Page 500 - Aircraft Stuctures for Engineering Student

11.5 Constraint of open section beams 481

Substituting for CT? from Eq. (1 1.69) we have

d2 u d2v
Mx =fi(z) IC tyds - 2ARtyds - EGIC txydS - .-Ic d? ty’ds
We have seen in the derivation of Eqs (11.60) and (11.61) that Jc2ARtyds = 0. Also
since

tyrds=O! jcfxyds=Ixy, jcty2ds=I,
d2u d2v
M, = -E- I - ED (11.71)
I,,
dz2 x”
Similarly
d2u d2v
c2tx& = -Es I& - E- I . (1 1.72)
dz2 x’
Equations (1 1.71) and (11.72) are identical to Eqs (9.19) so that from Eqs (9.17)

E- d2u = MxIxs -M y2 I xx, E- d2v = -My Iyy + My IxJ (1 1.73)
dz2 Ixx Iyy - I,, dz2 Ixx Iyy - I$

The first differential, d2$/dz2, of the rate of twist in Eq. (1 1.69) may be isolated by
multiplying throughout by 2ARt and integrating around the section. Thus

=fi(z)
lc~z2ARtds

As before
I
jc 2ARt ds = 0, jc 2ARtx ds = 2ARty ds = 0

and

or
d2B ( 1 1.74)
-=-Ic 0,2A~t dS
dz2 ErR
Substituting in Eq. (1 1.69) from Eqs (1 1.70), (1 1.73) and (1 1.74), we obtain