Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

390 CHAPTER 13 Airframe Loads

Asafirstapproximation,weneglectthetailloadP.Therefore,fromEq.(13.12),sinceT=0,we
have

L≈nW (i)

Hence

CL=

L

1

2 ρV

(^2) S≈

4.5× 8000

1

2 ×1.223×^60

(^2) ×14.5=1.113
FromFig.13.8(a),α=13.75◦andCM,CG=0.075.Thetailarml,fromFig.13.8(b),is
l=4.18cos(α− 2 )+0.31sin(α− 2 ) (ii)
Substitutingtheabovevalueofαgivesl=4.123m.InEq.(13.14)thetermsLa−Db−M 0 areequivalent
totheaircraftpitchingmomentMCGaboutitsCG.Equation(13.14)maythereforebewritten
MCG−Pl= 0
or
Pl=

1

2

ρV^2 ScCM,CG (iii)

wherec=wingmeanchord.SubstitutingPfromEq.(iii)intoEq.(13.12)wehave

L+

1

2 ρV

(^2) ScCM,CG
l
=nW
ordividingthroughby^12 ρV^2 S

CL+

c

l

CM,CG=

nW

1

2 ρV

(^2) S (iv)
WenowobtainamoreaccuratevalueforCLfromEq.(iv)
CL=1.113−

1.35

4.123

×0.075=1.088

givingα=13.3◦andCM,CG=0.073.SubstitutingthisvalueofαintoEq.(ii)givesasecondapproxi-
mationforl,namelyl=4.161m.
Equation(iv)nowgivesathirdapproximationforCL:CL=1.099.Sincethethreecalculatedvalues
ofCLareallextremelyclose,furtherapproximationswillnotgivevaluesofCLverymuchdifferent
fromthoseabove.Therefore,weshalltakeCL=1.099.FromFig.13.8(a),CD=0.0875.
Thevaluesoflift,tailload,drag,andforwardinertiaforcethenfollow:

LiftL=

1

2

ρV^2 SCL=

1

2

×1.223× 602 ×14.5×1.099=35000N

TailloadP=nW−L=4.5× 8000 − 35000 =1000N

DragD=

1

2

ρV^2 SCD=

1

2

×1.223× 602 ×14.5×0.0875=2790N

ForwardinertiaforcefW=D(FromEq.(13.13))=2790N