7.9 Torsion of Thin-Walled Multiple-Cell Closed Sections
245
To Find the Twist (θ)
For continuity, the twist of each cell should be the same.
We have
θ =
q
2 AG
dS
t
or
Gθ =
1
A
qdS
t
Let
a 1 =
dS
t
for Cell 1 including the web
a 2 =
dS
t
for Cell 2 including the web
a 12 =
dS
t
for the web only
Then for Cell 1
2Gθ =
1
A 1
(a 1 q 1 − a 12 q 2 )
(7.29)
For Cell 2
2Gθ =
1
A 2
(a 2 q 2 − a 12 q 1 )
(7.30)
Equations (7.28), (7.29) and (7.30) are sufficient to solve for q 1 , q 2 and θ .
7.10 Numerical Examples
Example 7.1 A hollow aluminium tube of rectangular cross-section shown in
Fig. 7.11 is subjected to a torque of 56,500 mN along its longitudinal axis. Determine
the shearing stresses and the angle of twist. Assume G = 27.6 × 10
9 N/m
2 .
Solution The above figure shows the membrane surface ABCD
Now, the applied torque = M t = 2q A
56,500 = 29(0.5 × 0.25)
56,500 = 0.25q
hence, q = 226,000 N/ m.
Now, the shearing stresses are