Question #151944
A hollow shaft is 225mm out side diameter and 150mm inside diameter . calculate,
1. the maximum power this shaft can transmit at 150rev/min if the maximum shear stress in not to exceed 70MN/m^2....?
2. The diameter of a solid shaft of the same material which would transmit the same maximum power at the same speed with the same stress
1
Expert's answer
2020-12-22T05:07:29-0500

J=π2(r24r14)=π2(0.112540.0754)=0.0002(m4)J=\frac{\pi}{2}(r_2^4-r_1^4)=\frac{\pi}{2}(0.1125^4-0.075^4)=0.0002(m^4)


T=Jτr2=0.0002701060.1125=124.444(kNm)T=\frac{J\tau}{r_2}=\frac{0.0002\cdot70\cdot10^6}{0.1125}=124.444(kN\cdot m)


1) At 110(rev/min)110 (rev/min) the power generated is


12444423.1411060=1.43(MW)124444\cdot 2\cdot3.14\cdot \frac{110}{60}=1.43(MW) . Answer


2) J=πr42=3.14r42=1.57r4J=\frac{\pi r^4}{2}=\frac{3.14 r^4}{2}=1.57r^4


Jτr22πν=1.57r4τr2πν\frac{J\tau}{r_2}\cdot2\pi\nu= \frac{1.57r^4\tau}{r}\cdot2\pi\nu


Jr2=1.57r4rr=(J1.57r2)1/3=(0.00021.570.1125)1/3=0.104(m)\frac{J}{r_2}= \frac{1.57r^4}{r}\to r=(\frac{J}{1.57r_2})^{1/3}=(\frac{0.0002}{1.57\cdot 0.1125})^{1/3}=0.104(m)


=(143000023.14(110/60)1.5740106)1/3=0.1255(m)=(\frac{1430000}{2\cdot3.14\cdot(110/60)\cdot1.57\cdot40\cdot10^6})^{1/3}=0.1255(m)


d=20.104=0.208(m)d=2\cdot0.104=0.208(m) . Answer




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Comments

mkhenso
11.11.23, 23:32

thankyou. this calculation helped me so much.

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