Page 359 - Aircraft Stuctures for Engineering Student

340 Open and closed, thin-walled beams

37.

Fig. 9.56 Shear flow distribution Wmm in walls of the beam section of Example 9.14.

q34 = -34.3N/mm = q56
q45 = -37.9N/=
and
qgl = 17.0N/~

giving the shear flow distribution shown in Fig. 9.56.

9.9.4 Torsion of open and closed section beams

No direct stresses are developed in either open or closed section beams subjected to a
pure torque unless axial constraints are present. The shear stress distribution is there-
fore unaffected by the presence of booms and the analyses presented in Sections 9.5
and 9.6 apply.

9.9.5 Alternative method for the calculation of shear
flow distribution

Equation (9.73) may be rewritten in the form
apr
q2-41 =-& (9.81)
in which P, is the direct load in the rth boom. This form of the equation suggests an
alternative approach to the determination of the effect of booms on the calculation of
shear flow distributions in open and closed section beams.
Let us suppose that the boom load varies linearly with z. This will be the case for a
length of beam over which the shear force is constant. Equation (9.81) then becomes
42 - ql = -APr (9.82)
in which APr is the change in boom load over unit length of the rth boom. AP, may be
calculated by first determining the change in bending moment between two sections of
a beam a unit distance apart and then calculating the corresponding change in boom
stress using either of Eqs (9.6) or (9.7); the change in boom load follows by multiply-
ing the change in boom stress by the boom area B,. Note that the section properties
contained in Eqs (9.6) and (9.7) refer to the direct stress carrying area of the beam