250
7 Torsion of Prismatic Bars
2Gθ =
1
A 2
(a 2 q 2 − a 12 q 1 )
=
1
4a 2
8a
t
q 2 −
2a
t
q 1
=
2a
4a 2 t
[4q 2 − q 1 ]
∴ 2Gθ =
1
2at
[4q 2 − q 1 ]
(b)
Equating (1) and (2), we get,
2
πat
[(π + 2)q 1 − 2q 2 ] =
1
2at
[4q 2 − q 1 ]
or
2
π
[(π + 2)q 1 − 2q 2 ] =
1
2
[4q 2 − q 1 ]
4
π
[(π + 2)q 1 − 2q 2 ] = [4q 2 − q 1 ]
∴
4(π + 2)
π
q 1 −
8
π
q 2 − 4q 2 + q 1 = 0
4(π + 2)
π
+ 1
q 1 −
8
π
+ 4
q 2 = 0
4(π + 2) + π
π
q 1 −
8 + 4π
π
q 2 = 0
or
(4π + 8 + π )q 1 = (8 + 4π )q 2
∴ q 2 =
5π + 8
4π + 8
q 1
But the torque due to shear flows should be equal to the applied torque.
i.e.
T = 2q 1 A 1 + 2q 2 A 2
(c)
Substituting the values of q 2 , A 1 and A 2 in (c), we get,
T = 2q 1
πa
2
2
+ 2
5π + 8
4π + 8
q 1 .4a
2