Page 69 - Aircraft Stuctures for Engineering Student

54 Torsion of solid sections
On the ends of the bar the direction cosines of the normal to the surface have the
values I = 0, m = 0 and n = 1. The related boundary conditions, from Eqs (1.7), are
then
x = rz.y
-
Y = rzy
Z=O

We now observe that the forces on each end of the bar are shear forces which are
distributed over the ends of the bar in the same manner as the shear stresses are
distributed over the cross-section. The resultant shear force in the positive direction
of the x axis, which we shall call S,, is then

S, = /IXdxdy = jjrzxdxdy

or, using the relationship of Eqs (3.2)

S, = 1 $ dx dy = j dx I$ dy = 0

as q5 = 0 at the boundary. In a similar manner, Sy, the resultant shear force in they
direction, is

It follows that there is no resultant shear force on the ends of the bar and the forces
represent a torque of magnitude, referring to Fig. 3.3

in which we take the sign of T as being positive in the anticlockwise sense.

Fig. 3.3 Derivation of torque on cross-section of bar.