Question

3. A solid steel shaft has to transmit 75 kW at 200 r.p.m., taking allowable shear stress as 70 N/mm². Find the diameter for the shaft, if maximum torque transmitted at each revolution exceeds the mean by 30%​

Answer 1

Answer:

solid steel shaft has to transmit 75 kW at 200 r.p.m., taking allowable shear stress as 70 N/mm². Find the diameter for the shaft, if maximum torque transmitted at each revolution exceeds the mean by 30%

Solution :

F = 75 kw

= 75 × 10³ Wt

.Now , n = 300 R.D.M

z = 70 N/mm²

P = 2πNT max / 60

Putting the eq in Formula :

Tmax = 35Q/N-m

Answer 2

Answer:

arunan , Abiram, Arvind, Lokesh, Gokul,nishanth

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