256
7 Torsion of Prismatic Bars
∴ A 1 =
π
2
(127)
2
= 25,322 mm
2
A 2 = 254 × 254 = 64,516 mm
2
A 3 = 64,516 mm
2
Now, From the figure,
q 1 = q 2 + q 4
q 2 = q 3 + q 5
q 3 = q 5
or
q 1 = τ 1 t 1 = τ 2 t 2 + τ 4 t 4
q 2 = τ 2 t 2 = τ 3 t 3 + τ 5 t 5
q 3 = τ 3 t 3 = τ 6 t 6
(1)
where τ 1 , τ 2 , τ 3 , τ 4 , τ 5 and τ 6 are the shear stresses in the various walls of the tube.
Now, the applied torque is
M t = 2 A 1 q 1 + 2 A 2 q 2 + 2 A 3 q 3
= 2(A 1 τ 1 t 1 + A 2 τ 2 t 2 + A 3 τ 3 t 3 )
i.e.,
113 × 10 6 = 2[(25,322τ 1 × 0.8) + (64,516τ 2 × 0.8) + (64,516 × 0.8)]
∴ τ 1 + 3.397(τ 2 + τ 3 ) = 3718
(2)
Now, considering the rotations of the cells and S 1 , S 2 , S 3 , S 4 , S 5 and S 6 as the
length of cell walls,
We have,
τ 1 S 1 + τ 4 S 4 = 2Gθ A 1
−τ 4 S 4 + 2τ 2 S 2 + τ 5 S 5 = 2Gθ A 2
−τ 5 S 5 + 2τ 3 S 3 + τ 6 S 6 = 2Gθ A 3
(3)
Here
S 1 = (π × 127) = 398 mm
S 2 = S 3 = S 4 = S 5 = S 6 = 254 mm
7 Torsion of Prismatic Bars
∴ A 1 =
π
2
(127)
2
= 25,322 mm
2
A 2 = 254 × 254 = 64,516 mm
2
A 3 = 64,516 mm
2
Now, From the figure,
q 1 = q 2 + q 4
q 2 = q 3 + q 5
q 3 = q 5
or
q 1 = τ 1 t 1 = τ 2 t 2 + τ 4 t 4
q 2 = τ 2 t 2 = τ 3 t 3 + τ 5 t 5
q 3 = τ 3 t 3 = τ 6 t 6
(1)
where τ 1 , τ 2 , τ 3 , τ 4 , τ 5 and τ 6 are the shear stresses in the various walls of the tube.
Now, the applied torque is
M t = 2 A 1 q 1 + 2 A 2 q 2 + 2 A 3 q 3
= 2(A 1 τ 1 t 1 + A 2 τ 2 t 2 + A 3 τ 3 t 3 )
i.e.,
113 × 10 6 = 2[(25,322τ 1 × 0.8) + (64,516τ 2 × 0.8) + (64,516 × 0.8)]
∴ τ 1 + 3.397(τ 2 + τ 3 ) = 3718
(2)
Now, considering the rotations of the cells and S 1 , S 2 , S 3 , S 4 , S 5 and S 6 as the
length of cell walls,
We have,
τ 1 S 1 + τ 4 S 4 = 2Gθ A 1
−τ 4 S 4 + 2τ 2 S 2 + τ 5 S 5 = 2Gθ A 2
−τ 5 S 5 + 2τ 3 S 3 + τ 6 S 6 = 2Gθ A 3
(3)
Here
S 1 = (π × 127) = 398 mm
S 2 = S 3 = S 4 = S 5 = S 6 = 254 mm