Question

a hollow shaft is required to transmit 600kw at 110 rpm, the maximum torque being 20% greater than the mean. the shear stress is not to exceed 63 mpa and twist in a length of 3 meters not t exeed 1.4 degrees Calculate the minimum external diameter satisfying these conditions.

Answer 1

Answer:

175.5 mm

Explanation:

a hollow shaft of diameter ratio 3/8 is required to transmit 600kw at 110 rpm, the maximum torque being 20% greater than the mean. the shear stress is not to exceed 63 mpa and twist in a length of 3 meters not t exeed 1.4 degrees Calculate the minimum external diameter satisfying these conditions. G = 80 GPa

Let

D = external diameter of shaft

Given that:

d = internal diameter of the shaft = 3/8 × D = 0.375D,

Power (P) = 600 Kw, Speed (N) = 110 rpm, Shear stress (τ) = 63 MPa = 63 × 10⁶ Pa, Angle of twist (θ) = 1.4⁰, length (l) = 3 m, G = 80 GPa = 80 × 10⁹ Pa

The torque (T) is given by the equation:

$T=\frac{60 *P}{2\pi N}\\ Substituting:\\T=\frac{60*600*10^3}{2\pi*110} =52087Nm$

The maximum torque ($T_{max$) = 1.2T = 1.2 × 52087 =62504 Nm

Using Torsion equation:

$\frac{T}{J} =\frac{\tau}{R}\\ J=\frac{T.R}{\tau} \\\frac{\pi}{32}[D^4-(0.375D)^4]=\frac{62504*D}{2(63*10^6)} \\D^3(0.9473)=0.00505\\D=0.1727m=172.7mm$

$\theta=1.4^0=\frac{1.4*\pi}{180}rad$

From the torsion equation:

$\frac{T}{J} =\frac{G\theta }{l}\\ J=\frac{T.l}{G\theta} \\\frac{\pi}{32}[D^4-(0.375D)^4]=\frac{62504*3}{84*10^9*\frac{1.4*\pi}{180} } \\D=0.1755m=175.5mm$

The conditions would be satisfied if the external diameter is 175.5 mm