Page 490 - Aircraft Stuctures for Engineering Student

11.5 Constraint of open section beams 471

Therefore, from Eq. (1 1.57)

Now
dY d
sin$=--, -(UR)=PR
ds ds
and the above expression may be integrated by parts, thus

The first term on the right-hand side vanishes as si 2ARt ds is zero at each open edge
of the beam, leaving

y2A~t ds = 0

Again integrating by parts

The integral in the first term on the right-hand side of the above equation may be
recognized, from Chapter 9, as being directly proportional to the shear flow produced
in a singly symmetrical open section beam supporting a shear load Sy. Its value is
therefore zero at each open edge of the beam. Hence

(11.60)

Similarly, for the horizontal component S, to be zero

(1 1.61)

Equations (11.60) and (11.61) hold if the centre of twist coincides with the shear
centre of the cross-section. To summarize, the centre of twist of a section of an
open section beam carrying a pure torque is the shear centre of the section.
We are now in a position to calculate rR. This may be done by evaluating sc 4Ait ds
in which 2AR is given by Eq. (1 1 S6). In general, the calculation may be lengthy unless
the section has flat sides in which case a convenient analogy shortens the work
considerably. For the flat-sided section in Fig. 11.29(a) we first plot the area 2AR:o
swept out from the point 1 where we choose s = 0 (Fig. 11.29(b)). The swept area
AR,O increases linearly from zero at 1 to (1/2)pI2dl2 at 2 and so on. Note that move-
ment along side 23 produces no increment of 2kfR,o as p23 = 0. Further, we adopt a
sign convention for p such that p is positive if movement in the positive s direction
of the foot of p along the tangent causes anticlockwise rotation about R. The
increment of 2AR.0 from side 34 is therefore negative.