Page 67 - Aircraft Stuctures for Engineering Student

52 Torsion of solid sections

and that the torque is resisted solely by shear stresses in the plane of the cross-section
giving
rXy = 0
To verify these assumptions we must show that the remaining stresses satisfy the
conditions of equilibrium and compatibility at all points throughout the bar and,
in addition, fulfil the equilibrium boundary conditions at all points on the surface
of the bar.
If we ignore body forces the equations of equilibrium, (lS), reduce, as a result of
our assumptions, to

The first two equations of Eqs (3.1) show that the shear stresses T,~ and T~~ are functions
of x and y only. They are therefore constant at all points along the length of the bar
which have the same x and y coordinates. At this stage we turn to the stress function
to simplify the process of solution. Prandtl introduced a stress function 4 defined by

which identically satisfies the third of the equilibrium equations (3.1) whatever form
4 may take. We therefore have to find the possible forms of 4 which satisfy the
compatibility equations and the boundary conditions, the latter being, in fact, the
requirement that distinguishes one torsion problem from another.
From the assumed state of stress in the bar we deduce that

E, = = E, = yxy = 0 (see Eqs (1.42) and (1.46))
Further, since T~~ and ry2 and hence yxz and yyz are functions of x and y only then the
compatibility equations (1.21)-( 1.23) are identically satisfied as is Eq. (1.26). The
remaining compatibility equations, (1.24) and (1.25), are then reduced to
a
- ( -- arYr I =O
ax ax ay
-(--%) a?, =o
a
aryz
ay ax
Substituting initially for yyz and rxz from Eqs (1.46) and then for T~~(= ryz) and
r,,(= r,.) from Eqs (3.2) gbes

or