Page 267 - Aircraft Stuctures for Engineering Student

248 Airworthiness and airframe loads

Hence
L
cL=---- 4.5 x 8000 = 1.113
N
ipV2S i x 1.223 x 602 x 14.5
From Fig. 8.10(a), a = 13.75" and CM,cG = 0.075. The tail arm I, from Fig. 8.10(b), is
1 =4.18cos(a-2)+0.31sin(a-2) (ii)
Substituting the above value of a gives 1 = 4.123m. In Eq. (8.14) the terms
La - Db - Mo are equivalent to the aircraft pitching moment MCG about its centre
of gravity. Thus, Eq. (8.14) may be written
MCG - P1 0
or
PI = ipv2sccMM,CG (iii)
where c = wing mean chord. Substituting P from Eq. (iii) into Eq. (8.12) we have

=nW

or dividing through by 4pV2S

We now obtain a more accurate value for CL from Eq. (iv)
1.35
CL = 1.113 -- x 0.075 = 1.088
4.123
giving a = 13.3" and CM,cG = 0.073.
Substituting this value of a into Eq. (ii) gives a second approximation for I, namely
1 = 4.161 m.
Equation (iv) now gives a third approximation for CL, i.e. CL = 1.099. Since the
three calculated values of CL are all extremely close further approximations will
not give values of CL very much different to those above. Therefore, we shall take
CL = 1.099. From Fig. 8.10(a) CD = 0.0875.
The values of lift, tail load, drag and forward inertia force then follow:
Lift L = ipV2SCL = 4 x 1.223 x 602 x 14.5 x 1.099 = 35000N

Tailload P=nW-L=4.5~8000-35000= l000N
Drag D = ipV2SCD = i x 1.223 x 602 x 14.5 x 0.0875 = 2790N
Forward inertia force fW = D (from Eq. (8.13)) = 2790 N

In Section 8.4 we determined aircraft loads corresponding to a given manoeuvre load
factor n. Clearly it is necessary to relate this load factor to given types of manoeuvre.