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
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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:
MRK ME BRAINILIEST
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