Introduction to Aircraft Structural Analysis (Elsevier Aerospace Engineering)

518 CHAPTER 17 Torsion of Beams

Fig.17.12

Channel section of Example 17.3.

sothat

wO2=− 2 ×

1

2

×8.04s 1 ×

10 × 103

25000 ×316.7

=−0.01s 1 (i)

thatis,thewarpingdistributionislinearinO2and

w 2 =−0.01× 25 =−0.25mm

Inthewall21

AR=

1

2

×8.04× 25 −

1

2

× 25 s 2

inwhichtheareasweptoutbythegeneratorinthewall21providesanegativecontributiontothetotal
sweptareaAR.Thus,

w 21 =− 25 (8.04−s 2 )

10 × 103

25000 ×316.7

or

w 21 =−0.03(8.04−s 2 ) (ii)

Again,thewarpingdistributionislinearandvariesfrom−0.25mmat2to+0.54mmat1.Examination
of Eq. (ii) shows thatw 21 changes sign ats 2 =8.04mm. The remaining warping distribution follows
fromsymmetry,andthecompletedistributionisshowninFig.17.13.Inunsymmetricalsectionbeams,
the position of the point of zero warping is not known but may be found using the method for the