315 Power Applied and Removed at Shaft Gears

Problem 315
A 5-m steel shaft rotating at 2 Hz has 70 kW applied at a gear that is 2 m from the left end where 20 kW are removed. At the right end, 30 kW are removed and another 20 kW leaves the shaft at 1.5 m from the right end. (a) Find the uniform shaft diameter so that the shearing stress will not exceed 60 MPa. (b) If a uniform shaft diameter of 100 mm is specified, determine the angle by which one end of the shaft lags behind the other end. Use G = 83 GPa.

Solution 315

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$T = \dfrac{P}{2\pi f}$

$T_A = T_C = \dfrac{-20(1000)}{2\pi (2)} = -1591.55 \, \text{N}\cdot\text{m}$

$T_B = \dfrac{70(1000)}{2\pi (2)} = 5570.42 \, \text{N}\cdot\text{m}$

$T_D = \dfrac{-30(1000)}{2\pi (2)} = -2387.32 \, \text{N}\cdot\text{m}$

Part (a)
$\tau_{max} = \dfrac{16T}{\pi d^3}$

For AB
$60 = \dfrac{16(1591.55)(1000)}{\pi d^3}$

$d = 51.3 \, \text{mm}$

For BC
$60 = \dfrac{16(3978.87)(1000)}{\pi d^3}$

$d = 69.6 \, \text{mm}$

For CD
$60 = \dfrac{16(2387.32)(1000)}{\pi d^3}$

$d = 58.7 \, \text{mm}$

Use d = 69.6 mm answer

Part (b)
$\theta = \dfrac{TL}{JG}$

$\theta_{D/A} = \dfrac{1}{JG} \Sigma TL$

$\theta_{D/A} = \dfrac{1}{\frac{1}{32}\pi (100^4)(83\,000)} \, [ \, -1591.55(2) + 3978.87(1.5) + 2387.32(1.5) \, ] \, (1000^2)$

$\theta_{D/A} = 0.007\,813 \, \text{rad}$

$\theta_{D/A} = 0.448^\circ$ answer

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