7.10 Numerical Examples
253
Fig. 7.15 Two-cell tube
Equating (iii) and (iv), we get
6aT
16a 4 Gt
=
32T
Gπa 4
6a
16t
=
32
π
∴ t =
6πa
16 × 32
∴ t =
3
4
πa
64
Example 7.5 A two-cell tube as shown in Fig. 7.15 is subjected to a torque of 10
kNm. Determine the shear stress in each part and angle of twist per metre length.
Take modulus of rigidity of the material as 83 kN/mm
2 .
Solution For Cell 1.
Area of the Cell = A 1 = 150 × 100 = 15,000 mm
2
a 1 =
ds
t
(including web)
=
150
5
+
100
5
+
150
2.5
+
100
5
= 130
For Cell 2
Area of the cell = A 2 =
1
2
× 150 ×
(125)
2
− (75)
2
= 7500 mm
2
253
Fig. 7.15 Two-cell tube
Equating (iii) and (iv), we get
6aT
16a 4 Gt
=
32T
Gπa 4
6a
16t
=
32
π
∴ t =
6πa
16 × 32
∴ t =
3
4
πa
64
Example 7.5 A two-cell tube as shown in Fig. 7.15 is subjected to a torque of 10
kNm. Determine the shear stress in each part and angle of twist per metre length.
Take modulus of rigidity of the material as 83 kN/mm
2 .
Solution For Cell 1.
Area of the Cell = A 1 = 150 × 100 = 15,000 mm
2
a 1 =
ds
t
(including web)
=
150
5
+
100
5
+
150
2.5
+
100
5
= 130
For Cell 2
Area of the cell = A 2 =
1
2
× 150 ×
(125)
2
− (75)
2
= 7500 mm
2