Page 70 - Aircraft Stuctures for Engineering Student

3.1 Prandtl stress function solution 55

Rewriting this equation in terms of the stress function 4

Integrating each term on the right-hand side of this equation by parts, and noting
again that 4 = 0 at all points on the boundary, we have

We are therefore in a position to obtain an exact solution to a torsion problem if a
stress function 4(x, y) can be found which satisfies Eq. (3.4) at all points within the
bar and vanishes on the surface of the bar, and providing that the external torques
are distributed over the ends of the bar in an identical manner to the distribution
of internal stress over the cross-section. Although the last proviso is generally
impracticable we know from St. Venant’s principle that only stresses in the end
regions are affected; therefore, the solution is applicable to sections at distances
from the ends usually taken to be greater than the largest cross-sectional dimension.
We have now satisfied all the conditions of the problem without the use of stresses
other than rzy and T,,, demonstrating that our original assumptions were justified.
Usually, in addition to the stress distribution in the bar, we require to know the
angle of twist and the warping displacement of the cross-section. First, however,
we shall investigate the mode of displacement of the cross-section. We have seen
that as a result of our assumed values of stress
&x = Ey = E, = TXY = 0
It follows, from Eqs (1.18) and the second of Eqs (1.20), that
- --=-=-+-=o
au - av aw av au
ax ay az ax ay
which result leads to the conclusions that each cross-section rotates as a rigid body
in its own plane about a centre of rotation or twist, and that although cross-
sections suffer warping displacements normal to their planes the values of this
displacement at points having the same coordinates along the length of the bar are
equal. Each longitudinal fibre of the bar therefore remains unstrained, as we have
in fact assumed.
Let us suppose that a cross-section of the bar rotates through a small angle 6 about
its centre of twist assumed coincident with the origin of the axes Oxy (see Fig. 3.4).
Some point P(r, a) will be displaced to P’(r, a + e), the components of its displace-
ment being
u = -resina, v = &osa
or
u= -ey, v= ex (3.9)
Referring to Eqs (1.20) and (1.46)