254
7 Torsion of Prismatic Bars
∴ a 2 =
ds
t
(including web)
=
150
2.5
+
125
2.5
+
125
2.5
∴ a 2 = 160
For the web,
a 12 =
150
2.5
= 60
For Cell (1)
2Gθ =
1
A 1
(a 1 q 1 − a 12 q 2 )
∴ 2Gθ =
1
15,000
(130q 1 − 60q 2 )
(a)
For Cell (2)
2Gθ =
1
A 2
(a 2 q 2 − a 12 q 1 )
=
1
7500
(160q 2 − 60q 1 )
(b)
Equating (a) and (b), we get
1
15,000
(130q 1 − 60q 2 ) =
1
7500
(160q 2 − 60q 1 )
Solving,
q 1 = 1.52q 2
(c)
Now, the torque due to shear flows should be equal to the applied torque.
i.e.,
M t = 2q 1 A 1 + 2q 2 A 2
10 × 10
6
= 2q 1 (15,000) + 2q 2 (7500)
( d)
Substituting (c) in (d), we get
10 × 10
6
= 2 × 15,000(1.52q 2 ) + 2q 2 (7500)
∴ q 2 = 165.02N
7 Torsion of Prismatic Bars
∴ a 2 =
ds
t
(including web)
=
150
2.5
+
125
2.5
+
125
2.5
∴ a 2 = 160
For the web,
a 12 =
150
2.5
= 60
For Cell (1)
2Gθ =
1
A 1
(a 1 q 1 − a 12 q 2 )
∴ 2Gθ =
1
15,000
(130q 1 − 60q 2 )
(a)
For Cell (2)
2Gθ =
1
A 2
(a 2 q 2 − a 12 q 1 )
=
1
7500
(160q 2 − 60q 1 )
(b)
Equating (a) and (b), we get
1
15,000
(130q 1 − 60q 2 ) =
1
7500
(160q 2 − 60q 1 )
Solving,
q 1 = 1.52q 2
(c)
Now, the torque due to shear flows should be equal to the applied torque.
i.e.,
M t = 2q 1 A 1 + 2q 2 A 2
10 × 10
6
= 2q 1 (15,000) + 2q 2 (7500)
( d)
Substituting (c) in (d), we get
10 × 10
6
= 2 × 15,000(1.52q 2 ) + 2q 2 (7500)
∴ q 2 = 165.02N