17.1 Torsion of Closed Section Beams 513TheshearstrainsareobtainedfromEq.(17.1)andare
γa=T
2 abGta, γb=T
2 abGtbfromwhich
θ=TL
4 a^2 b^2 G(
a
ta+
b
tb)
Thetotalangleoftwistfromendtoendofthebeamis2θ,therefore,2 θ
L=
TL
4 a^2 b^2 G(
2 a
ta+
2 b
tb)
or
dθ
dz=
T
4 A^2 G
∮
ds
tasinEq.(17.4).
Substitutingforθineitheroftheexpressionsfortheaxialdisplacementofthecorner1givesthe
warpingw 1 at1.Thus,
w 1 =a
2b
LTL
4 a^2 b^2 G(
a
ta+
b
tb)
−
T
2 abGtaa
2thatis,
w 1 =T
8 abG(
b
tb−
a
ta)
as before. It can be seen that the warping of the cross section is produced by a combination of the
displacements caused by twisting and the displacements due to the shear strains; these shear strains
correspondtotheshearstresseswhosevaluesarefixedbystatics.Theangleoftwistmustthereforebe
suchastoensurecompatibilityofdisplacementbetweenthewebsandcovers.
17.1.2 Condition for Zero Warping at a Section
Thegeometryofthecrosssectionofaclosedsectionbeamsubjectedtotorsionmaybesuchthatno
warpingofthecrosssectionoccurs.FromEq.(17.5),weseethatthisconditionariseswhen
δOs
δ=
AOs
Aor
1
δ∫s0ds
Gt=
1
2 A
∫s0pRds (17.6)