Page 472 - Aircraft Stuctures for Engineering Student

11.3 Thin-walled rectangular section beam 453

where
2 8Gtbt,
= BE(bt, + atb)
The differential equation (1 1.15) is of standard form and its solution is

T b a
w= Ccoshpz+Dsinhpz+- --- (11.16)
8abG (tb to)
in which the last term is seen to be the free warping displacement wo of the top right-
hand corner boom. The constants C and D in Eq. (1 1.16) are found from the bound-
ary conditions of the beam. In this particular case the warping w = 0 at the built-in
end and the direct strain dw/dz = 0 at the free end where there is no direct load.
From the first of these

and from the second

D = wo tanhpL
Thus
w = wo( 1 - cosh pz + tanh pL sinh pz) ( 1 I. 17)
or rearranging
1
cash p(L - Z) (11.18)
cosh pL
The variation of direct stress in the boom is obtained from a, = Edw/dz and Eq.
(11.18), i.e.
sinh p(L - z)
aZ = ~EwO (11.19)
cosh pL
and the variation of boom load P is then
sinhp(L - z)
P = BCT, B~EwO (1 1.20)
cosh pL
Substituting for w in Eqs (1 1.12) and rearranging, we obtain the shear stress distribu-
tion in the covers and webs. Thus
qa
7 --=- T (bt, - atb) coshp(L - Z) 1 (11.21)
a- ta 2abt, [' + (bt, + atb) cosh pL

(1 1.22)

Inspection of Eqs (1 1.21) and (1 1.22) shows that the shear stress distributions each
comprise two parts. The first terms, T/2abta and T/2abtb, are the shear stresses
predicted by elementary theory (see Section 9.9, while the hyperbolic second terms
represent the effects of the warping restraint. Clearly, for an anticlockwise torque
and bt, > atb, the effect of this constraint is to increase the shear stress in the