Page 322 - Aircraft Stuctures for Engineering Student

9.4 Shear of closed section beams 303

9.4.1 Twist and warping of shear loaded closed section beams

Shear loads which are not applied through the shear centre of a closed section beam
cause cross-sections to twist and warp; that is, in addition to rotation, they suffer out
of plane axial displacements. Expressions for these quantities may be derived in terms
of the shear flow distribution qs as follows. Since q = rt and r = (see Chapter 1)
then we can express qs in terms of the warping and tangential displacements ti' and ut
of a point in the beam wall by using Eq. (9.26). Thus

(9.39)

Substituting for aV,/dz from Eq. (9.30) we have
dw
de du
dv
- = - +p- + -cos + + -sin+ (9.40)
qs
Gt ds dz dz dz
Integrating Eq. (9.40) with respect to s from the chosen origin for s and noting that G
may also be a function of s, we obtain
dwp
-ds= -&+- de pds+- dtr cos+&+- dv sin$ds
1; tt jo ds dzlo dzjo dz lo
or

which gives
d6'
du
I$&= (W,-W~)+~AO~-+-((~,-XO)+~;~~-~O)
(9.41)
dv
dz dz
where Aos is the area swept out by a generator, centre at the origin of axes, 0, froin
the origin for s to any point s around the cross-section. Continuing the integration
completely around the cross-section yields, from Eq. (9.41)
fgds = 2A- d6'
dz
from which

(9.42)

Substituting for the rate of twist in Eq. (9.41) from Eq. (9.42) and rearranging, we
obtain the warping distribution around the cross-section
du dv
- - - - (x, - xO) - -bs -yo) (9.43)
-ds
AT f zt dz dz
Using Eqs (9.31) to replace du/dz and dv/dz in Eq. (9.43) we have