Question

A solid stainless-steel shaft is required to transmit 100 kN at 3000 rpm. The supported length
of the shaft is 3 m. It carries 2 pulleys of weight 1500 N each and supported at a distance of 1
m from the ends respectively. Assuming safe value of stress as 60 MPa, determine the torque
(T) in the shaft

Answer 1

Answer:

Given :

 

N = 200 r.p.m. ;

 

P = 20 kW = 20 × 103 W; τ= 42 MPa = 42 N/mm2

Let d = Diameter of the shaft.

We know that torque transmitted by the shaft,

         = P∗ 60

2πN 3

= 20 ∗ 10 ∗60 2π∗ 200

 

= 5 N-m

 

= 955 × 103 N-mm

           We also know that torque transmitted by the shaft ( T ),

955 × 103    =  π *

τ*d3  

         16

d3      = 955 × 103 / 8.25

         = 115 733 or d

         = 48.7 say 50 mm

Explanation:

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