## 1

By comparison with the pure torsion case we deduce that dfl_ 1 dz 2ArG

in which qb has previously been determined. There are N equations of the type (10.28) so that a further equation is required to solve for the N + 1 unknowns. This is obtained by considering the moment equilibrium of the i?th cell in Fig. 10.24.

The moment Mq R produced by the total shear flow about any convenient moment centre O is given by

Fig. 10.24 Moment equilibrium of ffih cell.

Substituting for qR in terms of the 'open section' shear flow qb and the redundant shear flow qs.o,R, we have

MgiR = qbPo di + 2ARqsfiiR

The sum of the moments from the individual cells is equivalent to the moment of the externally applied loads about the same point. Thus, for the wing section of Fig. 10.22

SxT]o ~ Syt„ = Y^ Mq,R = IbPo às + 2ARlsfi,R (10-29)

If the moment centre is chosen to coincide with the point of intersection of the lines of action of Sx and Sy, Eq. (10.29) becomes

Example 10.8

The wing section of Example 10.6 (Fig. 10.17) carries a vertically upward shear load of 86.8 kN in the plane of the web 572. The section has been idealized such that the booms resist all the direct stresses while the walls are effective only in shear. If the shear modulus of all walls is 27 600 N/mm2 except for the wall 78 for which it is three times this value, calculate the shear flow distribution in the section and the rate of twist. Additional data are given below.

Wall |
Length (mm) |
Thickness (mm) |
Cell area (mm2) |

12, 56 |

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