## Ii

j<7

z--

Fig. 11.12 Top cover of the beam of Fig. 11.10.

central boom need be considered as long as the symmetry is allowed for in the assumed directions of the panel shear flows q, as shown in Fig. 11.12. Further, the origin for z may be taken to be at either the free or built-in end. A marginally simpler solution is obtained if the origin is taken to be at the free end, in which case the solution represents that for an infinitely long panel. Considering the equilibrium of an element of an edge boom (Fig. 11.13), in which we assume that the boom load is positive (tension) and increases with increasing z, we have az

Similarly, for an element of the central boom (Fig. 11.14)

8Pa dz

Now considering the overall equilibrium of a length z of the cover (Fig. 11.15), we have

S Jib

Fig. 11.13 Equilibrium of boom element.

Fig. 11.16 Compatibility condition.

agrees with the assumed directions of the shear flows in Fig. 11.12 and that the shear strain increases with z. From Fig. 11.16(b)

(l+eB)Sz = (l+eA)Sz + d^Sz in which eb and eA are the direct strains in the elements of boom. Thus, rearranging and noting that 7 is a function of z only when the section is completely idealized, we have

We now select the unknown to be determined initially. Generally, it is simpler mathematically 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. (11.28) for q from Eq. (11.25) and for PB from Eq. (11.26). Hence

2 dz2

Rearranging, we obtain

GtSyz dEBh or d2PA

GtSv

in which A2 = Gt(2B + A)/dEAB. The solution of Eq. (11.29) is of standard form and is

The constants C and D are determined from the boundary conditions of the cover of the beam namely, PA = 0 when z = 0 and 7 — q/Gt = -{dPA/dz)/2Gt = 0 when z = L (see Eq. (11.25)). From the first of these C = 0 and from the second

Thus

St.A

The direct stress distribution crA(= Pa/A) follows, i.e.

Sy ( sinhAz

The distribution of load in the edge booms is obtained by substituting for PA from Eq. (11.30) in Eq. (11.26), thus

A sinhAz \

whence

h{2B +A) V" ' 2BX coshAL J Finally, from either pairs of Eqs (11.25) and (11.30) or (11.24) and (11.32)

0 0