Page 412 - Aircraft Stuctures for Engineering Student

10.3 Wings 393

M, = 1.65 kNm at the larger cross-section; the shear load is applied in the plane of
the internal spar web. If booms 1 and 6 lie in a plane which is parallel to the yz
plane calculate the forces in the booms and the shear flow distribution in the walls
at the larger cross-section. The booms are assumed to resist all the direct stresses
while the walls are effective only in shear. The shear modulus is constant throughout,
the vertical webs are all l.0mm thick while the remaining walls are all 0.8 mm thick.

~oom areas: B~ = B~ = B~ = B6 = 600mm2, B2 = B~ = 9OOmm'

At the larger cross-section
I,, = 4 x 600 x 902 + 2 x 900 x 902 = 34.02 x 106mm4

The direct stress in a boom is given by Eq. (9.6) in which Ixv = 0 and MJ = 0, i.e.

whence

or
1.65 x 106y B
p,,, = 34.02 x lo6 = 0.08yrB,.

The value of P,,, is calculated from Eq. (i) in column 0 Table 10.5; P.y,r and Py.,
of
The
follow from Eqs (10.10) and (10.9) respectively in columns @ and 0. axial load
+
in
P, is given by [a2 O2 + a2I1l2 column 0 and has the same sign as PZ,, (see Eq.
(10.12)). The moments of P,,r and P,,,, columns @ and 0, calculated for a
are
moment centre at the mid-point of the internal web taking anticlockwise moments
as positive.
From column @

(as would be expected from symmetry).

Table 10.5

1 2619.0 0 0.0417 0 109.2 2621.3 400 90 0 43680
2 3928.6 0.0833 0.0417 327.3 163.8 3945.6 0 90 -29457 0
3 2619.0 0.1250 0.0417 327.4 109.2 2641.6 200 90 -29466 21840
4 -2619.0 0.1250 -0.0417 -327.4 109.2 -2641.6 200 90 -29466 21840
5 -3928.6 0.0833 -0.0417 -327.3 163.8 -3945.6 0 90 -29457 0
6 -2619.0 0 -0.0417 0 109.2 -2621.3 400 90 0 -43680