17.2 Torsion of Open Section Beams 519Fig.17.13
Warping distribution in channel section of Example 17.3.
restrainedwarpingofanopensectionbeam.Thus,wecanseethat
2 A′R=∫
sect∫^2 AR,Otds
secttds(17.21)
in whichAR,Ois the area swept out by a generator rotating about the center of twist from some
convenientorigin,andA′RisthevalueofAR,Oatthepointofzerowarping.Asanillustration,weshall
applythemethodtothebeamsectionofExample17.3.
Supposethatthepositionofthecenteroftwist(i.e.,theshearcenter)hasalreadybeencalculated,
andsupposealsothatwechoosetheoriginforstobeatthepoint1.Then,inFig.17.14,
∫
secttds= 2 ×1.5× 25 +2.5× 50 =200mm^2Inthewall12,
A 12 =1
2
× 25 s 1 (AR,Oforthewall12) (i)fromwhich
A 2 =1
2
× 25 × 25 =312.5mm^2Also,
A 23 =312.5−1
2
×8.04s 2 (ii)