Page 354 - Aircraft Stuctures for Engineering Student

9.9 Effect of idealization 335

S

4.8 kN

1200 mm

Fig. 9.51 Idealized channel section of Example 9.13.
At the outside of boom 1, qs = 0. As boom 1 is crossed the shear flow changes by an
amount given by

Aql = x 300 x 200 = -6N/mm
Hence qI2 = -6N/mm since, from Eq. (i), it can be seen that no further changes in
shear flow occur until the next boom (2) is crossed. Hence
q23 = -6 - lop4 x 300 x 200 = -12N/mm
Similarly

q34 = -12 - lop4 x 300 x (-200) = -6N/mm
while, finally, at the outside of boom 4 the shear flow is

-6 - x 300 x (-200) = 0
as expected. The complete shear flow distribution is shown in Fig. 9.52.

12 N/mm

Fig. 9.52 Shear flow in channel section of Example 9.1 3.