Torsion of thin, open sections
shear stress distribution
τ max
torque,T
a
a
a
a
b
τ max =
J eff
t
(t << b)
Tt
J eff
1
= bt 3
3
T = GJ eff φ ′
twist/ unit length
Torsion of a Thin Closed Section
(single cell)
Ω = area contained within the
centerline of the cross section
T = v∫ qr⊥ ds
C
= q v∫ r⊥ ds = 2qΩ
q=
T
2Ω
(1)
C
q
r⊥
T
P
φ′ =
=
1
q
∫C t ds
2GΩ v
T
4GΩ 2
1. If T is known, q follows directly from Eq. (1),
φ' is found from Eq.(2)
2. If φ' is known, T follows from Eq.(2),
and q is then found from Eq. (1)
ds
vC∫ t
T = GJ eff φ ′
where J eff
(2)
4Ω 2
=
ds
vC∫ t
Torsion of a Thin Closed Section
(single cell)
The shear stress is not quite uniform across the thickness for thin
closed sections
yields uniform stress
t
The difference looks much like that for an open section
t
so as a small correction factor:
J eff
4Ω 2 1 3
=
+ ∫ t ( s ) ds
ds 3 v
vC∫ t C
Torsion of a Thin Closed Section
(multiple cells)
T = 2Ω1q1 + 2Ω 2 q2
cell 2
cell 1
q1
q2
(1)
φ′ =
1
q
ds
v
∫
2GΩ1 C1 t
(2)
φ′ =
1
2GΩ 2
(3)
q
vC∫ t ds
2
( the q in Eqs.(2) and (3) is the total q flowing in
a given cross section, i.e it is q1 – q2 flowing
in the vertical section)
1. If the torque T is known, then q1 and q2 are first found in terms
of the unknown φ' from Eqs. (2) and (3). These qm 's are then
placed into Eq.(1) which is solved for the unknown φ' . Once φ'
is known in this manner, the qm 's are completely determined.
2. If φ' is known, Eqs.(2) and ( 3) can be solved directly for the
qm 's and then Eq.(1) can be used to find the torque, T