Page 72 - Aircraft Stuctures for Engineering Student

3.1 Prandtl stress function solution 57

-x

Fig. 3.5 Lines of shear stress.

Consider now the line of constant q5 in Fig. 3.5. If s is the distance measured along
this line from some arbitrary point then
_- -o=--+--
dq5dy dq5 dx
84
dS dy ds dx ds
Using Eqs (3.2) and (3.6) we may rewrite this equation as

(3.14)

From Fig. 3.5 the normal and tangential components of shear stress are
r2, = rZx1 + rzyrn, rzs = rzyl - rzxm (3.15)
Comparing the first of Eqs (3.15) with Eq. (3.14) we see that the normal shear stress is
zero so that the resultant shear stress at any point is tangential to a line of constant 4.
These are known as lines of shear stress or shear lines.
Substituting q5 in the second of Eqs (3.15) we have
dq5
r- =--/--m 84
-s ax ay
which may be written, from Fig. 3.5, as

84
r-= dq5d.x dq5dy =-- (3.16)
-' ax dn dy dn dn
where, in this case, the direction cosines I and rn are defined in terms of an elemental
normal of length Sn.
Thus we have shown that the resultant shear stress at any point is tangential to the
line of shear stress through the point and has a value equal to minus the derivative of
4 in a direction normal to the line.

Example 3.1
To illustrate the stress function method of solution we shall now consider the
torsion of a cylindrical bar having the elliptical cross-section shown in Fig. 3.6.