17.1 Torsion of Closed Section Beams 513
TheshearstrainsareobtainedfromEq.(17.1)andare
γa=
T
2 abGta
, γb=
T
2 abGtb
fromwhich
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
t
asinEq.(17.4).
Substitutingforθineitheroftheexpressionsfortheaxialdisplacementofthecorner1givesthe
warpingw 1 at1.Thus,
w 1 =
a
2
b
L
TL
4 a^2 b^2 G
(
a
ta
+
b
tb
)
−
T
2 abGta
a
2
thatis,
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
A
or
1
δ
∫s
0
ds
Gt
=
1
2 A
∫s
0
pRds (17.6)