Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

13.1 Aircraft Inertia Loads 381

elementδminadirectionperpendiculartorandintheoppositesensetotheangularacceleration.This
inertiaforcehascomponentsαrδmcosθandαrδmsinθ,i.e.αxδmandαyδm,intheyandxdirections,
respectively.Thus,theresultantinertiaforces,FxandFy,aregivenby

Fx=

∫

αydm=α

∫

ydm

and

Fy=−

∫

αxdm=−α

∫

xdm

forαinthedirectionshown.Then,asbefore

Fx=α ̄ym (13.3)

and

Fy=α ̄xm (13.4)

Also,iftheCGliesonthexaxis, ̄y=0andFx=0. Similarly, if the CG lies on theyaxis, ̄x=0and
Fy=0.
The torque about the axis of rotation produced by the inertia force corresponding to the angular
accelerationontheelementδmisgivenby

δTO=αr^2 δm

Thus,forthecompletemass

TO=

∫

αr^2 dm=α

∫

r^2 dm

Theintegralterminthisexpressionisthemomentofinertia,IO,ofthemassabouttheaxisofrotation.
Thus,

TO=αIO (13.5)

Equation (13.5) may be rewritten in terms ofICG, the moment of inertia of the mass about an axis
perpendiculartotheplaneofthemassthroughtheCG.Hence,usingtheparallelaxestheorem

IO=m( ̄r)^2 +ICG

wherer ̄isthedistancebetweenOandtheCG.Then

IO=m[(x ̄)^2 +( ̄y)^2 ]+ICG

and

TO=m[(x ̄)^2 +( ̄y)^2 ]α+ICGα (13.6)