Question

A line shaft rotating at 200 r.p.m. is to transmit 20 kW. The allowable

shear stress for the material of the shaft is 42 MPa. If the shaft carries

a central load of 900 N and is simply supported between bearing 3

metre apart, determine the diameter of the shaft. The maximum tensile

or compressive stress is not to exceed 56 MPa. [Ans. 50 mm]
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Answer 1

Answer:

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Answer 2

Given:

N = 200 r.p.m.

P = 20 kW = 20 × 103 W

   = 42 MPa

    = 42 N/mm2

Explanation:

Let d = Diameter of the shaft

We know that torque transmitted by the shaft

T = P × 60 / 2πN

  = (20 × 10³ × 60) / 2π × 200

  =955 -m = 955 × 10³ N-mm

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

955 × 10³ = (π/16) × τ × d³

=  π/16 × 42 × d³

= 8.25 d³

d³ = 955 × 10³/8.25

    = 115 733 or d = 48.7 say 50mm

Answer = 50mm