Engineering, 22.04.2020 17:29, stdias
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.
Answers: 1
) = 1.2T = 1.2 × 52087 =62504 Nm![\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](/tpl/images/0618/3830/1fc9a.png)
![\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](/tpl/images/0618/3830/c8828.png)
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