Page 478 - Aircraft Stuctures for Engineering Student

11.4 Shear lag 459

agrees with the assumed directions of the shear flows in Fig. 11.12 and that the shear
strain increases with z. From Fig. 1 1.16(b)

in which and are the direct strains in the elements of boom. Thus, rearranging and
noting that y is a function of z only when the section is completely idealized, we have

(1 1.27)
Now

PB
EB = - &A=-, PA 7=- 4
BE ’ AE Gt
so that Eq. (11.27) becomes

(11.28)

We now select the unknown to be determined initially. Generally, it is simpler math-
ematically to determine either of the boom load distributions, PB or PA, rather than
the shear flow q. Thus, choosing PA, say, as the unknown, we substitute in Eq. (1 1.28)
for q from Eq. (1 1.25) and for PB from Eq. (1 1.26). Hence
1 #PA - Gt ( PA
PA)
S,Z
- - - - - - - - -
-
-
2 dz2 dE 2B 2Bh A
Rearranging, we obtain
d2PA Gt(2B + A) GtS,, z
--
a9 dEAB P --
A - dEBh
or
a2pA
-- ~2p - GtS,z (11.29)
a22 A-- dEBh
in which X2 = Gt(2B + A)/dEAB. The solution of Eq. (11.29) is of standard form
and is
S,A
PA =CcoshXz+DsinhXz-
h(2B + A)
The constants C and D are determined from the boundary conditions of the cover of
the beam namely, PA = 0 when z = 0 and y = q/Gt = -(aPA/&)/2Gt = 0 when
z = L (see Eq. (1 1.25)). From the iirst of these C = 0 and from the second
S,, A
D=
Xh(2B + A) cosh XL
Thus
(.- sinh Xz
PA = - (1 1.30)
h(2B + A) X cosh XL