Question

A solid steel shaft has to transmit 100 KW at 160 r.p.m. Taking allowable shear stress as 70 MPa,
find the suitable diameter of the shaft. The maximum torque transmitted in each revolution exceeds the
mean by 20%.

Answer 1

Step-by-step explanation:

Given A solid steel shaft has to transmit 100 KW at 160 r.p.m. Taking allowable shear stress as 70 MPa,find the suitable diameter of the shaft. The maximum torque transmitted in each revolution exceeds the  mean by 20%.

• Now power transmitted by shaft  
•                                   P = 2πNT / 60
•                                 100  = 2 x 3.14 x 160 x T / 60
•                                  6000 = 1004.8 T
•                                     T = 6000 / 1004.8
•                                     T = 5.97 kN-m
• Maximum torque T max = 1.2 x T
•                                        = 1.2 x 5.97
•                                       = 7.164 kN-m
• Now we need to find diameter D.
•                    So T max = π/16 x τ x D^3
•                            7.164 = π/16 x 70 x D^3
•                                       = 13.7375 D^3
•                               D^3 = 7.164 / 13.7375
•                                 D^3 = 0.5214
•                           Or D = 0.8049 m

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