246
7 Torsion of Prismatic Bars
0.25
0.5
t 1
t 3
0.006
t 2 =
0.006
t 4 =
0.012
All Dimensions in metre
Membrane Surface
A
B
C
p
D
Shear Flow
q=
0.01
Fig. 7.11 Hollow aluminium tube
τ 1 =
q
t 1
=
226,000
0.012
= 18.833 × 10
6 N/m
2
τ 2 =
q
t 2
=
226,000
0.006
= 37.667 × 10
6 N/m
2
τ 3 =
226,000
0.01
= 22.6 × 10
6 N/m
2
Now, the angle of twist per unit length is
θ =
q
2G A
ds
t
Therefore,
θ =
226,000
2 × 27.6 × 10 9 × 0.125
0.25
0.012
+
0.5
0.006
(2) +
0.25
0.01
or
θ = 0.00696014 rad/m.
Example 7.2 Figure 7.12 shows a two-cell tubular section as formed by a conventional airfoil shape, and having one interior web. An external torque of 10,000 Nm is
acting in a clockwise direction. Determine the internal shear flow distribution. The
areas of cell-1 and cell-2 are 680 cm
2 and 2000 cm
2 respectively. The peripheral
lengths are indicated in Fig. 7.12.
Solution For Cell 1,
7 Torsion of Prismatic Bars
0.25
0.5
t 1
t 3
0.006
t 2 =
0.006
t 4 =
0.012
All Dimensions in metre
Membrane Surface
A
B
C
p
D
Shear Flow
q=
0.01
Fig. 7.11 Hollow aluminium tube
τ 1 =
q
t 1
=
226,000
0.012
= 18.833 × 10
6 N/m
2
τ 2 =
q
t 2
=
226,000
0.006
= 37.667 × 10
6 N/m
2
τ 3 =
226,000
0.01
= 22.6 × 10
6 N/m
2
Now, the angle of twist per unit length is
θ =
q
2G A
ds
t
Therefore,
θ =
226,000
2 × 27.6 × 10 9 × 0.125
0.25
0.012
+
0.5
0.006
(2) +
0.25
0.01
or
θ = 0.00696014 rad/m.
Example 7.2 Figure 7.12 shows a two-cell tubular section as formed by a conventional airfoil shape, and having one interior web. An external torque of 10,000 Nm is
acting in a clockwise direction. Determine the internal shear flow distribution. The
areas of cell-1 and cell-2 are 680 cm
2 and 2000 cm
2 respectively. The peripheral
lengths are indicated in Fig. 7.12.
Solution For Cell 1,