13.4.Theinertiaload
actingattheCGofthefuselageaftofAAism 1 a,wherem 1 isthemassofthefuselageaftofAA.Then,
m 1 a=
4.5
g
3 g=13.5kN
Resolvingforcesparalleltotheaxisofthefuselage
N−T+m 1 acos10◦−4.5sin10◦= 0
thatis,
N−137.1+13.5cos10◦−4.5sin10◦= 0
fromwhich
N=124.6kN
Nowresolvingforcesperpendiculartotheaxisofthefuselage
S−m 1 asin10◦−4.5cos10◦= 0
thatis,
S−13.5sin10◦−4.5cos10◦= 0
sothat
S=6.8kN
Notethat,inadditiontotheaxialloadandshearloadatthesectionAA,therewillalsobeabending
moment.
Finally,fromelementarydynamics
v^2 =v^20 + 2 as
Fig.13.4
Shear and axial loads at the section AA of the aircraft of Example 13.1.