Page 493 - Aircraft Stuctures for Engineering Student

474 Structural constraint

rR follows from the moment of inertia of the ‘wire’ about an axis through its centre of
gravity. Hence

which simplifies to

td3h2 2h+d
rR =- 12 (-) h+2d (ii)

Equation (1 1.59), that is
de d3 8
T=GJ--EE~RT
dz dz
may now be solved for dO/dz. Rearranging and writing p2 = GJ/ErR we have
d38 2d8 T
-- z=-p (iii)
dz3
The solution of Eq. (iii) is of standard form, i.e.
de T
-
--- + Acoshpz + Bsinhpz
dz GJ
The constants A and B are found from the boundary conditions.
(1) At the built-in end the warping w = 0 and since w = -2ARd8/dz then
dO/dz = 0 at the built-in end.
(2) At the free end gr = 0, as there is no constraint and no externally applied direct
load. Therefore, from Eq. (1 1.54), d’O/d2 = 0 at the free end.
From (1)
A = -T/GJ

From (2)
B = (T/GJ) tanh pL
so that
d0 T
(1
- = - - coshpz + tanhpLsinh pz)
dz GJ
or
cosh pL 1
dz GJ
The first term in Eq. (iv) is seen to be the rate of twist derived from the St. Venant torsion
theory. The hyperbolic second term is therefore the modification introduced by the
axial constraint. Equation (iv) may be integrated to find the distribution of angle of
twist 8, the appropriate boundary condition being 8 = 0 at the built-in end. Thus
sinh p(L - z) - sinh pL 1 (4
GJ p cosh pL p cosh pL