# A hollow shaft of 1m length is designed to transmit a power of 30 kW at 700 rpm. The maximum permissible angle of twist in the shaft is 1°. The inner

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A hollow shaft of 1m length is designed to transmit a power of 30 kW at 700 rpm. The maximum permissible angle of twist in the shaft is 1°. The inner diameter of the shaft is 0.7 times the outer diameter. The modulus of rigidity is 80 GPa. The outside diameter (in mm) of the shaft is _______

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Concept:

The power transmitted by the shaft is given by,

$P = T.\omega = \frac{{2\pi NT}}{{60}}$

Torsion equation,

$\frac{T}{{{J}}} = \frac{\tau }{{{r}}}=\frac{{Gθ }}{l}\;$

Calculation:

Given:

P = 30 kW, N = 700 rpm, θ = 1° , G = 80 GPa and di = 0.7 do

from  $P = T.\omega = \frac{{2\pi NT}}{{60}}$

$30 \times 1000 = T \times \frac{{2π \times 700}}{{60}}$

∴ T = 409.256 N – m

From Torsion equation,

$\frac{T}{{{J}}} = \frac{{Gθ }}{l}\;$

L = 1 m, θ = 1° = π/180 rad

$\frac{{409.256}}{{\frac{π }{{32}}\left( {1 - {{0.7}^4}} \right)d_0^4}} = \frac{{80 \times {{10}^9}}}{1} \times \frac{π }{{180}}$

on Solving, we get

d0 = 44.52 mm