7.10 Numerical Examples
255
∴ q 1 = 1.52 × 165.02 = 250.83N
Shear flow in the web = q 3 = (q 1 − q 2 ) = (250.83 − 165.02)
∴ q 3 = 85.81N
∴ τ 1 =
q 1
t 1
=
250.83
5
= 50.17 N/mm
2
τ 2 =
q 2
t 2
=
165.02
2.5
= 66.01 N/mm
2
τ 3 =
q 3
t 3
=
85.81
2.5
= 34.32 N/mm
2
Now, the twist θ is computed by substituting the values of q 1 and q 2 in equation
(a).
i.e.,
2Gθ =
1
15,000
[130 × 250.83 × 60 × 165.02]
∴ θ =
1
15,000
×
22,706.7
83 × 1000
= 1.824 × 10
−5 rad/mm length
or
θ = 1.04
◦
/m length
Example 7.6 A tubular section having three cells as shown in Fig. 7.16 is subjected
to a torque of 113 kNm. Determine the shear stresses developed in the walls of the
section.
Solution Let q 1 , q 2 , q 3 , q 4 , q 5 , q 6 be the shear flows in the various walls of the tube
as shown in the figure. A 1 , A 2 , and A 3 be the areas of the three cells.
Fig. 7.16 Three cell with tubular section
255
∴ q 1 = 1.52 × 165.02 = 250.83N
Shear flow in the web = q 3 = (q 1 − q 2 ) = (250.83 − 165.02)
∴ q 3 = 85.81N
∴ τ 1 =
q 1
t 1
=
250.83
5
= 50.17 N/mm
2
τ 2 =
q 2
t 2
=
165.02
2.5
= 66.01 N/mm
2
τ 3 =
q 3
t 3
=
85.81
2.5
= 34.32 N/mm
2
Now, the twist θ is computed by substituting the values of q 1 and q 2 in equation
(a).
i.e.,
2Gθ =
1
15,000
[130 × 250.83 × 60 × 165.02]
∴ θ =
1
15,000
×
22,706.7
83 × 1000
= 1.824 × 10
−5 rad/mm length
or
θ = 1.04
◦
/m length
Example 7.6 A tubular section having three cells as shown in Fig. 7.16 is subjected
to a torque of 113 kNm. Determine the shear stresses developed in the walls of the
section.
Solution Let q 1 , q 2 , q 3 , q 4 , q 5 , q 6 be the shear flows in the various walls of the tube
as shown in the figure. A 1 , A 2 , and A 3 be the areas of the three cells.
Fig. 7.16 Three cell with tubular section