7.10 Numerical Examples
251
a
.
2a
t
a
(a)
(b)
Fig. 7.14 Thin-walled box section
= πa
2 q 1 + 8a
2
5π + 8
4π + 8
q 1
∴ T =
a
2
π
2
+ 12π + 16
(π + 2)
q 1
∴ q 1 =
(π + 2)T
a 2
π 2 + 12π + 16
Now, from Eq. (7.1), we have,
2Gθ =
2
πat
(π + 2)
(π + 2)T
a 2
π 2 + 12π + 16
− 2
5π + 8
4π + 8
(π + 2)T
a 2
π 2 + 12π + 16
Simplifying, we get the twist as θ =
(2π+3)T
2Ga 3 t(π 2 +12π+16)
.
Example 7.4 A thin-walled box section having dimensions 2a × a × t is to be
compared with a solid circular section of diameter as shown in Fig. 7.14. Determine
the thickness t so that the two sections have:
(a) Same maximum shear stress for the same torque.
(b) The same stiffness.
Solution
(a) For the box section, we have
T = 2q A
= 2.τ.t.A
T = 2.τ.t.2a × a
∴ τ =
T
4a 2 t
(i)
Now, for solid circular section, we have
251
a
.
2a
t
a
(a)
(b)
Fig. 7.14 Thin-walled box section
= πa
2 q 1 + 8a
2
5π + 8
4π + 8
q 1
∴ T =
a
2
π
2
+ 12π + 16
(π + 2)
q 1
∴ q 1 =
(π + 2)T
a 2
π 2 + 12π + 16
Now, from Eq. (7.1), we have,
2Gθ =
2
πat
(π + 2)
(π + 2)T
a 2
π 2 + 12π + 16
− 2
5π + 8
4π + 8
(π + 2)T
a 2
π 2 + 12π + 16
Simplifying, we get the twist as θ =
(2π+3)T
2Ga 3 t(π 2 +12π+16)
.
Example 7.4 A thin-walled box section having dimensions 2a × a × t is to be
compared with a solid circular section of diameter as shown in Fig. 7.14. Determine
the thickness t so that the two sections have:
(a) Same maximum shear stress for the same torque.
(b) The same stiffness.
Solution
(a) For the box section, we have
T = 2q A
= 2.τ.t.A
T = 2.τ.t.2a × a
∴ τ =
T
4a 2 t
(i)
Now, for solid circular section, we have