Question

The turbine develops 300 kW of power, which is transmitted to the gears such that both B and C receive an equal amount. If the rotation of the 100-mm-diameter A992 steel shaft is v = 600 rev>min., determine the absolute maximum shear stress in the shaft and the rotation of end D of the shaft relative to A. The journal bearing at D allows the shaft to turn freely about its axis.

Answer 1

Answer:

2387350 Nm

0.929 degrees

Explanation:

w = 2*pi*f = 2*pi*600/60 = 62.831 rad/s

Given data:

P = 300*10^3 W ; G = 75 GPa ; d = 0.1m ; r = 0.05m

$J = \frac{pi * c^4}{2} = \frac{pi * r^4}{2} = \frac{pi * 0.05^4}{2} = 9.8175*10^(-6) m^4\\\\T = \frac{P}{w} = \frac{300,000}{62.831} = 4774710 Nm\\\\TC = TB = 0.5*T = 2387350 Nm$

Tmax is the torque at any point of part AB and is equal to T generated at point A, and from B to D the power and torque will be removed by gears.

$t_{max} = \frac{T*r}{J} = \frac{4774710*0.05}{9.8175*10^(-6)}=24.3MPa\\\\Q_{D/A} = sum (\frac{T*L}{J*G}) = \frac{TAB*LAB}{J*G} + \frac{TBC*LBC}{J*G} \\\\= \frac{4774710*1.5}{(9.8175*10^(-6)) * 79 * 10^9} + \frac{2387350*2}{(9.8175*10^(-6)) * 79 * 10^9}\\\\= 0.01621 rad\\\\Q_{D/A} = 0.01621 * \frac{180}{pi} = 0.929 degrees$